Class 12

Odisha State Board CHSE Odisha Class 12 Psychology Solutions Unit 2 Long Answer Questions Part-2.

CHSE Odisha 12th Class Psychology Unit 2 Long Answer Questions Part-2

Long Questions With Answers

Question 1.
What are the Signs and Symptoms of Stress?
Answer:
The way we respond to stress varies depending upon our personality, early upbringing and life experiences. Everyone has their own pattern of stress response. So the warning signs may vary, as may their intensity. Some of us know our pattern of stress response and can gauge the depth of the problem by the nature and severity of our own symptoms or changes in behavior.

These symptoms of stress can be physical, emotional and behavioral. Any of the symptoms can indicate a degree of stress which, if left unresolved, might have serious implications, that are often unavoidable such as air pollution, crowding, noise, the heat of the summer, winter cold, etc. Another group of environmental stresses are catastrophic events or disasters such as fire, earthquake, floods, etc.

Question 2.
Discuss the Types of Stress.
Answer:
The three major types of stress, viz. physical and. environmental, psychological, and social are listed. It is important to understand that all these types of stress are interrelated.

Physical and Environmental Stress :
Physical stresses are demands that change the state of our body. We feel strained when, we overexert ourselves physically, lack a nutritious diet, suffer an injury, or fail to get enough sleep. Environmental stresses are aspects of our surroundings.

Psychological Stress :
These are stresses that we generate ourselves in our minds. These are personal and unique to the person experiencing them and are internal sources of stress. We worry about problems, feel anxiety, or become depressed. These are, not only symptoms of stress, but they cause further stress for us. Some of the important sources of psychological stress are frustration, conflicts, internal and social pressures, etc.

Frustration:
Frustration results from the blocking of needs and motives by something “or someone that hinders us from achieving the desired goal. There could be a number of causes of frustration such as social discrimination, interpersonal hurt, low grades in school, etc.

Conflicts:
Conflicts may occur between two or more incompatible needs or motives, e.g. whether to study dance or psychology. You may want to continue your studies or take up a job. There may be a conflict of values when you are pressurised to take any action that may be against the values held by you.

Internal pressures:
Internal pressures stem from beliefs based upon expectations from inside us to ourselves such as, ‘I must do everything perfectly. Such expectations can only lead to disappointment. Many of us drive ourselves ruthlessly towards achieving unrealistically high standards in achieving our goals.

Social pressures:
Social pressures may be brought about from people who make excessive demands on us. This can cause even greater pressure when we have to work with them. Also, there are people with whom we face interpersonal difficulties, ‘a personality clash’ of sorts.

Social Stress :
These are induced externally and result from our interaction with other people. Social events like death or illness in the family, strained relationships, and trouble with neighbors are some Examples of social stresses. These social stresses vary widely from person to person. Attending parties may be stressful for a person who likes to spend quiet evenings at home while an outgoing person may find staying at home in the evenings stressful.

Question 3.
Write the Sources of Stress.
Answer:
A wide range of events and conditions Can generate stress. Among the most important of these are major stressful life events, such as the death of a loved one or personal.
Life Events:
Changes, both big and small, sudden and gradual affect our life from the moment we are born. We learn to cope with small, everyday changes but major life events can be stressful because they disturb our routine and cause upheaval. If several of these life events that are planned (e.g. moving into a new house) or unpredicted (e.g. break-up of a long-term relationship) occur within a short period of time, we find it difficult to cope with them and will be more prone to the symptoms of stress.
A Measure of Stressful Life Events

Life Events Mean Stress Score
Death of a close family member	66
Unexpected accident or trauma	53
Illness of a family member	52
Break-up with friend	47
Appearing for examinations	43
Change in eating habits	27

The mean number of stressful life events experienced over a period of one year. Holmes and Rahe developed a life event measure of stress. A measure of stressful life events based on the above scale known as the Presumptive Stressful Life Events Scale has been developed for the Indian population by Singh, Kaur and Kaur. It is a self-rating questionnaire made up of fifty-one life changes, that a person may have experienced. Each of these life events is assigned a numerical value in terms of their severity.

For example, the death of one’s spouse is assigned 95, personal illness or injury 56, failure in examination 43, appearing for examination or interview 43, and change in sleeping habits 33, as the mean stress score. Both positive and negative events are taken, believing that both kinds of changes cause stress. The respondent’s stress score is the weighted sum of all the items/life change events in the past one year checked by her/him. Some sample items of without producing overt physical or mental illness is approximately two.

However, the correlations between life events and susceptibility to any particular illness is low, indicating a weak association between life events and stress. It has been argued as to whether life events have caused some stress-related illness or whether stress caused life events and illness. The impact of most life events varies from person to person.

Factors such as age at which the event was first experienced, frequency of occurrence, duration of the stressful event and social support must be studied in evaluating the relationship between stressful life events and the subsequent illness episode, injury, the annoying frequent hassles of everyday life and traumatic events that affect our lives.

Hassles :
These are the personal stresses we endure as individuals, due to the happenings in our daily life, such as noisy surroundings, commuting, quarrelsome neighbors, electricity and water shortage, traffic snarls and so on. Attending to various emergencies are daily hassles experienced by a housewife. There are some jobs in which daily hassles are very frequent. These daily hassles may sometimes have devastating consequences for the individual who is often the one coping alone with them as others may not even be aware of them as outsiders. The more stress people report as a result of daily hassles, the poorer is their psychological well-being.

Traumatic Events :
These include being involved in a variety of extreme events such as a fire, train or road accident, robbery, earthquake, tsunami, etc. The effects of these events may occur after some lapse of time and sometimes persist as symptoms of anxiety, flashbacks, dreams and intrusive thoughts, etc. Severe trauma can also strain relationships. Professional help will be needed to cope with them especially if they persist for many months after the event is over.

Question 4.
What is the Effects of Stress on Psychological Functioning and Health,
Answer:
Many of the effects are physiological in nature, however, other changes also occur inside stressed individuals. There are four major effects of stress associated with the stressed state, viz. emotional, physiological, cognitive, and behavioral.

Emotional Effects :
Those who suffer from stress are far more likely to experience mood swings and show erratic behavior that may alienate them from family and friends. In some cases this can start a vicious circle of decreasing confidence, leading to more serious emotional problems. Some examples are feelings of anxiety and depression, increased physical tension, increased psychological tension and mood swings. Box 3.2 presents the phenomenon of ‘Examination Anxiety’.

Physiological Effects :
When the human body is placed under physical or psychological stress, it increases the production of certain hormones, such as adrenaline and cortisol. These hormones produce marked changes in heart rate, blood pressure levels, metabolism and physical activity. Although this physical reaction will help us to function more effectively when we are under pressure for short periods of time, it can be extremely damaging to the body in the long-term effects. Examples of physiological effects are release of epinephrine and norepinephrine, slowing down of the digestive system, expansion of air passages in the lungs, increased heart rate, and constriction of blood vessels.

Cognitive Effects :
If pressures due to stress continue, one may suffer from mental overload. This suffering from high level of stress can rapidly cause individuals to lose their ability to make sound decisions. Faulty decisions made at home, in career, or at the workplace may lead to arguments, failure, financial loss or even loss of job. Cognitive effects of stress are poor concentration and reduced short-term memory capacity.

Behavioral Effects:
Stress affects our behavior in the form of eating less nutritional food, increasing intake of stimulants such as caffeine, excessive consumption of cigarettes, alcohol and other drugs such as tranquilizers etc. Tranquilizers can be addictive and have side effects such as loss of concentration, poor coordination and dizziness. Some of the typical behavioral effects of stress seen are disrupted sleep patterns, increased absenteeism, and reduced work performance.

Stress and Health :
You must have often observed that many of your friends (maybe including yourself as well!) fall sick during examination time. They suffer from stomach upsets, body aches, nausea, diarrhea and fever etc. You must have also noticed that people, who are unhappy in their personal lives fall sick more often than those who are happy and enjoy life. Chronic daily stress can divert an individual’s attention from caring for herself or himself.

When stress is prolonged, it affects physical health and impairs psychological functioning.People experience exhaustion and attitudinal problems when the stress due to demands from the environment and constraints are too high and little support is available from family and friends. Physical exhaustion is seen in the signs of chronic fatigue, weakness and low energy. Mental exhaustion appears in the form of irritability, anxiety, feelings of helplessness and hopelessness.

This state of physical, emotional and psychological exhaustion is known as burnout. There is also convincing evidence to show that stress can produce changes in the immune system and increase the chances of someone becoming ill. Stress has been implicated in the development of cardiovascular disorders, high blood pressure, as well as psychosomatic disorders including ulcers, asthma, allergies and headaches. Researchers estimate that stress plays an important role in fifty to seventy percent of all physical illnesses. Studies also reveal that sixty percent of medical visits are primarily for stress-related symptoms.

General Adaptation Syndrome :
What happens to the body when stress is prolonged? Selye studied this issue by subjecting animals to a variety of stressors such as high temperature, X-rays, and insulin injections, in the laboratory over a long period of time. He also observed patients with various injuries and illnesses in hospitals. Selye noticed a similar pattern of bodily response in all of them. He called this pattern the General Adaptation Syndrome (GAS). According to him, GAS involves three stages: alarm reaction, resistance and exhaustion.

Alarm reaction stage :
The presence of a noxious stimulus or stressor leads to the activation of the adrenal pituitary- cortex system. This triggers the release of hormones producing the stress response. Now the individual is ready for fight or flight.

Resistance stage :
If stress is prolonged, the resistance stage begins. The parasympathetic nervous system calls for more cautious use of the body’s resources. The organism makes efforts to cope with the threat, as through confrontation.

Exhaustion stage:
Continued exposure to the same stressor or additional stressors drains the body of its resources and leads to the third stage of exhaustion. The physiological systems involved in alarm reaction and resistance become ineffective and susceptibility to stress-related diseases such as high blood pressure becomes more likely. Selye’s model has been criticized for assigning a very limited role to psychological factors in stress. Researchers have reported that the psychological appraisal of events is important for the determination of stress. How people respond to stress is substantially influenced by their perceptions, personalities and biological constitutions.

Stress and the Immune System:
Stress ean cause illness by impairing the workings of the immune system. The immune system guards the body against attackers, both from within and outside.

Psychoneuroimmunology:
Psychoneuroimmunology focuses on the links between the mind, the brain and the immune system. It studies the effects of stress on the immune system. How does the immune system work? The white blood cells (leucocytes) within the immune (antigens) such as viruses. It also leads to the production of antibodies. There are several kinds of white blood cells or leucocytes within the immune system, including T cells, B cells and natural killer cells. T cells destroy invaders, and T-helper cells increase immunological activity. It is these T-helper cells that are attacked by the Human Immuno Deficiency Virus (HIV), the virus causing Acquired Immuno Deficiency Syndrome (AIDS). B cells produce antibodies.

Natural killer cells are involved in the fight against both viruses and tumors, Stress can affect natural killer cell cytotoxicity, which is of major importance in the defense against various infections and cancer. Reduced levels of natural, killer cell cytotoxicity have been found in people who are highly stressed, including students facing important examinations, bereaved persons, and those who are severely depressed. Studies reveal that immune functioning is better in individuals receiving social support. Also, changes in the immune system will have more effect on health among those whose immune systems are already weakened.

Figure 3.4 depicts this sequence comprising negative emotions, release of stress hormones which lead to the weakening of the immune system, thereby affecting mental and physical health. Psychological stress is accompanied by negative emotions and associated behaviors, including depression, hostility, anger and aggression. Negative emotional states are of particular concern to the study of the effects of stress on health. The incidence of psychological disorders, such as panic attacks and obsessive behavior increases with the build-up of long-term stress. Worries can reach such a level that they surface as a frightening, painful physical sensation, which can be mistaken for a heart attack.

People under prolonged stress are more prone to irrational fears, mood swings and phobias, and may experience fits of depression, anger and irritability. These negative emotions appear to be related to the function of the immune system. Our ability to interpret our world and to invest that interpretation with personal meaning and emotion have a powerful and direct effect on the body. Negative moods have been associated with poorer health outcomes. Feelings of hopelessness are related to the worsening of disease and increased risk of injury and death due to various causes.

Lifestyle :
Stress can lead, to unhealthy lifestyles or health-damaging behavior. Lifestyle is th,e overall pattern of decisions and behaviors that determine a person’s health and quality of life. Stressed individuals may be more likely to expose themselves to pathogens, which are agents causing physical illness. People who are stressed have poor nutritional habits, sleep less and are likely to engage in other health-risking behaviors like smoking and alcohol abuse. Such health-impairing behaviors develop gradually and are accompanied by pleasant experiences temporarily.

However, we tend to ignore their long-term damaging effects and underestimate the risk they pose to our lives. Studies have revealed that health-promoting behavior like a balanced diet, regular exercise, family support, etc. play an important role in good health. Adhering to a lifestyle that includes a balanced low-fat diet, regular exercise and continued activity along with positive thinking enhances health and longevity. The modem lifestyle of excesses in eating, drinking and the so called fast-paced good life has led to violation of basic principles of health in some of us, as to what we eat, think or do with our lives.

Question 5.
How to cope up with Stress?
Answer:
In recent years the conviction has grown that it is how we cope with stress and not the stress one experiences that influence our psychological well-being, social functioning and health. Coping is a dynamic situation-specific reaction to stress. It is a set of concrete responses to stressful situations or events that are intended to resolve the problem and reduce stress. The way we cope with stress often depends on rigid deep-seated beliefs, based on experience, e.g. when caught in a traffic jam we feel angry because we believe that the traffic ‘should’ move faster.

To manage stress we often need to reassess the way we think and learn coping strategies. People who cope poorly with stress have an impaired immune response and diminished activity of natural killer cells. Individuals show consistent individual differences in the coping strategies they use to handle stressful situations. These can include both overt and covert activities.

The three coping strategies given by Endler and Parker are:

Task-oriented Strategy :
This involves obtaining information about the stressful situation and about alternative courses of action and their probable outcome; it also involves deciding priorities and acting so as to deal directly with the stressful situation. For example, schedule my time better, or think about how I have solved similar problems.

Emotion-oriented Strategy:
This can involve efforts to maintain hope and to control one’s emotions; it can also involve venting feelings of anger and frustration, or deciding that nothing can be done to change things. For example, tell myself that it is not really happening to me, or worry about what I am going to do.

Avoidance-oriented Strategy:
This involves denying or minimizing the seriousness of the situation; it also involves conscious suppression of stressful thoughts and their replacement by self-protective thoughts. Examples of this are watching TV, phone up a friend, or try to be with other people. Lazarus and Folkman has conceptualized coping as a dynamic process rather than an individual trait. Coping refers to constantly changing cognitive and behavioral efforts to master, reduce or tolerate the internal or external demands that are created by the stressful transaction.

Coping serves to allow the individual to manage or alter a problem and regulate the emotional response to that problem. According to them coping responses can be divided into two types of responses, problem-focused and emotion-focused. Problem-focused -strategies attack the problem itself, with behaviors designed to gain information, to alter the event, and to alter belief and commitments. They increase the person’s awareness, level of knowledge, and range of behavioral and cognitive coping options. They can act to reduce the threat value of the event.

For example “I made a plan of action and followed it”. Emotion-focused strategies call for psychological changes designed primarily to limit the degree of emotional disruption caused by an event, with minimal effort to alter the event itself. For example “I did some things to let it out of my system”. While both problem-focused and emotion-focused coping are necessary when facing stressful situations, research suggests that people generally tend to use the former more often than the latter.

Stress Management Techniques:
Stress is a silent killer. It is estimated to play a significant role in physical illness and disease. Hypertension, heart disease, ulcers, diabetes and even cancer are linked to stress. Due to lifestyle changes stress is on the increase. Therefore, schools, other institutions, offices and communities are concerned about knowing techniques to manage stress. Some of these techniques are.

Relaxation Techniques :
It is an active skill that reduces symptoms of stress and decreases the incidence of illnesses such as high blood pressure and heart disease. Usually, relaxation starts from the lower part of the body and progresses up to the facial muscles in such a way that the whole body is relaxed. Deep breathing is used along with muscle relaxation to calm the mind and relax the body.

Meditation Procedures :
The yogic method of meditation consists of a sequence of learned techniques for refocusing of attention that brings about an altered state of consciousness. It involves such a thorough concentration that the meditator becomes unaware, of any outside stimulation and reaches a different state of consciousness.

Biofeedback :
It is a procedure to monitor and reduce the physiological aspects of stress by providing feedback about current physiological activity and is often accompanied by relaxation training. Biofeedback training involves three stages: developing an awareness of the particular physiological response, e.g. heart rate, learning ways of controlling that physiological response in quiet conditions; and transferring that control into the conditions of everyday life.

Creative Visualisation:
It is an effective technique for dealing with stress. Creative visualization is a subjective experience that uses imagery and imagination. Before visualizing one must set oneself a realistic goal, as it helps build confidence. It is easier to visualize if one’s mind is quiet, body relaxed and eyes are closed. This reduces the risk of interference from unbidden thoughts and provides the creative energy needed for turning an imagined scene into reality.

Cognitive Behavioural Techniques :
These techniques aim to inoculate people against stress. Stress inoculation training is one effective method developed by Meichenbaum. The essence of this approach is to replace negative and irrational thoughts with positive and rational ones.

There are three main phases in this: assessment, stress reduction techniques, and application and follow-through. Assessment involves discussing the nature of the problem and seeing it from the viewpoint of the person/client. Stress reduction involves learning the techniques of reducing stress such as relaxation and self-instruction.

Exercise :
Exercise can provide an active outlet for the physiological arousal experienced in response to stress. Regular exercise improves the efficiency of the heart, enhances the function of the lungs, maintains good circulation, lowers blood pressure, reduces fat in the blood and improves the body’s immune system. Swimming, walking, running, cycling, skipping, etc. help to reduce stress. One must practice these exercises at least four times a week for 30 minutes at a time. Each session must have a warm-up, exercise and cool-down phases.

Question 6.
What is Stress-Resistant Personality?
Answer:
Recent studies by Kobasa have shown that people with high levels of stress but low levels of illness share three characteristics, which are referred to as the personality traits of hardiness. It consists of ‘the three Cs i.e. commitment, control and challenge. Hardiness is a set of beliefs about oneself, the world, and how they interact. It takes shape as a Sense of personal commitment to what you are doing, a sense of control over your life and a feeling of challenge.

Stress-resistant personalities have control which is a sense of purpose and direction in life; commitment to work, family, hobbies and social life and challenge, that is, they see changes in life as normal and positive rather than as a threat. Everyone does not have these characteristics, many of us have to relearn specific life skills in areas such as rational thinking and assertiveness to equip ourselves better to cope with the demands of everyday life, etc.

Life Skills :
Life skills are abilities for adaptive and positive behavior that enable individuals to deal effectively with the demands and challenges of everyday life. Our ability to cope depends on how well we are prepared to deal with and-counterbalance everyday demands and keep equilibrium in our lives. These life skills can be learned and even improved upon. Assertiveness, time management, rational thinking, improving relationships, self-care and overcoming unhelpful habits such as perfectionism, and procrastination, etc. are some life skills that will help to meet the challenges of life.

Assertiveness :
Assertiveness is a behavior or skill that helps to communicate, clearly and confidently, our feelings, needs, wants and thoughts. It is the ability to say no to a request, to state an opinion without being self-conscious, or to express emotions such as love, anger, etc. openly. If you are assertive, you feel confident and have high self-esteem and a solid sense of your own identity.

Time Management:
The way you spend your time determines the quality of your life. Learning how to plan time and delegate can help to relieve the pressure. The major way to reduce time stress is to change one’s perception of time. The central principle of time management is to spend your time doing the things that you value, or that help you to achieve your goals. It depends on being realistic about what you know and that you must do it within a certain time period, knowing what you want to do and organizing your life to achieve a balance between the two.

Rational Thinking :
Many stress-related problems occur as a result of distorted thinking. The way you think and the way you feel are closely connected. When we are stressed, we have an inbuilt selective bias to attend to negative thoughts and images from the past, which affect our perception of the present and the future. Some of the principles of rational thinking are: challenging your distorted thinking and irrational beliefs, driving out potentially intrusive negative anxiety-provoking thoughts and making positive statements.

Improving Relationships:
The key to a sound lasting relationship is communication. This consists of three essential skills: listening to what the other person is saying, expressing how you feel and what you think and accepting the other person’s opinions and feelings, even if they are different from your own. It also requires us to avoid misplaced jealousy and sulking behavior.

Self-care :
If we keep ourselves healthy, fit and relaxed, we are better prepared physically and emotionally to tackle the stresses of everyday life. Our breathing patterns reflect our state of mind and emotions. When we are stressed or anxious, we tend towards rapid and shallow breathing from high in the chest, with frequent sighs. The most relaxed breathing is slow, stomach-centered breathing from the diaphragm, i.e. a dome-like muscle between the chest and the abdominal cavity. Environmental stresses like noise, pollution, space, light, color, etc. can all exert an influence on our mood. These have a noticeable effect on our ability to cope with stress and well-being.

Overcoming Unhelpful Habits:
Unhelpful habits such as perfectionism, avoidance, procrastination, etc. are strategies that help to cope in the short-term but which make one more vulnerable to stress. Perfectionists are persons who have to get everything just right. They have difficulty in varying standards according to factors such as time available, consequences of not being able to stop work, and the effort needed. They are more likely to feel tense and find it difficult to relax, are critical of self and others and may become inclined to avoid challenges.

Avoidance is to put the issue under the carpet and refuse to accept or face it. Procrastination means putting off what we know we need to do. We all are, guilty of saying “I will do it later”. People who procrastinate are deliberately avoiding confronting their fears of failure or rejection. Various factors have been identified which facilitate the development of positive health.

Health is a state of complete physical, mental, social and spiritual well-being and not merely the absence of disease or infirmity. Positive health comprises the following constructs: “a healthy body high quality of personal relationships; a sense of purpose in life; self-regard, mastery of life’s tasks and resilience to stress, trauma, and change”. Box 3.3 presents the relationship between resilience and health. Specifically, factors that act as stress buffers and facilitate positive health are diet, exercise, positive attitude, positive thinking and social support.

Diet:
A balanced diet can lift one’s mood, give more energy, feed muscles, improve circulation, prevent illness, strengthen the immune system and make one feel better to cope with stresses of life. The key to healthy living is to eat three main meals a day and eat a varied well-balanced diet. How much nutrition one needs depends on one’s activity level, genetic makeup, climate and health history. What people eat and how much do they weigh involve behavioral processes. Some people are able to maintain a healthy diet and weight while others become obese. When we are stressed, we seek ‘comfort foods that are high in fats, salt and sugar.

Exercise :
A large number of studies confirm a consistently positive relationship between physical fitness and health. Also, of all the measures an individual can take to improve health, exercise is the lifestyle change with the widest popular approval. Regular exercise plays an important role in managing weight and stress and is shown to have a positive effect on reducing tension, anxiety and depression.Physical exercises that are essential for good health are stretching exercises such as yogic asanas and aerobic exercises such as jogging, swimming, cycling, etc. Whereas stretching exercises have a calming effect, aerobic exercises increase the arousal level of the body. The health benefits of exercise work as a stress buffer. Studies suggest that fitness permits individuals to maintain general mental and physical well-being even in the face of negative life events.

Positive Attitude:
Positive health and well-being can be realized by having a positive attitude. Some of the factors leading to a positive attitude are: having a fairly accurate perception of reality; a sense of purpose in life and responsibility; acceptance and tolerance for different viewpoints of others and taking credit for success and accepting blame for failure. Finally, being open to new ideas and having a sense of humor with the ability to laugh at oneself help us to remain centered Mid see things in a proper perspective.

Positive Thinking:
The power of positive thinking has been increasingly recognized in reducing and coping with stress. Optimism, which is the inclination to expect favorable life outcomes, has been linked to psychological and physical well-being. People differ in the manner in which they cope. For example, optimists tend to assume that adversity can be handled successfully whereas pessimists anticipate disasters. Optimists use more problem-focused coping strategies, and seek advice and help from others. Pessimists ignore the problem or source of stress and use strategies such as giving up the goal with which stress is interfering or denying that stress exists.

Social Support:
Social support is defined as the existence and availability of people on whom we can rely upon, people who let us know that they care about, value and love us. Someone who believes that she/he belongs to a social network of communication and mutual obligation experiences social support. Perceived support, i.e. the quality of social support is positively related to health and well-being. In contrast, social network, i.e. the quantity of social support is unrelated to well-being because it is very time-consuming and demanding to maintain a large social network.

Studies have revealed that women exposed to life event stresses, who had a close friend, were less likely to be depressed and had lesser medical complications during pregnancy. Social support can help to provide protection against stress. People with high levels of social support from family and friends may experience less stress when they confront a stressful experience, and they may cope with it more successfully. Social support may be in the form of tangible support or assistance involving material aid, such as money, goods, services, etc.

For example, a child gives notes to her/his friend, since she/he was absent from school due to sickness. Family and friends also provide informational support about stressful events. For example, a student facing a stressful event such as a difficult board examination, if provided information by a friend who has faced a similar one, would not only be able to identify the exact procedures involved, but also it would facilitate in determining what resources and coping strategies could be useful to successfully pass the examination.

During times of stress, one may experience sadness, anxiety, and loss of self-esteem. Supportive friends and family provide emotional support by reassuring the individual that she/he is loved, valued, and cared for. Research has demonstrated that social support effectively reduces psychological distress such as depression or anxiety, during times of stress. There is growing evidence that social support is positively related to psychological well-being. Generally, social support leads to mental health benefits for both the giver and the receiver.

Question 7.
Effect of noise on Child health.
Answer:
Noise :
Children’s reading abilities, cognitive development, physiological indicators, and motivational tasks are affected by exposure to noise. The most common noises that children are exposed to are transportation (e.g. cars, airplanes), music and other people. Evans’ research reveals significant reading delays for children living near airports and exposed to airport noise. He and his colleagues found these delays in reading to occur alnoise levels far below those required to produce hearing damage or loss.

Chronic and acute noise exposure also affects cognitive development, particularly long-term memory, especially if the task is complex. Short-term memory appears to be less affected, but this is dependent upon volume of noise. One way that children adapt to chronic noise is by disregarding or ignoring auditory input. A consequence of this coping strategy is that children also tune out speech, which is a basic and required component of reading. As a result, not only are children’s reading abilities affected, but also their abilities at tasks that require speech perception.

Noise levels also indirectly influence children’s cognitive development via their effect on the adults and teachers who interact with children. Teachers in noisy schools are more fatigued, annoyed and less patient than teachers in quieter schools. Teachers in noisy schools also losC instruction time due to noise distractions and have a compromised teaching style. Children exposed to chronic loud noise also experience a rise in blood pressure, and stress hormones. And children as young as four are less motivated to perform on challenging language and pre-reading tasks under conditions of exposure to chronic noise.

Question 8.
Short notes :
Answer:
Crowding:
Research demonstrates that crowding has an effect on interpersonal behaviors, mental health, motivation, cognitive development and biological measures. Family size has not been found to be a critical factor in crowding. Rather, Evans identifies density, or number of people per room, as the crucial variable for measuring effects of crowding on children’s development. Regarding child development, Evans has found that 10-12-year-old children are more likely to withdraw in overcrowded situations. Children may engage in withdrawal behavior as a means of coping with an overstimulating environment.

Evans’ research also reveals that a highly concentrated number of children in an activity area results in more distractions and less constructive play among preschool-aged children. Overcrowding also influences parenting behaviors. Parents in crowded homes are less responsive to young children. Evidence of parental unresponsiveness begins early— before a child is one year old and occurs at all income levels. Overcrowding also strains parent-child Relationships. Parents in overcrowded homes are more likely to engage in punitive parenting, which in turn, affects the level of children’s distress.

Evans’ research shows that strained parent-child relationships negatively influence social, emotional and biological measures (e.g. elevated blood pressure) in 10- to 12-year-old children. Children’s mental health status may be affected by overcrowding. Elementary school-aged children who live in more crowded homes display higher levels of psychological distress and they also have higher levels of behavior difficulties in school.Evans has found that overcrowding produces psychological distress among 3rd and 4th-grade students as reported by both the children and their teachers. These effects are intensified if children reside in large, multifamily structures. Effects were also intensified among a group of 8- to 10-year-olds if the family home was chaotic.

Chronic overcrowding influences children’s motivation to perform tasks. Independent of household income, children aged 6-12 show declines in motivational behavior and also demonstrate a level of learned helplessness—a belief that they have no control over their situation and therefore do not attempt to change it—although they have the power to do so. But there are gender differences: Evans found the link between overcrowding and learned helplessness among 10-to- 12-year-olds to exist for girls, but not for boys.

Evans’ studies find several effects of overcrowding on both objective and subjective measures of children’s cognitive development. Elementary school children living in more crowded homes score lower on standardized reading tests and they see themselves as less scholastically competent than their classmates.  Parenting behaviors directly related to children’s cognitive and language development are also affected by density level.Evans found that parents in crowded homes speak less to their infants and rise fewer complicated words during the period from infancy up to age two and a half. Research demonstrates that the quality and sophistication of speech as well as the quantity of words spoken by parents to their children are significant factors in the amount and types of words children produce.

Biological measures implicate the effects of overcrowding on children’s physiology. In one study, Evans found gender differences in measures of blood pressure among 10-12- year-old children with males in higher residential crowding situations demonstrating elevations in blood pressure, but not females. However, higher overnight levels of the stress hormones epinephrine and norepinephrine were found in both 8- to 10-year-old male and female children living in high-density apartments. This finding was especially relevant when chaos and disorder was present in the family.

Housing and Quality of Neighborhood:
Housing quality and the neighborhoods in which houses are situated have also been investigated in relation to children’s socio-emotional development. For example, families living in high-rise housing, as opposed to single-family residences, have fewer relationships with neighbors, resulting in less social support. Studies on housing and the quality of neighborhoods have also examined the role of chaos in children’s environments finding an association between chaotic home environments and levels of psychological distress among middle school children.

Research has identified the physical characteristics of neighborhoods that significantly influence children’s development. These characteristics include residential instability, housing quality, noise, crowding, toxic exposure, quality of municipal services, retail services, recreational opportunities, including natural settings, street traffic, accessibility of transportation and the physical quality of both educational and health facilities. Perhaps not surprisingly, Evans’ research findings support the therapeutic effects of children’s exposure to natural settings.

Natural settings are preferred by children and allow them to exercise gross motor abilities as well as engage in social interactions. In addition, these settings also alleviate the adverse effects of children’s exposure to chronic stress. The research outlined above demonstrates both the direct and indirect effects of the physical environment on children’s development. Direct effects include cognitive, social, emotional and biological outcomes.Indirect effects include interactions with parents and teachers, which in turn, influence developmental outcomes such as learning and language development.

Although in several studies Evans demonstrates these effects for children at all income levels, low-income children experience excessive exposure to noise, overcrowding and unfavorable housing and neighborhood conditions.Exposure to these poor-quality physical conditions is linked to other psychological and social aspects of the environment, especially poverty. Using a building block analogy, low-income children have more blocks stacked one on top of the other than children of other income levels. Thus, children living in poverty experience multiple exposures, rather than a single exposure to risk.

What You Can Do to check the impact of the environment on human behavior?

Guard against additional, interior noise sources:
Individuals living in noisy environments often habituate or become accustomed to the noise level. Aim to reduce the existing noise instead of adding other sources of noise. Check the volume level on your child’s music devices (e g., iPod, Walkman; it is too loud if someone else can hear the music). If he listens to his favorite music too loudly, make proper volume adjustments. Also monitor the volume level on computers, televisions, and other electronic devices, keeping them as low as possible.

Engage your child:
Children ignore and tune out speech as a way of coping with environmental overstimulation. Take notice if your child is not paying attention or listening to your speech and if so, intervene. Take your child to a quiet outdoor nature spot or a quiet indoor location such as the local library) This is especially important during the preschool and early elementary school years (ages 3-6 years) when children are learning to read.

Tune in instead of tuning out:
Parents living under high noise exposure appear to withdraw, be less responsive and talk less to their children. The natural tendency is to disengage from speaking and reading to children so as not to compete with the noise. These coping strategies negatively affect children’s reading and cognitive abilities. Be alert to the occurrence of these behaviors and counter them by talking to your child, reading aloud to her, engaging her in discussions, and actively listening to what she has to say to you.

Modify your environment:
If your budget permits, consider purchasing extra noise attenuation devices for your child’s room for use during homework activities and sleeping. Earplugs are a low-cost alternative.

Consider your child’s school environment:
If you have a choice, send your child to a quiet, less chaotic school. This is particularly desirable if your home environment is also noisy. Be active in your community. The noisiest environmental conditions occur in low-income and ethnic minority communities. One way to counteract this is to be active and involved. Ask your representative why it is noisier, in these communities.

Seek information:
If a major source of noise in your community is road traffic, check with your local planning department. Note that traffic volume is closely aligned with traffic noise levels. The busiest streets are usually the noisiest.

Examination Anxiety:
Examination anxiety is a fairly common phenomenon that involves feelings of tension or uneasiness that occur before, during, or after an examination. Many people experience feelings of anxiety around, examinations and find it helpful in some ways, as it can be motivating and create the pressure that is needed to stay focused on one’s performance. Examination nerves, worry, or fear of failure are normal for even the most talented student. However, the stress of formal examination results in such high degrees of anxiety in some students that they are unable to perform at a level that matches the potential they have shown in less stressful classroom situations.

Examination stress has been characterized as “evaluative apprehension” or “evaluative stress” and produces debilitating behavioral, cognitive and physiological effects no different from those produced by any other stressor. High stress can interfere with the student’s preparation, concentration and performance. Examination stress can cause test anxiety which adversely affects test performance. Persons who are high in test anxiety tend to perceive evaluative situations as personally threatening; in test situations, they are often tense, apprehensive, nervous and emotionally aroused.Moreover, the negative self-centered cognitions which they experience distract their attention and interfere with concentration during examinations.

High-test anxious students respond to examination stress with intense emotional reactions, negative thoughts about themselves, feelings of inadequacy, helplessness and loss of status and esteem that impair their performance. Generally, the high-test anxious person instead of plunging into a task plunges inward, that is, either neglect or misinterprets informational cues that may be readily available to her/him, or experiences attentional blocks.While preparing for examinations, one must spend enough time for study, overviewing and weighing one’s strengths and weaknesses, discuss difficulties with teachers and classmates, plan a revision timetable, condensing notes, space out revision periods and most importantly on the examination day concentrate on staying palm.

Question 9.
What do you understand by the term ‘environment’? Explain the different perspectives to understand the human-environment relationship.
Answer:
‘Environment’ refers to all that is around us. Literally, it means everything that surrounds us including the physical, Social world and cultural environment. In general, it includes all the forces outside the human beings to which they respond in Some way. There is more than one way of looking at the human-environment relationship.
A psychologist named Stokols (1990) describes three approaches that may be adopted to describe the human-environment relationship.

• The minimalist perspective assumes that the physical environment has minimal or negligible influence on human behavior, health and well-being. The physical environment and human beings exist as parallel components.
• The instrumental perspective suggests that the physical environment exists mainly for use by human beings for their comfort and well-being. Most of the human influences on the environment reflect the instrumental perspective.
• The spiritual perspective refers to the view of the environment as. something to be respected and valued rather than exploited. It implies that human beings recognize the interdependent relationship between themselves and the environment, i.e. human beings will exist and will be happy only as long as the environment is kept healthy and natural.
  The traditional Indian view about the environment supports the spiritual perspective.

Question 10.
“Human beings affect and are affected by the environment”. Explain this statement with the help of examples.
Answer:
Human beings exert their influence on the natural environment for fulfilling their physical needs and other purposes. The human-environment relationship can be appreciated fully by understanding that the two influence each other and depend on each other for their survival and maintenance. Some aspects of the environment influence human perception.

• Environmental influences on perception :
  Some aspects of the environment influence human perception. For example, a tribal society of Africa lives in circular huts, that is, in houses without angular walls. They show less error in a geometric illusion (the Muller-Lyer illusion) than people from cities, who live in houses with angular walls.
• Environmental influences on emotions:
  The environment affects our emotional reactions as well. Watching nature iri any form, whether it is a quietly flowing river, a smiling flower, or a tranquil mountain top, provides a kind of joy that cannot be matched by any other experience. Natural disasters, such as floods, droughts, landslides, and quakes on the earth or under the ocean, can affect people’s emotions to such an extent that they experience deep depression and sorrow, a sense of complete helplessness and lack of control over their lives.
• Ecological influences on occupation, living style, and attitudes :
  The natural environment of a particular region determines whether people living in that region rely on agriculture (as in the plains), or on other occupations such as hunting and gathering (as in a forest, mountainous or desert regions), or on industries (as in areas that are not fertile enough for agriculture). In turn, occupation determines the lifestyle and attitudes of the residents of a particular geographical region.

Question 11.
What is noise? Discuss the effects of noise on human behavior.
Answer:
Any sound that is annoying or irritating and felt to be unpleasant is said to be noise. From common experience, it is known that noise, especially for long periods of time, is uncomfortable and puts people in an unpleasant mood.

Effects of noise on human behavior:

• When the task being performed is a simple mental task, such as the addition of numbers, noise does not affect overall performance, whether it is loud or soft.
  In such situations, people adapt, or ‘get used to noise.
• If the task being performed is very interesting, then, too, the presence of noise does not affect performance. This is because the nature of the task helps the individual to pay full attention to the task and ignore the noise. This may also be one kind of adaptation.
• When the noise comes at intervals, and in an unpredictable way, it is experienced as more disturbing than if the noise is continuously present.
• When the task being performed is difficult or requires full concentration, then intense, unpredictable and uncontrollable noise reduces the level of task performance.
• When tolerating or switching off the noise is within the control of the person, the number of errors in task performance decreases.
• In terms of emotional effects, noise above a certain level causes annoyance, and can also lead to sleep disturbance. These effects are also reduced if the noise is controllable, or is necessary as a part of the person’s occupation. However, continued exposure to uncontrollable and annoying noise can have harmful effects on mental health.

Question 12.
What are the salient features of crowding? Explain the major psychological consequences of crowding.
Answer:
Crowding to a feeling of discomfort because there are too many people or things around us, giving us the experience of physical restriction and sometimes the lack of privacy. Crowding is the person’s reaction to the presence of a large number of persons within a particular area or space. When this number goes beyond a certain level, it causes stress to individuals caught in that situation. In this sense, crowding is another example of an environmental stressor.

The experience of crowding has the following features :

• Feeling of discomfort,
• Loss or decrease in privacy,
• Negative view of the space around the person, and
• The feeling of loss of control over social interaction.

1. Crowding and high density may lead to abnormal behavior and aggression. This was shown many years ago in a study of rats. These animals were placed in an enclosure, initially in small numbers. As their population increased within this enclosed space, they started showing aggressive and unusual behavior, such as biting the tails of other rats. This aggressive behavior increased to such an extent that ultimately the animals died in large numbers, thus decreasing the population in the enclosure. Among human beings also, an increase in population has sometimes been found to be accompanied by an increase in violent crime.
2. Crowding leads to lowered performance on difficult tasks that involve cognitive processes and has adverse effects on memory and the emotional state. These negative effects are seen to a smaller extent in people who are used to crowded surroundings.
3. Children growing up in very crowded households show lower academic performance. They also show a weaker tendency to continue working on a task if they are unsuccessful at it, compared to children growing up in non-crowded households. They experience greater conflict with their parents and get less support from their family members.

Question 13.
Why is the concept of ‘personal space’ important for human beings? Justify your answer with the help of an example.
Answer:
Personal space or the comfortable physical space one generally likes to maintain around oneself, is affected by a high-density environment. In a crowded context, there is a restriction on personal space and this can also be a cause of negative reactions to crowding.

For example: In social situations, human beings like to maintain a certain physical distance from die person with whom they are interacting. This is called interpersonal al physical distance and is a part of a broader concept called personal space, i.e. the physical space we like to have all around us. One reason for the negative reactions to crowding, as described earlier, is the decrease in personal space.

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 1 Relation and Function Ex 1(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Exercise 1(a)

Question 1.
If A = {a,b,c,d} mention the type of relations on A given below, which of them are equivalence relations?
(i) {(a, a), (b, b)}
(ii) {(a, a), (b, b), (c, c), (d, d)}
(iii) {(a, b), (b, a), (b, d), (d, b)}
(iv) {(b, c), (b, d), (c, d)}
(v) {(a, a), (b, b), (c, c), (d, d), (a, d), (a, c), (d, a), (c, a), (c, d), (d, c)}
Solution:
(i) Symmetric and transitive but not reflexive.
(ii) Reflexive, symmetric as well as transitive. Hence it is an equivalence relation.
(iii) Only symmetric
(iv) Only transitive
(v) Reflexive, symmetric and transitive. Hence it is an equivalence relation.

Question 2.
Write the following relations in tabular form and determine their type.
(i) R = {(x, y) : 2x – y = 0] on A = {1,2,3,…, 13}
(ii) R = {(x, y) : x divides y} on A = {1,2,3,4,5,6}
(iii) R = {(x, y) : x divides 2 – y} on A = {1,2,3,4,5}
(iv) R = {(x, y) : y ≤, x ≤, 4} on A = {1,2,3,4,5}.
Solution:
(i) R = {(x, y) : 2x- y = 0} on A
= {(x, y) : y = 2x} on A
= {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10), (6, 12)}
R is neither reflexive nor symmetric nor transitive.

(ii) R = {(1, 1), (1, 2), (1, 3), (1, 4), (1,5), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (5,5), (6, 6)}
R is reflexive transitive but not symmetric.

(iii) R = {(x, y) : x divides 2 – y} on A
= {1, 2, 3, 4, 5}
= {(x, y) : 2-y is a multiple of x}
= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 4), (3, 2), (3, 5), (4, 2), (5, 2)}
R is neither reflexive nor symmetric nor transitive.

(iv) R = {(x, y) : y ≤ x ≤ 4} on A
= {1, 2, 3, 4, 5}
= {(1, 1), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (4, 4)}
R is neither reflexive nor symmetric but transitive.

Question 3.
Test whether the following relations are reflexive, symmetric or transitive on the sets specified.
(i) R = {(m,n) : m-n ≥ 7} on Z.
(ii) R = {(m,n) : 2|(m+n)} on Z.
(iii) R = {(m,n) : m+n is not divisible by 3} Z.
(iv) R = {(m,n) : is a power of 5} on Z – {0}.
(v) R = {(m,n) : mn is divisible by 2} on Z.
(vi) R = {(m,n) : 3 divides m-n} on {1,2,3…,10}.
Solution:
(i) R = {{m, n) : m- n ≥ 7} on Z
Reflexive:
∀ m ∈ Z, m – m = 0 < 7
⇒ (m, m) ∉ R
Thus, R is not reflexive.
Symmetry:
Let (m, n) ∈ R
⇒ m – n ≥ 7
⇒ n – m < 7
∴ (n, m) ∉ R
⇒ R is not symmetric.
Transitive:
Let (m, n), (n, p) ∈ R
m – n ≥ 7
and n – p > 7
⇒ m – p ≥ 7
⇒ (m, p) ∈ R
⇒ R is transitive.

(ii) R = {(m, n) : 2 | (m + n)} on Z
Reflexive:
∀ m ∈ Z, m + m = 2m
which is divisible by 2.
⇒ 2 | (m + m)
⇒ (m, m) ∈ R
⇒ R is reflexive.
Symmetry:
Let (m, n) ∈ R
⇒ 2 | (m + n)
⇒ 2 | (n + m)
(n, m) ∈ R
⇒ R is symmetric.
Transitive:
Let (m, n), (n, p), ∈ R
⇒ 2 | (m + n) and 2 | (n + p)
⇒ m + n = 2k₁
⇒ n + p = 2k₂
⇒ m + 2n + p = 2k₁ + 2k₂
⇒ m + p = 2(k₁ + k₂ – 1)
⇒ 2 | (m + p)
⇒ (m, p) ∈ R
⇒ R is transitive.
Thus, R is an equivalence relative.

(iii) R = {(m, n) : m + n is not divisible by 3} on Z
Reflexive:
As 3 + 3 is divisible by 3
we have (3, 3) ∉ R
⇒ R is not reflexive.
Symmetric:
Let (m, n) ∈ R
⇒ m + n is not divisible by 3
⇒ n + m is not divisible by 3
⇒ (n, m) ∈ R
⇒ R is symmetric.
Transitive:
(3, 1), (1, 6) ∈ R
But (3, 6) ∉ R
⇒ R is not transitive.

(iv) R = {(m, n) : \(\frac{m}{n}\) is a power of 5} on Z – {0}
Reflexive:
∀ m ∈ Z – {0}
\(\frac{m}{m}\) = 1 = 5°
⇒ (m, m) ∈ R
⇒ R is reflexive.
Symmetric:
Let (m, n) ∈ R
\(\frac{m}{n}\) = 5^k
\(\frac{n}{m}\) = 5^-k
⇒ (n, m) ∈ Z
⇒ R is symmetric.
Transitive:
Let (m, n), (n, p) ∈ R
⇒ \(\frac{m}{n}\) = 5^k1 , \(\frac{n}{p}\) = 5^k2
⇒ \(\frac{m}{n}\) . \(\frac{n}{p}\) = 5^k1 . 5^k2
⇒ \(\frac{m}{p}\) = 5 ^k1+k2
⇒ (m, p) ∈ R
⇒ R is transitive.
Thus R is an equivalence relation.

(v) R = {(m, n) : mn is divisible by 2} on Z
Reflexive:
3 ∈ Z
3 x 3 = 9
which is not divisible by 2.
∴ (3, 3) ∉ R
⇒ R is not reflexive.
Symmetric:
Let (m, n) ∈ R
⇒ mn is divisible by 2
⇒ nm is divisible by 2
⇒ (n, m) ∈ R
⇒ R is symmetric.
Transitive:
⇒ (3, 2), (2, 5) ∈ R
⇒ But 3 x 5 = 15,
⇒ which is not divisible by 2.
⇒ (3, 5) ∉ R
R is not transitive.

(vi) R = {(m, n) : 3 divides m-n} on A = {1, 2, 3……,10}
Reflexive:
Clearly ∀ m ∈ A, m – m = 0
which is divisible by 3
⇒ (m, m) ∈ R
⇒ R is reflexive
Symmetric:
Let (m, n) ∈ R
⇒ m – n is divisible by 3
⇒ n – m is also divisible by 3
⇒ (n, m) ∈ R
⇒ R is symmetric
Transitive:
Let (m, n), (n, p) ∈ R
⇒ m – n and n – p are divisible by 3
⇒ m – n + n – p is also divisible by p.
⇒ m – p is divisible by p.
⇒ (m, p) ∈ R
⇒ R is transitive.
Thus R is an equivalence relation.

Question 4.
List the members of the equivalence relation defined by the following partitions on X= {1,2,3,4}. Also find the equivalence classes of 1,2,3 and 4.
(i) {{1},{2},{3, 4}}
(ii) {{1, 2, 3},{4}}
(iii) {{1,2, 3, 4}}
Solution:
(i) The equivalence relation is
R = {(1, 1), (2, 2), (3, 3), (4, 4), (3, 4), (4, 3)}
[1] = {1}, [2] = {2}, [3] = {3, 4} and [4] = {3, 4}

(ii) The equivalence relation is
R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)}
[1] = [2] = [3] = {1, 2, 3}
[4] = {4}

(iii) The equivalence relation is
R = A x A, [1] = [2] = [3] = [4] = A

Question 5.
Show that if R is an equivalence relation on X then dom R = rng R = X.
Solution:
Let R is an equivalence relation on X.
⇒ R is reflexive
⇒ (x, x) ∈ R ∀ x ∈ X
⇒ Dom R = Rng R = X

Question 6.
Give an example of a relation which is
(i) reflexive, symmetric but not transitive.
(ii) reflexive, transitive but not symmetric.
(iii) symmetric, transitive but not reflexive.
(iv) reflexive but neither symmetric nor transitive.
(v) transitive but neither reflexive nor symmetric.
(vi) an empty relation.
(vii) a universal relation.
Solution:
(i) The relation R = {(a, b), (b, a), (a, c), (c, a), (a, a), (b, b), (c, c)} defined on the set {a, b, c} is reflexive, symmetric but not transitive.
(ii) “The relation x ≤ y on z” is reflexive, transitive but not symmetric.
(iii) The relation R = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)} defined on the set {a, b, c, d} is symmetric, transitive but not reflexive.
(iv) The relation R = {(a, a), (b, b), (c, c), (a, b), (b, c)} defined on the set A = {a, b, c} is reflexive but neither symmetric nor transitive.
(v) R = {(a, b), (b, c), (a, c)} on A = {a, b, c} is transitive but neither reflexive nor symmetric.
(vi) On N the relation R= {(x, y) : x + y = – 5} is an empty relation.
(vii) On N the relation R = {(x, y) : x + y > 0} is an universal relation.

Question 7.
Let R be a relation on X, If R is symmetric then xRy ⇒ yRx. If it is also transitive then xRy and yRx ⇒ xRx. So whenever a relation is symmetric and transitive then it is also reflexive. What is wrong in this argument?
Solution:
Let R is a relation on X.
If R is symmetric then xRy ⇒ yRx
If R is also transitive then xRy and yRx ⇒ xRx
⇒ Whenever a relation is symmetric and transitive, then it is reflexive. This argument is wrong because the symmetry of R does not imply dom R = X and for reflexive xRx ∀ x ∈ X.

Question 8.
Suppose a box contains a set of n balls (n ≥ 4) (denoted by B) of four different colours (may have different sizes), viz. red, blue, green and yellow. Show that a relation R defined on B as R={(b1, b2): balls b1 and b2 have the same colour} is an equivalence relation on B. How many equivalence classes can you find with respect to R?
[Note: On any set X a relation R={(x, y): x and y satisfy the same property P} is an equivalence relation. As far as the property P is concerned, elements x and y are deemed equivalent. For different P we get different equivalence relations on X]
Solution:
On B, R = {(b1, b2) : balls b1 and b2 have the same colour}

Reflexive:
∀ b ∈ B, b and b are of same colour
⇒ (b, b) ∈ R
⇒ R is reflexive.

Symmetric:
Let (b1, b2) ∈ R
⇒ b1 and b2 are of same colour
⇒ b2 and b1 are of same colour
⇒ (b2, b1) ∈ R
⇒ R is symmetric.

Transitive :
Let (b1, b2) and (b2, b3) ∈ R
⇒ b1 and b2 are of same colour
b2 and b3 are of same colour
⇒ b1, b3 are of same colour
⇒ (b1, b3) ∈ R
⇒ R is transitive
∴ R is an equivalence relation.
As there are 4 types of balls there are 4 equivalence relations with respect to R.

Question 9.
Find the number of equivalence relations on X={1,2,3}. [Hints: Each partition of a set gives an equivalence relation.]
Solution:
Method – 1: Number of equivalence relations on a set A with | A | = n.
= The number of distinct partitions of A
= B_n
where B_n+1 = \(\sum_{k=0}^n \frac{n !}{k !(n-k) !} \mathrm{B}_k\)
with B₀ = 1
Here n = 3
B₁ = 1
B_{2 =} \(\frac{1 !}{0 ! 1 !}\) B₀ + \(\frac{1 !}{1 ! 1 !}\) B₁
= 1 + 1 = 2
B₃ = \(\frac{2 !}{0 ! 2 !}\) B₀ + \(\frac{2 !}{1 ! 1 !}\) B₁ + \(\frac{2 !}{2 ! 0 !}\) B₂
= 1 + 2 + 2 = 5
Thus there are 5 equivalence relations.

Method – 2:
X= {1, 2, 3}
Number of equivalence relations = number of distinct partitions.
Different partitions of X are
{{1} {2}, {3}}
{{1}, {2, 3}}, {{2}, {1,3}},
{{3}, {1,2}} and {{1, 2,3}}
Thus number of equivalence relations = 5.

Question 10.
Let R be the relation on the set R of real numbers such that aRb iff a-b is an integer. Test whether R is an equivalence relation. If so find the equivalence class of 1 and ½ w.r.t. this equivalence relation.
Solution:
The relation R on the set of real numbers is defined as
R = {(a, b) : a – b ∈ Z}

Reflexive:
∀ a ∈ R (set of real numbers)
a – a = 0 ∈ Z
⇒ (a, a) ∈ R
⇒ R is reflexive.

Symmetric:
Let (a, b) ∈ R
⇒ a – b ∈ Z
⇒ b – a ∈ Z
⇒ (b, a) ∈ R
⇒ R is symmetric.

Transitive:
Let (a, b), (b, c) ∈ R
⇒ a – b and b – c ∈ Z
⇒ a – b + b – c ∈ Z
⇒ a – c ∈ Z
⇒ (a, c) ∈ R
⇒ R is transitive.
Thus R is an equivalence relation.
[1] = {x ∈ R : x -1 ∈ Z} = Z
\(\begin{aligned}
{\left[\frac{1}{2}\right] } &=\left\{x \in \mathrm{R}: x-\frac{1}{2} \in \mathrm{Z}\right\} \\
&=\left\{x \in \mathrm{R}: x=\frac{2 k+1}{2}, k \in \mathrm{Z}\right\}
\end{aligned}\)

Question 11.
Find the least positive integer r such that
(i) 185 ∈ [r]₇
(ii) – 375 ∈ [r]₁₁
(iii) -12 ∈ [r]₁₃
Solution:
(i) 185 ∈ [r]₇
⇒ 185 – r = 7k, k ∈ z and r < 7
⇒ r = 3
(ii) – 375 ∈ [r]₇
⇒ – 375 – r = 11k, k ∈ z and r < 11
⇒ r = 10
(iii) – 12 ∈ [r]₁₃
⇒ – 12 – r = 13k, k ∈ z and r < 13
⇒ r= 1

Question 12.
Find least non negative integer r such that
(i) 7 x 13 x 23 x 413 r (mod 11)
(ii) 6 x 18 x 27 x (- 225) = r (mod 8)
(iii) 1237(mod 4) + 985 (mod 4) = r (mod 4)
(iv) 1936 x 8789 = r (mod 4)
Solution:
(i) 7 x 13 x 23 x 413 ≡ r (mod 11)
Now 7 x 13 ≡ 3 mod 11
23 ≡ 1 mod 11
413 ≡ 6 mod 11
∴ 7 x 13 x 23 x 413 ≡ 3 x 1 x 6 mod 11
≡ 18 mod 11
≡ 7 mod 11
∴ r = 7

(ii) 6 x 18 x 27 x – 225 ≡ r (mod 8)
Now 6 x 18 ≡ 108 = 4 mod 8
27 ≡ 3 mod 8
– 225 ≡ 7 mod 8
⇒ 6 x 18 x 27 x – 225 ≡ 4 x 3 x 7 mod 8
≡ 84 mod 8
≡ 4 mod 8
∴ r = 4

(iii) 1237 (mod 4) + 985 (mod 4) r (mod 4)
Now 1237 ≡ 1 mod 4
985 ≡ 1 mod 4
⇒ 1237 (mod 4) + 985 (mod 4)
≡ (1 + 1) mod 4
≡ 2 mod 4
⇒ r = 2

(iv) 1936 x 8789 ≡ r (mod 4)
1936 x 8789 ≡ 0 mod 4
∴ r = 0

Question 13.
Find least positive integer x satisfying 276x + 128 ≡ (mod 7)
[Hint: 276 ≡ 3, 128 ≡ 2 (mod 7)]
Solution:
Now 128 ≡ 2 mod 7
Now 176 x + 128 ≡ 4 mod 7
⇒ 176 x ≡ (4 – 2) mod 7
⇒ 176 x ≡ 2 mod 7
176 x x ≡ 2 mod 7,
But 276 ≡ 3 mod 7
Thus x = 3.

Question 14.
Find three positive integers x_i, i =1, 2, 3 satisfying 3x ≡ 2 (mod 7)
[Hint: If X₁ is a solution then any member of [X₁] is also a solution]
Solution:
3x ≡ 2 mod 7
Least positive value of x ≡ 3
Each member of [3] is a solution
∴ x = 3, 10, 17 …..

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 1 Relation and Function Ex 1(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Exercise 1(b)

Question 1.
Let X={x, y} and Y={u, v}. Write down all the functions that can be defined from X to Y. How many of these are (i) one-one (ii) onto and (ii) one-one and onto?
Solution:
The functions from X = {x, y} to y = {u, v} are:
f1 = {(x, u), (y, v)}
f2 = {(x, v), (y, u)}
f3 = {(x, u), (y, u)}
f4 = {(x, v), (y, v)}

Out of these 4 functions there are:
(i) 2 one-one functions
(ii) 2 onto functions
(iii) 2 one-one and onto function.

Question 2.
Let X and Y be sets containing m and n elements respectively.
(i) What is the total number of functions from X to Y.
(ii) How many functions from X to Y are one-one according as men, m > n and m = n?
Solution:
If | x | = m and | y | = n then
(i) Number of functions =n^m
(ii) If m < n then number of one-one functions = ⁿP_m.
If m > n then number of one-one functions = 0
If m = n then number of one-one functions = m!

Question 3.
Examine each of the following functions if it is
(i) injective (ii) surjective, (iii) bijective and (iv) none of the three
(a) f : R → R, f(x) = x²
(b) f : R → [-1, 1], f(x) = sin x
(c) f : R₊ → R + , f(x) = x + 1/x
where R₊ = {x ∈ R : x > 0}
(d) f : R → R, f(x) = x³ + 1
(e) f : (-1, 1) → R, f(x) = \(\frac{x}{1-x^2}\)
(f) f : R → R, f(x) = [x] = the greatest integer ≤ x.
(g) f : R → R, f(x) = | x |
(h) f : R → R, f(x) = sgn x
(i) f : R → R, f = id_R = the identity function on R.
Solution:
(a) f : R → R, f(x) = x²
for x₁, x₂ ∈ R
Let f(x₁) = f(x₂)
⇒ x₁² = x₂²
⇒ x₁  = ± x₂
∴ f is not one-one.
Hence f is not injective or bijective.
Rng f = [0, ∞) ≠ R
∴ f is not surjective.

(b) f : R → [-1, 1], f(x) = sin x
For x₁ , x₂ ∈ R
let f(x₁) = f(x₂) ⇒ sin x₁ = sin x₂ 
⇒ x₁ = nπ + (- 1)ⁿ  x₂
⇒ x₁ = x₂ (not always)
∴ f is not injective and also not bijective.
But f is onto, as ∀ y ∈ [-1, 1]
there is a x ∈ R such that f(x) = sin x.
i.e., f is surjective.

(c) f : R₊ → R₊ , f(x) = x + \(\frac{1}{x}\)
f(2) = 2 + \(\frac{1}{2}\) = \(\frac{5}{2}\)
f(\(\frac{1}{2}\)) = \(\frac{1}{2}\) + 2 = \(\frac{5}{2}\)
f(2) = f(\(\frac{1}{2}\))
but 2 ≠ \(\frac{1}{2}\)
∴ f is not injective (one-one).
Again, for 1 ∈ R₊ (domain)
⇒ there is no x ∈ R₊^(Dom)
such that x + \(\frac{1}{2}\) = 1
∴ f is not onto.

(d) f : R → R, f(x) = x³ + 1
for x₁, x₂ ∈ R
Let f(x₁) = f(x₂)
⇒ x₁³ = x₂³
⇒ x₁ = x₂
∴ f is injective.
Let f(x) = y ⇒ y = x³ + 1
⇒ x³ = y – 1
⇒ x = (y – 1)^1/3 which exists ∀ y ∈ R
∴ f is onto.
∴ f is bijective.

(e) f : (-1, 1) → R, f(x) = \(\frac{x}{1-x^2}\)
for x₁, x₂ ∈ (-1, 1)
Let f(x₁) = f(x₂) ⇒ \(\frac{x_1}{1-x_1^2}\) = \(\frac{x_2}{1-x_2^2}\)
⇒x₁ – x₁ x₂² = x₂ – x₁² x₂
⇒ x₁ – x₂ + x₁² x₂ – x₁ x₂² = 0
⇒ (x₁ – x₂) (1 + x₁ x₂) = 0
⇒ x₁ = x₂ (for x₁ x₂ ∈ (-1, 1) x₁ x₂ ≠ -1)
∴ f is injective.
Again let y = \(\frac{x}{1-x^2}\) ⇒ y – x²y = x
⇒ x²y + x – y = 0
⇒ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{-1 \pm \sqrt{1+4 y^2}}{2 y}\) ∉ (-1, 1) ∀ y ∈¸.
∴f is surjective.
∴f is bijective.

(f) f : R → R
f(x) = [x]
f(1.2) = f(1.5) =1
∴ f is not injective.
Rng f = Z ⊂ R
∴ f is not surjective.
∴ Hence it is not also bijective.

(g) f : R → R
f(x) = | x |
As f(-1) = f(1) = 1
∴ f is not injective.
Again Rng f = [0, ∞) ⊂ R
⇒ f is not surjective.
Thus f is not bijective.

(h) f : R → R
f(x) = Sgn (x) = \(\left\{\begin{array}{cc}
1, & x>0 \\
0, & x=0 \\
-1, & x<0
\end{array}\right.\)
As f(1) = f(2) = 1
We have f is not injective.
Again Rng f = {- 1, 0, 1} ≠ R
∴ R is not surjective.
⇒ R is not bijective.

(i) f : R → R
f = idx
∴ f(x) = x
for x₁, x₂ ∈ R
Let f(x₁) = f(x₂) where x₂ x₂ ∈ R
⇒ x₁ = x₂
∴ f is one-one.
Again Rng f = R (codomain)
∴ f is onto.
Thus f is a bijective function.

Question 4.
Show that the following functions are injective.
(i) f(x) = sin x on \(\left[0, \frac{\pi}{2}\right]\)
(ii) f(x) = cos x [0, π]
(iii) fix) = log_a x on (0, ∞), (a > 0 and a ≠ 1)
(iv) f(x) = a^x on R. (a > 0 and a ≠ 1)
Solution:
(i) f(x) = sin x . \(\left[0, \frac{\pi}{2}\right]\)
for α, β ∈ \(\left[0, \frac{\pi}{2}\right]\)
Let f(α) = f(β) ⇒ sin α = sin β
⇒ α = β, as α, β ∈ \(\left(0, \frac{\pi}{2}\right)\) and no other values of α is possible.
∴ f is one-one.

(ii) f(x) = cos x, [0, π]
for α, β ∈ [0, π]
Let f(α) = f(β) ⇒ cos α = cos β
⇒ α = β, since α, β ∈ [0, π] and cos x is +ve in 1st quadrant and -ve in 2nd quadrant.
∴ f is one-one.

(iii) f(x) = log_a x [1, ∞]
for α, β ∈ [1, ∞]
Let f(α) = f(β)
⇒ log_a α = log_a β ⇒ α = β
∴ f is one-one.

(iv) f(x) = a^x, (a > 0), x ∈ R
for x₁, x₂ ∈ R
Let f(x₁) = f(x₂) ⇒ a^x₁ = a^x₂
⇒ x₁ = x₂
∴ f is one-one.

Question 5.
Show that functions f and g defined by f(x) = 2 log x and g(x) = log x² are not equal even though log x² = 2 log x.
Solution:
f(x) = 2 log x
g(x) = log x²
Dom f(x) = (0, ∞)
Dom g(x) = R – {0}
As Dom f(x) ≠ Dom g(x) we have f(x) ≠ g(x), though log x² = 2 log x

Question 6.
Give an example of a function which is
(i) Surjective but not injective.
(ii) injective but not surjective.
(iii) neither injective nor surjective.
(iv) bijective
Solution:
(i) f(x) = sin x
from R → [-1, 1]
which is surjective but not injective.

(ii) f(x) = \(\frac{x}{2}\) from Z → R
is injective but not surjective.

(iii) f : (-1, 1) → R, f(x) = \(\frac{x}{1-x^2}\)
is neither surjective nor injective.
[Refer Q. No. 3(e)].

(iv) f(x) = x³ + 1, f : R → R
is bijective.
[Refer No. 3 (d)].

Question 7.
Prove that the following sets are equivalent:
{1, 2, 3, 4, 5, 6,…}
{2, 4, 6, 8, 10,…}
{1, 7, 5, 7, 9,…}
{1, 4, 9, 16, 25,…}
Solution:
Let A = {1, 2, 3, 4, 5, 6, ……..}
B = {2, 4, 6, 8, 10, …….}
C = {1, 3, 5, 7, 9, ……}
D = {1, 4, 9, 16, 25, ……}
Let f : A → B defined as f(x) = 2x
Clearly f is bijective.
There is a one-to-one correspondence between A and B.
⇒ A and B are equivalent.
Let g : A → C defined as g(x) = 2x – 1
Clearly f is bijective.
⇒ There is a one-to-one correspondence between A to C
∴ A and C are equivalent.
Let h : A → D defined as h(x) = x².
Clearly h is bijective.
⇒ There is a one-to-one correspondence between A to D.
⇒ A and D are equivalent.
Thus A, B, C and D are equivalent.

Question 8.
Let f = {(1, a), (2, b), (3, c), (4, d)} and g = {{a, x), (b, x), (c, y), (d, x)}. Determine gof and fog if possible. Test whether fog = gof.
Solution:
f = {(1, a), (2, b), (3, c), (4, d)}
g = {{a, x), (b, x), (c, y), (d, x)}
gof (1) = g(a) = x
gof (2) = g(b) = x
gof (3) = g(c) = y
gof (4) = g(a) = x
∴ gof = {(1, x), (2, x), (3, y), (4, x)} Here fog is not defined.

Question 9.
Let f = {(1, 3), (2, 4), (3, 7)} and g = {(3, 2), (4, 3), (7, 1)}. Determine gof and fog if possible. Test whether fog = gof.
Solution:
f = {(1, 3), (2, 4), (3, 7)}
g = {(3, 2), (4, 3), (7, 1)}
We have fog (3) = f (g(3)) = f(2) = 4
fog (4) = f (g(4)) = f(3) = 7
fog (7) = f (g(7)) = f(1) = 3
∴  fog = {(3, 4), (4, 7), (7, 3)}
Again gof (1) = g (f(1)) = g(3) = 2,
gof (2) = g (f(2)) = g(4) = 3,
gof (3) = g (f(3)) = g(7) = 1
∴ gof = {(1, 2), (2, 3), (3, 1)}
∴ gof ≠ fog
So the composition of functions is not necessarily commutative.

Question 10.
Let f(x) =√x and g(x) = 1 – x².
(i) Find natural domains of f and g.
(ii) Compute fog and gof and find their natural domains.
(iii) Find natural domain of h(x) = 1 – x.
(iv) Show that h = gof only on R₀ = {x ∈ R : x ≥ 0} and not on R.
Solution:
Let f(x) = √x, g(x) = 1 – x²
(i) ∴ Dom f = R₊ U_{0}, Dom g = R

(ii) fog (x) = f (g(x))
= f (1 – x²) = \(\sqrt{1-x^2}\)
∴ fog (x) exists when 1 – x² ≥ 0
⇒ x² ≤ 1 ⇒ -1 ≤ x ≤ 1 i.e., x ∈ [-1, 1]
∴ Dom fog = [-1, 1]
Again gof (x) = g (f(x))
= g( √x ) = 1 – ( √x )² = 1 – x
∴ Dom gof = R₀ = (0, ∞)

(iii) Domain of h(x) = 1 – x is R.

(iv) We have proved in (ii) that gof (x) = 1 – x.
∴ h(x) = gof (x) ⇒ h = gof only when x ∈ R₀ as dom f is R₀ = [0, ∞]

Question 11.
Find the composition fog and gof and test whether fog = gof when f and g are functions on R given by the following:
(i) f(x) = x³ + 1, g(x) = x² – 2
(ii) f(x) = sin x, g(x) = x⁵
(iii) f(x) = cos x, g(x) = sin x²
(iv) f(x) = g(x) = (1 – x³)^1/3
Solution:
(i) f(x) = x³ + 1, g(x) = x² – 2
∴ fog (x) = f (g(x)) = f(x² – 2)
= (x² – 2)³ + 1
gof (x) = g (f(x)) = g(x³ + 1)
= (x³ + 1)² – 1
fog ≠ gof

(ii) f(x) = sin x, g(x) = x⁵
∴ fog (x) = f (g(x)) = f(x⁵) = sin x⁵
∴ gof (x) = g (f(x))
= g(sin x) = (sin x)⁵ = sin⁵ x
fog ≠ gof

(iii) f(x) = cos x, g(x) = sin x²
∴ fog (x) = f (g(x))
= f( sin x²) = cos (sin x²)
and gof (x) = g (f(x)) = g(cos x)
= sin (cos x)² = sin (cos² x)
fog ≠ gof

(iv) f(x) = g(x) = (1 – x³)^1/3
fog (x) = f (g(x))
= (1 – (g(x))³)^1/3
= [1 – (1 – x³)]^1/3 = x
gof (x) = g (f(x))
= [1 – (f(x))³]^1/3 = x
⇒ fog = gof

Question 12.
(a) Let f be a real function. Show that h(x) = f(x) + f(-x) is always an even function and g(x) = f(x) – f(-x) is always an odd function.
(b) Express each of the following function as the sum of an even function and an odd function:
(i) 1 + x + x² , (ii) x², (iii) e^x, (iv) e^x + sin x
Solution:
(a) We have h(x) = f(x) + f(-x)
∴ h(-x) = f(-x) + f(x) = h(x)
∴ h is always an even function.
Further, g(x) = f(x) – f(-x)
∴ g(-x) = f(-x) – f(x)
= – [f(x) – f(-x)] = – g(x).
∴ g is always an odd function.

(b) (i) Let f(x) = 1 + x + x²
∴ f(-x) = 1 – x + x²
∴ g(x) = \(\frac{f(x)+f(-x)}{2}\)
= \(\frac{1+x+x^2+1-x+x^2}{2}\)
= x² + 1 and
g(-x) = (-x)² + 1 = x² + 1
∴ g is an even function.
h(x) = \(\frac{f(x)-f(-x)}{2}\)
= \(\frac{\left(1+x+x^2\right)-\left(1-x+x^2\right)}{2}\) = x
h(-x) = -x = -h(x)
⇒ h is an odd function.
∴ f(x) = g(x) + f(x)
where g is even and h is odd.

(ii) Let f(x) = x²
So that f(-x) = (-x)² = x²
∴ g(x) = \( \frac{f(x)+f(-x)}{2}\) = \( \frac{x^2+x^2}{2}\) = x²
g(-x) = g(x)
∴ g is an even function.
and h(x) = \( \frac{f(x)-f(-x)}{2}\) = \(\frac{x^2-x^2}{2}\) = 0
h(x) = 0 is both even and odd.
∴ f(x) = g(x) = f(x),
where g is even and h is odd.

(iii) Let f(x) = e^x
f(-x) = e^-x
g(x) = \( \frac{e^x+e^{-x}}{2}\)
g(-x) = g(x)
g is an even function.
and h(x) = \( \frac{e^x-e^{-x}}{2}\)
h(-x) = \( \frac{e^{-x}-e^x}{2}\) = -h²(x)
⇒ h is an odd function.
∴ f(x) = g(x) + h(x),
where g is even and h is odd.

(iv) Let f(x) = e^x + sin x
f(-x) = e^-x + sin (-x) = e^-x –  sin x
∴ g(x) = \( \frac{f(x)+f(-x)}{2}\)
= \( \frac{e^x+\sin x+e^{-x}-\sin x}{2}\)
= \( \frac{e^x+e^{-x}}{2}\) and h(x) = \( \frac{f(x)-f(-x)}{2}\)
= \( \frac{e^x+\sin x-e^{-x}+\sin x}{2}\)
= \( \frac{e^x-e^{-x}+2 \sin x}{2}\)
∴ f(x) = g(x) + h(x)
where g is even and g is odd.

Question 13.
Let X = {1, 2, 3, 4} Determine whether f : X → X defined as given below have inverses.
Find f^-1 if it exists:
(i) f = {(1, 4), (2, 3), (3, 2), (4, 1)}
(ii) f = {(1, 3), (2, 1), (3, 1), (4, 2)}
(iii) f = {(1, 2), (2, 3), (3, 4), (4, 1)}
(iv) f = {(1, 1), (2, 2), (2, 3), (4, 4)}
(v) f = {(1, 2), (2, 2), (3, 2), (4, 2)}
Solution.
(i) x = {1, 2, 3, 4}
f is bijective. Hence f^-1 exists.
f^-1 = {(4, 1), (3, 2), (2, 3), (1, 4)}

(ii) f(2) = f(3) = 1
⇒ f is not injective
∴ f is not invertible.

(iii) f is bijective. Hence f^-1 exists.
f^-1 = {(2,1 ), (3, 2), (4, 3), (1, 4)}

(iv) f is not a function as
f(2) = 2 and f(2) = 3

(v) f is not injective hence not invertible.

Question 14.
Let f : X → Y.
If there exists a map g : Y → X such that gof = id_x and fog = id_y, then show that
(i) f is bijective and (ii) g = f^-1
[Hint: Since id_x is a bijective function, gof = id_x is bijective. By Theorem 2(iv) f is injective. Similarly fog is bijective ⇒ f is surjective by Theorem 2(iii)]
Solution:
Let f : x → y and g : y – x
where gof = id_x and fog = id_y
we know that id_x and id_y are bijective functions.
⇒ gof and fog are both bijective functions.
⇒ f is a bijective function.

(ii) As f is bijective (by (i)) we have f^-1 exists.
and f^-1 : y → x where f^-1of = id_x and fof^-1 = id_y
But g : y → x with gof = id_x and fog = id_y
∴ g = f^-1

Question 15.
Construct an example to show that f(A ∩ B) ≠ f(A) ∩ f(B) where A ∩ B ≠ Ø
Solution:
Let f(x) = cos x.
Let A = \(\left\{0, \frac{\pi}{2}\right\}\), B = \(\left\{\frac{\pi}{2}, 2 \pi\right\}\).
∴ f(A) = \(\left\{\cos 0, \cos \frac{\pi}{2}\right\}\)
= {1, 0} = {0, 1}
∴ f(B) = \(\left\{\cos \frac{\pi}{2}, \cos 2 \pi\right\}\) = {0, 1}
∴ f(A) ∩ f(B) = {0, 1}
Again,
A ∩ B = \(\left\{\frac{\pi}{2}\right\}\) and f(A ∩ B) = cos \(\frac{\pi}{2}\) = {0}
∴ f(A ∩ B) ≠ f(A) ∩ f(B)

Question 16.
Prove that for any f : X → Y, foid_x = f = id_yof.
Solution:
Let f : X → Y, so that y = f(x), x ∈ X.
∴ foid_x = fof^-1 of (x) = fof^-1 (f(x))
= f(x) = y (∵ id_x = fof^-1‍)
Again, (id_yof)(x) = (fof^-1) of (x)
= (fof^-1)(y) = f (f^-1(y)) = f(x) = y  ….(2)
∴ From (1) and (2)
we have foid_x = f = id_yof

Question 17.
Prove that f : X → Y is surjective iff for all B ⊆ Y, f (f^-1(B)) = B.
Solution:
Let f : X → Y is surjective.
i.e. for all y ∈ Y, ∃ a x ∈ X such that
y = f(x).
∴ x = f^-1(y) ⇔ f(x)
= f (f^-1(y)) ∈ f (f^-1(B)).
for y = B ⊂ Y ⇔ y ∈ f (f^-1(B)).
∴ y ∈ f (f^-1(B)) ⇔ y ∈ B
∴ f (f^-1(B)) = B

Question 18.
Prove that f : X → Y is injective iff f (f^-1(A)) = A for all A ⊆ X.
Solution:
f : X → Y is injective.
Let x ∈ A ⇔ f(x) ∈ f(A) (∵ f is injective)
⇔ x ∈ f (f^-1(A))
∴ A = f (f^-1(A)) for all A ⊆ X.

Question 19.
Prove that f : X → Y is injective iff for all subsets A, B of X, f(A ∩ B) = f(A) ∩ f(B).
Solution:
f : X → Y is injective.
Let A and B are subsets of X.
Let f(x) ∈ f(A ∩ B)
⇔ x ∈ A ∩ B ⇔ x ∈ A ∧ x ∈ B
⇔ f(x) ∈ f(A) ∧ f(x) ∈ f(B) (∵ f is injective)
⇔ f(x) = f(A) ∩ f(B)
∴ f(A ∩ B) = f(A) ∩ f(B)

Conversely, suppose that
f(A ∩ B) = f(A) ∩ f(B)
Let f is not injective.
The if f(x) ∈ f(A ∩ B) ⇔ x ∈ A ∩ B
⇔ x ∈ A ∧ ⇔ x ∈ B
≠ f(x) ∈ f(A) ∧ ⇔ f(x) ∈ f(B)
⇔ f(x) ∈ f(A) ∩ f(B)
∴ f(A ∩ B) = f(A) ∩ f(B) is false.
so f must be injective.

Question 20.
Prove that f : X → Y is surjective iff for all A ⊆ X, (f(A))‘ ⊆ f(A‘), where A‘ denotes the complement of A in X.
Solution:
f : X → Y is surjective.
Then for all y ∈ Y ∃ x ∈ X
such that f(x) = y.
Let y ∈ [f(A)]‘ ⇒ y ∉ f(A)
⇒ f(x) ∉ f(A) ⇒ x ∉ A ⇒ x ∈ A‘
⇒ f(x) ∈ f(A‘) ⇒ y ∈ f(A‘)
∴ [f(A)]‘ ⊂ f(A‘)

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Exercise 3(a)

Question 1.
A merchant sells two models X and Y of TV with cost price ₹25000 and ₹50000 per set respectively. He gets a profit of ₹1500 on model X and ₹2000 on model Y. The sales cannot exceed 20 sets in a month. If he cannot invest more than 6 lakh rupees, formulate the problem of determining the number of sets of each type he must keep in stock for maximum profit.
Solution:
To get maximum profit let x TVs of model X and Y TVs of model Y must be kept in stock.
∴ Total profit = Z = 1500x + 2000y which is to be maximum.
According to the question, the sales cannot exceed 20 sets i.e. x + y ≤ 20.
Again total investment does not exceed 6 lakh.
25000x + 50000y ≤ 600000
⇒ x + 2y ≤ 24
∴ The LPP is:
maximise: Z = 1500x + 2000y
subject to: x + y ≤ 20
x + 2y ≤ 24
x, y ≥ 0

Question 2.
A company manufactures and sells two models of lamps L₁ and L₂, the profit being ₹15 and ₹10 respectively. The process involves two workers W, and W2 who are available for this kind of work 100 hours and 80 hours per month respectively, W1 assembles L₁ in 20 and L₂ in 30 minutes. W2 paints L₁ in 20 and L₂ in 10 minutes. Assuming that all lamps made can be sold, formulate the LPP for determining the production figures for maximum profit.
Solution:
Let x units of L₁ and y units of L₂: are to be produced to get maximum profit.
Total profit = Z = 15x + 10y
According to the question
20x + 30y ≤ 600
and 20x + 10y ≤ 480
=> 2x + 3y ≤ 600
2x + y ≤ 480
.-. The LPP is maximize: Z = 15x + 10y
subject to: 2x + 3y ≤ 600
2x + y ≤ 480
x, y ≥ 0.

Question 3.
A factory uses three different resources for the manufacture of two different products, 20 units of the resource A, 12 units of B and 16 units of C being available. One unit of the first product requires 2, 2 and 4 units of the resources and one unit of the second product requires 4, 2 and 0 units of the resources taken in order. It is known that the first product gives a profit of ₹20 per unit and the second ₹30 per unit. Formulate the LPP so as to earn maximum profit.
Solution:
Let to earn maximum profit the factory produces x units of first product and y units of the second product.
The given data can be summarised as below:

	Resource A	Resource B	Resource C	Profit per unit in ₹
Product – I	2	2	4	20
Product – II	4	2	0	30
Availability	20	12	16

Total profit = 20x + 30y
which is to be maximum.
According to the question
2x + 4y ≤ 20
2x + 2y ≤ 12
4x + 0y ≤ 16
.-. The LPP is
maximize: Z = 20x + 30y
subject to: x + 2y ≤ 10
x + y ≤ 6
x ≤ 4
x, y ≥ 0.

Question 4.
A man plans to start a poultry farm by investing at most ₹3000. He can buy old hens for ₹80 each and young ones for ₹140 each, but he cannot house more than 30 hens. Old hens lay 4 eggs per week and young ones lay 5 eggs per week, each egg being sold at ₹5. It costs ₹5 to feed an old hen and ₹8 to feed a young hen per week. Formulate his problem determining the number of hens of each type he should buy so as to earn a profit of more than ₹300 per week.
Solution:
Let to get maximum profit he has to purchase x old hens and y young hens.
Total cost = 80x + 140y ≤ 3000
⇒ 4x + 7y ≤ 150
Total number of hens x + y ≤ 30
Number of eggs per week = 4x + 5y
Total income per week = 20x + 25y
Total cost to feed per week = 5x + 8y
∴ Weekly profit = 15x + 17y
∴ 15x + 17y > 300
and also total profit = z = 15x + 17y is to be maximum.
∴ The LPP is
maximize: Z = 15x + 17y
subject to: 4x + 7y ≤ 150
x + y ≤ 30
15x + 17y ≥ 300
x, y ≥ 0.

Question 5.
An agro-based company produces tomato sauce and tomato jelly. The quantity of material, machine hour, labour (man-hour) required to produce one unit of each product and the availability of raw material ore given in the following table:

	Sauce	Jelly	availability
Man-hour	3	2	10
Machine hour	1	2.5	7.5
Raw material	1	1.2	4.2

Assume that one unit of sauce and one unit of jelly each yield a profit of ₹2 and ₹4 respectively. Formulate the LPP so as to yield maximum profit.
Solution:
Let the company produces x units of sauce and y units of jelly.
Total profit = 2x + 4y to be maximum.
Man hour = 3x + 2y ≤ 10
Machine hour = x + 2.5y ≤ 7.5
⇒ 2x + 5y ≤ 15
Raw material = x + 1.2y ≤ 4.2
⇒ 5x + 6y ≤ 21
∴ The LPP is
maximize: Z = 2x + 4y
subject to: 3x + 2y ≤ 10
2x + 5y ≤ 15
5x + 6y ≤ 21
x, y ≥ 0.

Question 6.
(Allocation Problem.) A farmer has 5 acres of land on which he wishes to grow two crops X and Y. He has to use 4 cart loads and 2 cartloads of manure per acre for crops X and Y respectively. But not more than 18 cartloads of manure is available. Other expenses are ₹200 and ₹500 per acre for the crops X and Y respectively. He estimates profit from crops X and Y at the rates ₹1000 and ₹800 per acre respectively. Formulate the LPP as to how much land he should allocate to each crop for maximum profit.
Solution:
Let x acres are allocated for crop X and y acres for crop Y.
Total profit = 100x + 800y to be maximum.
According to the question
x + y ≤ 5
Manure = 4x + 2y ≤ 1 8 ⇒ 2x + y ≤ 9
∴ The LPP is
maximize: Z = 1000x + 800y
subject to: x + y ≤ 5
2x + y ≤ 9
x, y ≥ 0.

Question 7.
(Transportation Problem) A company has two factories at locations X and Y. He has to deliver the products from these factories to depots located at three places A, B and C. The production capacities at X and Y are respectively 12 and 10 units and the requirements at the depots are 8, 8 and 6 units respectively. The cost of transportation from the factories to the depots per unit of the product is given below.

(Cost in ₹)
	To →	A	B	C
From	X	210	160	250
	Y	170	180	140

The company has to determine how many units of product should be transported from each factory to each depot so that the cost of transportation is minimum. Formulate this LPP.
Solution:
Let x units are transported from X to A and y units from X to B. The transportation matrix is

Total cost of transportation
= 210x + 160y + 250 (12 – x – y) +170 (8 – x) +180 (8 – y) + 140 (x + y – 6)
= 4960 – 70x – 130y to be minimum.
Now all costs of transportation are ≥ 0.
∴ x ≥ 0, y ≥ 0.
12 – x – y ≥ 0 ⇒ x + y ≤ 12
8 – x ≥ 0 ⇒ x ≤ 8
8 – y ≥ 0 ⇒ y ≤ 8
x + y – 6 ≥ 0 x + y ≥ 6
∴ The LPP is
Minimize: Z = 4960 – 70x – 130y
subject to: x + y ≤ 12
x ≤ 8
y ≤ 8
x + y ≥ 6
x, y ≥ 0

Question 8.
(Diet Problem) Two types of food X and Y are mixed to prepare a mixture in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. These vitamins are available in one kg of food as per the table given below.

	Vitamins
food	A	B	C
X	1	2	3
Y	2	2	1

One kg. of food X cost ₹16 and one kg. of food Y costs ₹20. Formulate the LPP so as to determine the least cost of the mixture containing the required amount of vitamins.
Solution:
Let x units of food X and y units of food Y are to be mixed to prepare the mixture.
Cost of the mixture = 16x + 20y to be minimum.
According to the question
Vitamin A content = x + 2y ≥ 10
Vitamin B content = 2x + 2y ≥ 12
Vitamin C content = 3x + y ≥ 8
∴ The LPP is
minimize: Z = 16x + 20y
subject to x + 2y ≥ 10
x + y ≥ 6
3x + y ≥ 8
x, y ≥ 0.

Question 9.
Special purpose coins each weighing 10gms are to be manufactured using two basic metals M₁ and M₂ and a mix of other metals M₃. M₁, M₂ and M₃ cost ₹500, ₹800 and ₹50 per gram respectively. The strength of a coin demands that not more than 7gm. of M₁ and a minimum of 3 gm of M₁ should be used. The amount of M₃ in each coin is maintained at 25% of that of M₁. Since the demand for the coin is related to its price, formulate the LPP to find the minimum cost of a coin.
Solution:
Let x gm of M₁ and y g of M₂ are used to make the coin. According to the demand of the coin \(\frac{x}{4}\) g of M₃ is to be mixed.
Cost of the coin = 500x + 800y + \(\frac{50 x}{4}\)
= (512.5)x + 800y
which is to be minimum.
Again weight of the coin = 10g
⇒ x + y + \(\frac{x}{4}\) = 10
⇒ 5x + 4y = 40
According to the question x ≤ 7, y ≤ 3.
Thus the L.P.P is
minimize: Z = (512.5)x + 800y
subject to: 5x + 4y = 40
x ≤ 7
x ≤ 3
x, y ≥ 0.

Question 10.
A company produces three types of cloth A, B and C. Three kinds of wool, say red, green and blue are required for the cloth. One unit length of type A cloth needs 2 metres of red and 3 metres of blue wool; one unit length of type B cloth needs 3 metres of red, 2 metres of green and 2 metres of blue wool and one unit length of type C cloth needs 5 metres of green and 4 metres of blue wool. The firm has a stock of only 80 metres of red, 100 metres of green and 150 metres of blue wool. Assuming that income obtained from one unit length of cloth is ₹30, ₹50 and ₹40 of types A, B and C respectively, formulate the LPP so as to maximize income.
Solution:
Let x units of cloth A, y units of cloth B and z units of cloth C are to be produced from the available materials to get the maximum income. The given data can be summarised as:

	Red wool	Green wool	Blue wool	Income
Cloth A	2	–	3	30
Cloth B	3	2	2	50
Cloth C	–	5	4	40
Availability	80	100	150

Total Income = 30x + 50y + 40z
which is to be maximum.
According to the question
2x + 3y ≤ 80
2y + 5z ≤ 100
3x + 2y + 4z ≤ 150
Thus the L.P.P. is
minimize: Z = 30x + 50y + 40z
subject to: 2x + 3y ≤ 80
2y + 5z ≤ 100
3x + 2y + 4z ≤ 150
x, y, z ≥ 0

Question 11.
A person wants to decide the constituents of a diet which will fulfil his daily requirements of proteins, fats and carbohydrates at minimum cost. The choice is to be made from three different types of food. The yields per unit of these foods are given in the following table.

food	yield/unit			cost/unit
	Protein	Fat	Carbonate
f1	3	2	6	45
f2	4	2	3	40
f3	8	7	7	85
Minimum Requirement	100	200	800

Formulate the LPP.
Solution:
Let the diet constitues x units of f1, y units of f2 and z units of f3.
Total cost = 45x + 40y + 85z, which is to be minimum.
According to the question
3x + 4y + 8z ≥ 1000
2x + 2y + 7z ≥ 200
6x + 3y + 7z ≥ 800
Thus the LPP is
Minimize: Z = 45x + 40y + 85z
Subject to: 3x + 4y + 8z ≥ 1000
2x-+2y + 7z ≥ 200
6x + 3y + 7z ≥ 800
x, y, z ≥ 0

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 1 Relation and Function Ex 1(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Exercise 1(c)

Question 1.
Show that the operation ∗ given by x ∗ y = x + y – xy is a binary operation on Z, Q and R but not on N.
Solution:
The operation ∗ given by
x ∗ y = x + y – xy
Clearly for all x, y ∈ Z
x + y – xy ∈ Z
⇒ x ∗ y ∈ Z
∴ ∗ is a binary operation on Z.
For all x, y ∈ Q
x + y – xy ∈ Q
⇒ x ∗ y ∈ Q
⇒ ∗ is a binary operation on Q for all x, y, ∈ R.
x + y – xy ∈ R
⇒ x ∗ y ∈ R
⇒ ∗ is a binary operation on R
Again 3, 4 ∈ N.
3 + 4 – 3 x 4 = 7 – 12 = – 5 ∉ N
i.e., x, y ∈ N
≠ x ∗ y ∈ N
∴ ∗ is not a binary operation on N.

Question 2.
Determine whether the following operations as defined by ∗ are binary operations on the sets specified in each case. Give reasons if it is not a binary operation.
(i) a ∗ b = 2a + 3b on Z.
(ii) a ∗ b = ma – nb on Q⁺ where m and n ∈ N.
(iii) a ∗ b = a + b (mod 7) on {0, 1, 2, 3, 4, 5, 6}
(iv) a ∗ b = min {a, b} on N.
(v) a ∗ b = GCD {a, b} on N.
(vi) a ∗ b = LCM {a, b} on N.
(vii) a ∗ b = LCM {a, b} on {0, 1, 2, 3, 4……, 10}
(viii) a ∗ b = \(\sqrt{a^2+b^2}\) on Q⁺
(ix) a ∗ b =a × b (mod 5) on {0, 1, 2, 3, 4}.
(x) a ∗ b = a² + b² on N.
(xi) a ∗ b = a + b – ab on R – {1}.
Solution:
(i) For all a, b ∈ Z
2a + 3b ∈ Z
⇒ a ∗ b ∈ Z
∗ is a binary operation on Z.

(ii) Let a = 1, b = 2
m = 1, n = 3
ma – nb = 1 – 6 = – 5 ∉ Q⁺
∴ a, b ∈ Q⁺ ≠ a ∗ b ∈ Q⁺
⇒ ∗ is not a binary operation on Q⁺

(iii) a ∗ b = a + b (mod 7) ∈ {0, 1, 2, 3, 4, 5, 6}
for a, b ∈ 7
∗ is a binary operation on the given set.

(iv) a, b ∈ N ⇒ min {a, b} ∈ N
∴ a ∗ b ∈ N
⇒ ∗ is a binary operation on N.

(v) for all a, b ∈ N, GCD [a, b] ∈ N
⇒ a ∗ b ∈ N
⇒ ∗ is a binary operation on N.

(vi) for all a, b ∈ N, LCM {a, b} ∈ N
⇒ a ∗ b ∈ N
⇒ ∗ is a binary operation on N.

(vii) Let A = {0, 1, 2, ….. 10}
4, 5 ∈ A but 4 ∗ 5 = LCM {4, 5}
= 20 ∉ A
⇒ ∗ is not a binary operation on A.

(viii) for all a, b ∈ Q⁺
a ∗ b = \(\sqrt{a^2+b^2}\) ∉ Q⁺
⇒ ∗ is not a binary operation on Q⁺.

(ix) For all a, b ∈ {0, 1, 2, 3, 4}
a ∗ b = a × b (mod 5) ∈ {0, 1, 2, 3, 4}
∴ ∗ is a binary operation on the given set.

(x) for all a, b ∈ N, a * b = a² + b² ∈ N
∴ ∗ is a binary operation on N.

(xi) For all a, b ∈ R – {1}
a ∗ b = a + b – ab ∈ R – {1}
∴ ∗ is a binary operation on R – {1}

Question 3.
In case ∗ is a binary operation in Q2 above, test whether it is (i) associative
(ii) commutative, Test further if the identity element exists and the inverse element for any element of the respective set exists.
Solution:
(i) On Z the binary operation is
a ∗ b = 2a + 3b
Commutative:
b ∗ a = 2b + 3a ≠ a ∗ b
∴ ∗ is not commutative.

Associative:
(a ∗ b) ∗ c = (2a + 3b) ∗ c
= 2 (2a + 3b) + 3c
= 4a + 6b + 3c
a ∗ (b ∗ c) = a ∗ (2b + 3c)
= 2a + 3 (2b + 3c)
= 2a + 6b + 9c
As (a ∗ b) ∗ c ≠ a ∗ (b ∗ c)
∗ is not associative.

Existance of identity:
Let e is the identity
∴ e ∗ a = a
⇒ 2e + 3a = a
⇒ e = -2a / 2 = -a
which depends on a.
∴ Identity element does not exist.

(iii) A = (0, 1, 2, 3, 4, 5, 6}
Commutative:
a ∗ b = a + b (mod 7)
= The remainder obtained when a + b is divided by 7.
b ∗ a = b + a (mod 7) = a + b (mod 7)
∴ ∗ is commutative.

Associative:
(a ∗ b) ∗ c = {a + b (mod 7)} ∗ c
= a + b + c (mod 7)
= The remainder obtained if a + b + c is divided by 7.
a ∗ (b ∗ c) = a ∗ {b + c (mod 7)}
= a + b + c (mod 7)
= The remainder obtained if a + b + c is divided by 7.
∴ (a ∗ b) ∗ c = a + (b ∗ c)
∴ ∗ is associative.

Existance of identity:
Let e is the identity
⇒ e ∗ a = a ∗ e = a
⇒ e + a mod 7 = a
⇒ e = 0
∴ 0 is the identity.

Existance of inverse:
Let a^-1 = the inverse of a
⇒ a ∗ a^-1 = a^-1 ∗ a = e = 0
⇒ a + a^-1 (mod 7) = 0
⇒ a + a^-1 is divisible by 7.
1^-1 = 6, 6^-1 = 1
2^-1 = 5, 5^-1 = 2
3^-1 = 4, 4^-1 = 3

(iv) a ∗ b = min {a, b} on N.
Commutative:
a ∗ b = min {a, b}
b ∗ a – min {b, a} = a ∗ b
∴ ∗ is commutative.
Associative:
(a ∗ b) ∗ c = min {a, b} ∗ c
= min {a, b, c}
a ∗ (b ∗ c)= a ∗ min {b, c}
= min {a, b, c}
⇒ a ∗ (b ∗ c) = (a ∗ b) ∗ c
∴ ∗ is associative.

Existance of Identity:
Let e is the identity
∴ For all a ∈ N
e ∗ a = a ∗ e = a
⇒ min {e, a} = a
No such element exists in N.
∴ ∗ has no identity element on N.

(v) a ∗ b = GCD {a, b} on N.
b ∗ a = GCD {b, a} = GCD {a, b} = a ∗ b
∴ ∗ is commutative.
Associative:
(a ∗ b) ∗ c = GCD {a, b} ∗ c
= GCD {a, b, c}
a ∗ (b ∗ c) = a ∗ GCD {b, c}
= GCD {a, b, c}
⇒ (a ∗ b) ∗ c = a ∗ (b ∗ c)
⇒ ∗ is associative.

Existance of Identity:
Let e is the identity
∴ a ∗ e = e ∗ a = a
⇒ GCD {e, a} = a
No such element exists in N
⇒ ∗ has no indentity element.

(vi) a ∗ b = LCM {a, b} on N
Commutative:
a ∗ b = LCM {a, b}
= LCM {b, a}
= b ∗ a
∴ ∗ is commutative.

Associative
(a ∗ b) ∗ c = LCM {a, b} ∗ c
= LCM {a, b, c}
a ∗ (b ∗ c) =» a ∗ LCM {b, c}
= LCM {a, b, c}
⇒ (a ∗ b) ∗ c = a ∗ (b ∗ c)
∴ ∗ is associative.

Existance of Identity:
Let e is the identity
∴ e ∗ a = a ∗ e = a
⇒ LCM {e, a} = a
⇒ e – 1
∴ 1 is the identity element.

Existance of inverse:
Let a^-1 is the inverse of a
⇒ a * a^-1 = e = 1
⇒ LCM [a, a^-1} = 1
a = a^-1 = 1
Only 1 is invertible with 1^-1 = 1.

(ix) a ∗ b = a × (mod 5) on {0, 1, 2, 3, 4}

Commutative:
a ∗ b = a x b (mod 5)
= Remainder on dividing a x b by 5
= Remainder on dividing b x a by 5
= b x a (mod 5)
= b x a
∴ ∗ is commutative.

Associative:
(a ∗ b) ∗ c=a x b (mod 5) ∗ c
= a x b x c (mod 5)
a and a ∗ {b ∗ c} = a ∗ {b x c (mod 5)}
= a x b x c (mod 5)
∴ (a ∗ b) ∗ c -=a ∗ (b ∗ c)
⇒ ∗ is associative.

Existance of identity:
Let e is the identity
∴ For all a ∈ {0, 1, 2, 3, 4}
a ∗ a = e ∗ a = a
a × e (mod 5) = a
⇒ e = 1
∴ 1 is the identity element.

Existance of inverse:
Let a^-1 is the inverse of a
∴ a ∗ a^-1= a^-1 ∗ a = e = 1
⇒ a x a^-1 (mod 5) = 1
⇒ 1^-1 = 1
2^-1 = 3, 3^-1 = 2, 4^-1 = 4
0 has no inverse.

(x) a ∗ b = a² + b² on N.
Commutative:
a ∗ b = a² + b²
b ∗ a = b² + a² = a² + b² = a ∗ b
∴ ∗ is commutative.

Associative:
(a ∗ b) ∗ c = (a² + b²) ∗ c
= (a² + b²)² + c²
a ∗ (b ∗ c) = a ∗ (b² + c²)
= a² + (a² + b²)²
(a ∗ b) ∗ c ≠ a ∗ (b ∗ c)
∴ ∗ is not associative.

Existance of Identity:
Let e is the identity
a ∗ e = e ∗ a = a
⇒ a² + e² = a
⇒ e = \( \sqrt{a-a^2}\) which depends on a
∴ Identity does not exist.

(xi) a ∗ b = a + b – ab on R – {1}
Commutative:
a ∗ b = a + b – ab
b ∗ a = b + a – ba
a ∗ b = b ∗ a
∴ ∗ is commutative.

Associative:
a ∗ (b ∗ c) = a ∗ (b + c – bc)
= a + (b + c – bc) – a (b + c – bc)
= a + b + c – bc – ab – ac + abc
(a ∗ b) ∗ c = (a + b – ab) ∗ c
= a + b – ab + c – (a + b – ab) c
= a + b + c – ab – bc – ca + abc
∴ (a ∗ b) ∗ c = a ∗ (b ∗ c)
⇒ ∗ is associative.

Existance of Identity:
Let e is the identity
∴ e ∗ a = a ∗ e = a
⇒ a + e – ae = a
⇒ e (1 – a) = 0
⇒ e = 0 (∵ a ≠ 1)
∴ 0 is the identity.

Existance of inverse:
Let a^-1 is the inverse of a
⇒ a ∗ a^-1 = a^-1 ∗ a = e
⇒ a + a^-1 – aa^-1 = 0
⇒ a^-1 (1 – a) = – a
⇒ a^-1 = \(\frac{a}{a-1}\) for a ∈ R – {1}

Question 4.
Construct the composition table/multiplication table for the binary operation ∗ defined on {0, 1, 2, 3, 4} by a ∗ b = a × b {mod 5). Find the identity element if any. Also find the inverse elements of 2 and 4.
[This operation is called multiplication moduls 5 and denoted by x⁵. In general, on a finite subset of N, x_m denotes the operation of multiplication modulo m where m is a fixed positive integer].
Solution:
A = {0. 1, 2, 3, 4}
a ∗ b = a × b mod 5

∗	0	1	2	3	4
0	0	0	0	0	0
1	0	1	2	3	4
2	0	2	4	1	3
3	0	3	1	4	2
4	0	4	3	2	1

As 3rd row is identical to the first row we have 1 is the identity clearly 2^-1 = 3 and 4^-1 = 4.

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Exercise 3(b)

Question 1.
Maximize Z = 5x₁+ 6x₂
Subject to: 2x₁ + 3x₂ ≤ 6
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraint as equation, we get 2x₁ + 3x₂ = 6
Step – 2 Let us draw the graph

x1	3	0
x2	0	0

Step – 3 Clearly (0,0) statisfies 2x₁ + 3x₂ ≤ 6
The shaded region is the feasible region with vertices 0(0,0), A(3,0), B(0,2).
Step – 4

Corner point	Z = 5x1+ 6x2
0(0.0)	0
A(3,0)	15 → maximum
B(0,2)	12

Z is maximum at A (3,0)
∴ The solution of LPP is x₁ = 3, x₂ = 0
Z_max = 15

Question 2.
Minimize: Z = 6x₁ + 7x₂
Subject to: x₁ + 2x₂ ≥ 4
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraint as equation we get x₁ + 2x₂ = 0
Step – 2 Let us draw the graph of x₁ + 2x₂ = 4

x1	0	4
x2	2	0

Step – 3 Clearly 0(0,0) does not satisfy
x₁ + 2x₂ > 4, x₁ > 0, x₂ > 0 is the first quadrant.
The feasible region is the shaded region with vertices A(4, 0), B(0, 2).
Step – 4 Z (4, 0) = 24
Z (0, 2) = 14 → minimum
Step – 5 As the feasible region is unbounded we cannot immediately decide Z is minimum at B (0, 2).
Let us draw the half-plane 6x₁ + 7x₂ < 14

x1	0	3.5
x2	2	-1

As this half-plane has no point common with the feasible region, we have Z is minimum for x₁= 0, x₂ = 2 and the minimum value of Z = 14.

Question 3.
Maximize Z = 20x₁+ 40x₂
Subject to: x₁ + x₂ ≤ 1
6x₁ + 2x₂ ≤ 3
x₁, x₂ ≥ 0.
Solution:
Step – 1 Treating the constraints as equations
x₁ + x₂ = 1    …. (1)
6x₁ + 2x₂ = 3   …. (2)
x₁, x₂ ≥ 0
Step – 2 Let us draw the graph:
Table – 1

x1	0	1
x2	1	0

Table – 2

x1	0	0.5
x2	1.5	0

Step – 3 As (0, 0) satisfies both the inequations the shaded region is the feasible region.
Step – 4 Solving
x₁ + x₂ = 1
6x₁ + 2x₂ = 3
we have x₁ = ¼ x₂ = ¾
The vertices are O(0, 0), A(0.5, 0), B(0,1) and C(¼, ¾)
Now Z(O) = 0
Z(A) = 10
Z(B) = 40
Z(C) = 20 × ¼ + 40 × ¾ = 35
∴ Z attains maximum at B for x₁= 0, x₂ = 1
Z_max = 40

Question 4.
Minimize: Z = 30x₁ + 45x₂
Subject to: 2x₁ + 6x₂ ≥ 4
5x₁ + 2x₂ ≥ 5
x₁, x₂ ≥ 0
Solution:
Step – 1 Consider the constraints as equations
2x₁ + 6x₂ = 4
5x₁ + 2x₂ = 5
Step – 2
Table – 1

x1	2	-1
x2	0	1

Table – 2

x1	1	0
x2	0	2.5

Step – 3 Clearly 0(0,0) does not satisfy 2x₁ + 6x₂ ≥ 4 and 5x₁ + 2x₂ ≥ 5.
Thus the shaded region is the feasible region.
Solving the equations we get
x₁ = \(\frac{11}{13}\), x₂ = \(\frac{5}{13}\).
∴ The vertices are A(2, 0)
B(\(\frac{11}{13}\), \(\frac{5}{13}\)) and C(0, \(\frac{5}{2}\)).
Step – 4 Z(A) = 60
Z(B) = \(\frac{555}{13}\) → minimum
Z(C) = \(\frac{225}{2}\)
Step – 5 As the feasible region is unbounded we cannot immediately decide Z is minimum at B(\(\frac{11}{13}\), \(\frac{5}{13}\))
Let us draw the half plane
30x₁ + 45x₂ < \(\frac{555}{13}\)

x1	\(\frac{11}{13}\)	0
x2	\(\frac{5}{13}\)	\(\frac{27}{39}\)

As this half plane and the feasible region has no point in common we have Z is minimum for x₁ = \(\frac{11}{13}\), x₂ = \(\frac{5}{13}\), and Z_min = \(\frac{555}{13}\)

Question 5.
Maximize: Z = 3x₁+ 2x₂
Subject to: -2x₁ + x₂ ≤ 1
x₁ ≤ 2
x₁+ x₂ ≤ 3
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
-2x₁ + x₂ = 1        …..(1)
x₁ = 2                   …..(2)
x₁+ x₂ = 3            …..(3)
Step – 2 Let us draw the lines.
Table – 1

x1	0	-1
x2	1	-1

Table – 2

x1	2	2
x2	0	1

Table – 3

x1	0	3
x2	3	0

Step – 3 (0, 0) satisfies all the constraints and x₁, x₂ > 0 is the 1st quadrant the shaded region is the feasible region.

Step – 4 Solving -2x₁ + x₂ = 1
x₁+ x₂ = 3
we have 3x₁ = 2
⇒ x₁ = \(\frac{2}{3}\), x₂ = 3 – \(\frac{2}{3}\) = \(\frac{7}{3}\)
From x₁+ x₂ = 3 and x₁ = 2 we have x₁ = 2, x₂ = 1
∴ The vertices are 0(0, 0), A(2, 0), B(2, 1), C(\(\frac{2}{3}\), \(\frac{7}{3}\)), D(0, 1)
Z(0) = 0, Z(A) = 6, Z(B) = 8, Z(C) = 3.\(\frac{2}{3}\) + 2.\(\frac{7}{3}\) = \(\frac{20}{3}\), Z(D) = 2
Z is maximum at B.
∴ The solution of given LPP is x₁ = 2, x₂ = 1, Z_(max) = 8.

Question 6.
Maximize: Z = 50x₁+ 60x₂
Subject to: x₁ + x₂ ≤ 5
x₁+ 2x₂ ≤ 4
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x₁ + x₂ = 5     ….(1)
x₁+ 2x₂ = 4    ….(2)
Step – 2 Let us draw the graph
Table – 1

x1	5	5
x2	0	0

Table – 2

x1	4	0
x2	0	2

Step – 3 0(0,0) satisfies x₁ + x₂ ≤ 5 and does not satisfy x₁+ 2x₂ ≤ 4
Thus the shaded region is the feasible region.
Step – 4 The corner points are A(4,0), B(5,0), C(0,5) , D(0,2)
∴

Corner point	z = 50x1+ 60x2
A(4,0)	200
B (5,0)	250 → maximum
C(0,5)	300
D(0,2)	120

Z is maximum for x₁ = 0, x₂ = 5, Z_(max) = 300.

Question 7.
Maximize: Z = 5x₁+ 7x₂
Subject to: x₁ + x₂ ≤ 4
5x₁+ 8x₂ ≤ 30
10x₁+ 7x₂ ≤ 35
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get,
x₁ + x₂ = 4           …. (1)
5x₁+ 8x₂ = 30      …. (2)
10x₁+ 7x₂ = 35    …. (3)
Step – 2 Let us draw the graph
Table – 1

x1	4	0
x2	0	4

Table – 2

x1	6	2
x2	0	2.5

Table – 3

x1	0	3.5
x2	5	0

Step – 3 0(0,0) satisfies all the constraints.
Thus the shaded region is the feasible region.
From (1) and (2) we get (\(\frac{2}{3}\), \(\frac{10}{3}\))
From (1) and (3) we get
x₁ = \(\frac{7}{3}\), x₁ = \(\frac{5}{3}\)
∴ The corner points are 0(0,0), A(\(\frac{7}{2}\), 0), B(\(\frac{7}{3}\), \(\frac{5}{3}\)), C(\(\frac{2}{3}\), \(\frac{10}{3}\)), D(0, \(\frac{15}{4}\))
Step – 4

Corner point	z = 5x1+ 7x2
0(0,0)	0
A(\(\frac{7}{2}\), 0)	\(\frac{35}{2}\)
B(\(\frac{7}{3}\), \(\frac{5}{3}\))	\(\frac{70}{3}\)
C(\(\frac{2}{3}\), \(\frac{10}{3}\))	\(\frac{80}{3}\)
D(0, \(\frac{15}{4}\))	\(\frac{105}{4}\)

Z attains its maximum value \(\frac{80}{3}\) for x₁ = \(\frac{2}{3}\) and x₂ = \(\frac{10}{3}\).

Question 8.
Maximize: Z = 14x₁ – 4x₂
Subject to: x₁ + 12x₂ ≤ 65
7x₁ – 2x₂ ≤ 25
2x₁+ 3x₂ ≤ 10
x₁, x₂ ≥ 0
Also find two other points which maximize Z.
Solution:
Step – 1 Treating the constraints as equations we get
x₁ + 12x₂ = 65   …. (1)
7x₁ – 2x₂ = 25    …. (2)
2x₁ + 3x₂ = 10   …. (3)
Step – 2 Let us draw the graph
Table – 1

x1	65	5
x2	0	5

Table – 2

x1	5	10
x2	5	22.5

Table – 3

x1	5	2
x2	0	2

Step – 3 Clearly 0(0,0) satisfies x₁ + 12x₂ ≤ 65 and 7x₁ – 2x₂ ≤ 25 but does not satisfy 2x₁+ 3x₂ ≤ 10. Thus shaded region is the feasible region.
Equation (1) and (2) meet at (5, 5).
From (2) and (3)

∴ The corner points of the feasible region are A(0, \(\frac{10}{3}\)), B(\(\frac{19}{5}\), \(\frac{4}{5}\)), C(5, 5), D(0, \(\frac{65}{12}\)).
Step – 4

Corner point	z = 14x1 – 4x2
A(0, \(\frac{10}{3}\))	\(\frac{-40}{3}\)
B(\(\frac{19}{5}\), \(\frac{4}{5}\))	50 → maximum
C(5, 5)	50 → maximum
D(0, \(\frac{65}{12}\))	\(\frac{65}{3}\)

Z is maximum for x₁ = \(\frac{19}{5}\), x₂ = \(\frac{4}{5}\) or x₁ = 5, x₂ = 5 and Z_max = 50
There is no other point that maximizes Z.

Question 9.
Maximize: Z = 10x₁ + 12x₂ + 8x₃
Subject to: x₁ + 2x₂ ≤ 30
5x₁ – 7x₃ ≤ 12
x₁ + x₂ + x₃ = 20
x₁, x₂ ≥ 0
[Hints: Eliminate x₃ from all expressions using the given equation in the set of constraints, so that it becomes an LPP in two variables]
Solution:
Eliminating x₃ this LPP can be written as Maximize Z = 2x₁ + 4x₂ + 160
Subject to: x₁ + 2x₂ ≤ 30
5x₁ – 7x₃ ≤ 12
x₁, x₂ ≥ 0
Step – 1 Treating the consraints as equations we get
x₁ + 2x₂ = 30    …..(1)
5x₁ – 7x₃ = 12   …..(2)
Step – 2 Let us draw the graph
Table – 1

x1	30	0
x2	0	15

Table – 2

x1	8	1
x2	8	20

Step – 3 Clearly 0(0,0) satisfies x₁ + 2x₂ ≤ 30 and does not satisfy 12x₁ + 7x₂ ≤ 152
∴ The shaded region is the feasible region.

Step – 4

Z is maximum for x₁ = 30, x₂ = 0 and Z_max = 220

Question 10.
Maximize: Z = 20x₁ + 10x₂
Subject to: x₁ + 2x₂ ≤ 40
3x₁ + x₂ ≥ 30
4x₁+ 3x₂ ≥ 60
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equalities we have:
x₁ + 2x₂ = 40   ….(1)
3x₁ + x₂ = 30   ….(2)
4x₁+ 3x₂ = 60  ….(3)
Step – 2 Let us draw the graph

Step – 3 (0, 0) satisfies x₁ + 2x₂ ≤ 40 and does not satisfy 3x₁ + x₂ ≥ 30 and 4x₁+ 3x₂ ≥ 60, x₁, x₂ ≥ 0 is the first quadrant.
∴ The shaded region is the feasible region.
Step – 4 x₁ + 2x₂ = 40 and 3x₁ + x₂ = 30

∴ The vetices are A(15, 0), B(10, 0), C(4, 18) and D(6, 12)
Z(A) = 300, Z(B) = 800
Z (C) = 20 x 4 + 10 x 18 = 260
Z (D) = 120 + 120 = 240
Z attains minimum at D(6 ,12).
∴ The required solution x₁ = 6, x₂ =12 and Z_min = 240

Question 11.
Maximize: Z = 4x₁ + 3x₂
Subject to: x₁ + x₂ ≤ 50
x₁ + 2x₂ ≥ 80
2x₁+ x₂ ≥ 20
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
x₁ + x₂ ≤ 50    ….(1)
x₁ + 2x₂ ≥ 80  ….(2)
2x₁+ x₂ ≥ 20   ….(3)
Step – 2 Let us draw the graph

Step – 3 (0, 0) satisfies x₁ + x₂ < 50, x₁ + 2x₂ < 80 but does not satisfy
2x₁ + x₂ > 20, x₁ > 0, x₂ > 0 is the 1st quadrant.
Hence the shaded region is the feasible region.
Step – 4 x₁ + x₂ = 50
x₁ + 2x₂ = 80
=> x₂ = 30, x₁ = 20
The vertices of feasible region are
A(10, 0), B(50, 0), C(20, 30), D (0, 40) and E (0, 20)

Point	Z = 4x1 + 3x2
A(10,0)	40
5(50,0)	200
C(20,30)	170
D(0,40)	120
E(0,120)	60

Question 12.
Optimize: Z = 5x₁ + 25x₂
Subject to: -0.5x₁ + x₂ ≤ 2
x₁ + x₂ ≥ 2
-x₁+ 5x₂ ≥ 5
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
-0.5x₁ + x₂ = 2   ….(1)
x₁ + x₂ = 2         ….(2)
-x₁+ 5x₂ = 5      ….(3)
Step – 2 Let us draw the graph.

Step – 3 (0, 0) satisfies -0.5x₁ + x₂ ≤ 2, but does not satisfy x₁ + x₂ ≥ 2 and -x₁+ 5x₂ ≥ 5, x₁ > 0, x₂ > 0 is the 1st quadrant.
The shaded region is the feasible region with vertices A(\(\frac{5}{6}\), \(\frac{7}{6}\)) and B(0, 2).
Step – 4 Z can be made arbitrarily large.
∴ Problem has no maximum.
But Z(A) = \(\frac{100}{3}\), Z(B) = 50
Z is minimum at A(\(\frac{5}{6}\), \(\frac{7}{6}\)).
But the feasible region is unbounded.
Hence we cannot immediately decide, Z is minimum at A.
Let us draw the half plane
5x₁ + 25x₂ < \(\frac{100}{3}\)
⇒ 3x₁ + 15x₂ < 20
As there is no point common to this half plane and the feasible region.
we have Z is minimum for x₁ = \(\frac{5}{6}\), x₂ = \(\frac{7}{6}\) and the minimum value = \(\frac{100}{3}\)

Question 13.
Optimize: Z = 5x₁ + 2x₂
Subject to: -0.5x₁ + x₂ ≤ 2
x₁ + x₂ ≥ 2
-x₁+ 5x₂ ≥ 5
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
-0.5x₁ + x₂ = 2   ….(1)
x₁ + x₂ = 2         ….(2)
-x₁+ 5x₂ = 5      ….(3)
Step – 2 Let us draw the graph.

Step – 3 The shaded regian is feasible region which is unbounded, thus Z does not have any maximum.

As Z can be made arbitrarily large, the given LPP has no maximum.
Z is minimum at B (0, 2). But we cannot immediately decide, Z is minimum at B.
Let us draw the half plane 5x₁ + 2x₂ < 4

x1	0	4/5
x2	2	0

As there is no point common to this half plane and the feasible region,
we have Z is minimum for x₁ = 0, x₂ = 2 and the minimum value of Z = 4.

Question 14.
Optimize: Z = -10x₁ + 2x₂
Subject to: -x₁ + x₂ ≥ -1
x₁ + x₂ ≤ 6
x₂ ≤ 5
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
-x₁ + x₂ = -1     ….(1)
x₁ + x₂ = 6        ….(2)
x₂ = 5                ….(3)
Step – 2 Let us draw the graph
Table – 1

x1	1	0
x2	0	-1

Table – 2

x1	6	0
x2	0	1

Step – 3 Clearly 0(0,0) satisfies all the constraints.
Thus the shaded region is the feasible region.
The vertices are 0(0,0) , A(1,0), B(\(\frac{7}{2}\), \(\frac{5}{2}\)) ,C(1, 5) and D (0, 5)
Step – 4 Z(O) = 0
Z(A) = -10
Z(B) = – 30
Z(C) = 0
Z(D) = 10
∴ Z is maximum for x₁= 0, x, = 5 and Z_(max) = 10
Z is minimum for x₁ = \(\frac{7}{2}\)  x₂ = \(\frac{5}{2}\) and Z_(min) = -30

Question 15.
Solve the L.P.P.s obtained in Exercise 3(a) Q.1 to Q. 9 by graphical method.
(1) Maximise: Z = 1500x + 2000y
Subject to: x + y < 20
x + 2y < 24
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x + y = 20
x + 2y = 24
Step – 2 Let us draw of graph.

Step – 3 Clearly 0(0,0) satisfies all the constraints.
Thus the shaded region is the feasible region.
From (1) and (2) we get
y = 14
x = 16
With vertices 0(0, 0), A(20, 0), B(16, 4), C(0, 12).
Step – 4 Z(0) = 0
Z(A) = 30,000
Z(B) = 32,000 → Maximum
Z(C) = 24000
Z is maximum for x = 16, y = 4 with Z = 32000
To get maximum profit he must keep 16 sets of model X and 4 sets of model Y.
Maximum profit = 1500 × 16 + 2000 × 4 = ₹32,000

(2) Maximize: 15x + 10y
Subject: x + 3y ≤ 600
2x + y ≤ 480
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
2x +3y = 600
2a + y = 480
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints.
The corner point are 0(0, 0), A (240, 0) B(210, 60),C(0, 200)
Step – 4 Z(0) = 6
Z(A) = 3600
Z(B) = 3150 + 600
= 3750 → maximum
Z(C) = 2000
Thus Z is maximum for x = 210 and y = 60
and Z_(max) = 3750

(3) Maximize: Z = 20x + 30y
Subject to: x + 2y ≤ 10
x + y ≤ 6
x ≤ 4
x, y ≥ 0.
Solution:
Step – 1 Treating the constraints as equations we get
x + 2y = 10       …(1)
x + y = 6           …(2)
x = 4
Step – 2 Let us draw the graph

Step – 3 As 0(0,0) satisfies all the constraints the shaded region is the feasible region.
Solving (1) and (2) we get x = 2, y = 4.
The vertices and 0(0, 0) , A(4, 0), B(4, 2), C(2, 4), D (0, 5).
Step – 4 Z(0) =0
Z(A) = 80
Z (B) =140
Z(C) = 1 60 → maximum
Z (D) = 150
∴ Z is Maximum when x = 2, y = 4 and Z_(max) = 160

(4) Maximize: Z = 15x + 17y
Subject to: 4x + 7y ≤ 150
x + y ≤ 30
15x + 17y > 300
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
4x + 7y = 150      ….(1)
x + y = 30            ….(2)
15x + 17y = 300  ….(3)
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints.
4x + 7y ≤ 150, x + y ≤ 30, but does not satisfy 15x + 17y ≥ 300.
∴ The shaded region is the feasible region.
From (1) and (2) we get

∴ Z is maximum for x = 20. y = 10 and Z_(max) = 470

(5) Maximize: Z = 2x + 4y
Subject to: 3x + 2y ≤ 10
2x + 5y ≤ 15
5x + 6y ≤ 21
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
3x + 2y = 10  …(1)
2x + 5y = 15  …(2)
5x + 6y = 21  …(3)
Step – 2 Let us draw the graph

Step – 3 As 0(0,0) satisfies all the constraints the shaded region is the feasible region.
From (1) and (3) we get

From (2) and (3) we get

Step-4 Z(O) = 0

(6) Maximize: Z = 1000x + 800y
Subject to: x + y ≤ 5
2x + y ≤ 9
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x + y = 5    ….(1)
2x + y = 9  ….(2)
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints.
∴ Thus the shaded region is the feasible region.
From (1) and (2) we get x = 4, y = 1.
∴ The vertices are A(0, 0), A(4.5, 0), B(4, 1) and C(0, 5).
Step – 4 Z(0) =0
Z (A) = 4500
Z (B) = 4800 → Maximum
Z (C) = 4000
Z is maximum for x = 4 and y = 1, Z_(max) = 4800

(7) Minimize: Z = 4960 – 70x – 130y
Subject to: x + y ≤ 12
x + y ≥ 6
x ≤ 8
y ≤ 8
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x + y = 12   ….(1)
x + y = 6     ….(2)
x = 8           ….(3)
y = 4           ….(4)
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints except x + y > 6.
The shaded region is the feasible region.
The vertices are A(6, 0), B(8, 0), C(8, 4), D(4, 8), E(0, 8) and F(0, 6).
Step – 4 Z (A) = 4540
Z (B) = 4400
Z (C) = 3880
Z (D) = 3640 → Minimum
Z (E) = 3920
Z (F) = 4180
∴ Z is maximum for x = 4 and y = 8 and Z_(min) = 3640.

(8) Minimize: Z = 16x + 20y
Subject to x + 2y ≥ 10
x + y ≥ 6
3x + y ≥ 8
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x + 2y = 10  ….(1)
x + y = 6      …(2)
3x + y = 8    …(3)
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints. Thus the shaded region is the feasible region.
From (1) and (2) we get y = 4, x = 2.
From (2) and (3) we get x = 1, y = 5.
The vertices are A(10, 0), B(2, 4), C(1, 5), D(0, 8).
Step – 4 Z (A) = 160
Z (B) = 112 → Minimum
Z (C) =116
Z (D) = 160
As the region is unbounded, let us draw the half plane Z < Z_(min)
⇒ 16x + 20y < 112
⇒ 4x + 5y < 28

x1	7	0
x2	0	5.6

There is no point common to the shaded region and the half plane 4x + 5y ≤ 28 other than B(2, 4).
∴ Z is minimum for x = 2, y = 4 and Z_(min) = 112.

(9) Minimize: Z = (512.5)x + 800y
Subject to: 5x + 4y = 40
x ≤ 7
x ≤ 3
x, y ≥ 0
Solution:
Step – 1 Let us draw the graph of
5x + 4y = 40
x = 7, y = 3

x1	8	0
x2	0	10

Step – 1 Let us draw the graph of
5x + 4y = 40
x = 7, y = 3
Step – 2 The line segment AB is the feasible region.
Step – 3 Z (A) = 3587.5 + 1000 = 4587.5
Z (B) = 2870 + 2400 = 5270
Clearly Z is minimum for
x = 7, y = \(\frac{5}{4}\) and Z_(min) = 4587.5

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 2 Inverse Trigonometric Functions Ex 2 Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Exercise 2

Question 1.
Fill in the blanks choosing correct answer from the brackets:
(i) If A = tan^-1 x, then the value of sin 2A = ________. (\(\frac{2 x}{1-x^2}\), \(\frac{2 x}{\sqrt{1-x^2}}\), \(\frac{2 x}{1+x^2}\))
Solution:
\(\frac{2 x}{1+x^2}\)

(ii) If the value of sin^-1 x = \(\frac{\pi}{5}\) for some x ∈ (-1, 1) then the value of cos^-1 x is ________. (\(\frac{3 \pi}{10}\), \(\frac{5 \pi}{10}\),\(\frac{3 \pi}{10}\))
Solution:
\(\frac{3 \pi}{10}\)

(iii) The value of tan^-1 x (2cos\(\frac{\pi}{3}\)) is ________. (1, \(\frac{\pi}{4}\), \(\frac{\pi}{3}\))
Solution:
\(\frac{\pi}{4}\)

(iv) If x + y = 4, xy = 1, then tan^-1 x + tan^-1 y = ________. (\(\frac{3 \pi}{4}\), \(\frac{\pi}{4}\), \(\frac{\pi}{3}\))
Solution:
\(\frac{\pi}{2}\)

(v) The value of cot^-1 2 + tan^-1 \(\frac{1}{3}\) = ________. (\(\frac{\pi}{4}\), 1, \(\frac{\pi}{2}\))
Solution:
\(\frac{\pi}{4}\)

(vi) The principal value of sin^-1 (sin \(\frac{2 \pi}{3}\)) is ________. (\(\frac{2 \pi}{3}\), \(\frac{\pi}{3}\), \(\frac{4 \pi}{3}\))
Solution:
\(\frac{\pi}{3}\)

(vii) If sin^-1 \(\frac{x}{5}\) + cosec^-1 \(\frac{5}{4}\) = \(\frac{\pi}{2}\), then the value of x = ________. (2, 3, 4)
Solution:
x = 3

(viii) The value of sin (tan^-1 x + tan^-1 \(\frac{1}{x}\)), x > 0 = ________. (0, 1, 1/2)
Solution:
1

(ix) cot^-1 \(\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right]\) = ________. (2π – \(\frac{x}{2}\), \(\frac{x}{2}\), π – \(\frac{x}{2}\))
Solution:
π – \(\frac{x}{2}\)

(x) 2sin^-1 \(\frac{4}{5}\) + sin^-1 \(\frac{24}{25}\) = ________. (π, -π, 0)
Solution:
π

(xi) if Θ = cos^-1 x + sin^-1 x – tan^-1 x, x ≥ 0, then the smallest interval in which Θ lies is ________. [(\(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\)), [0, \(\frac{\pi}{2}\)), (0, \(\frac{\pi}{2}\)])
Solution:
(0, \(\frac{\pi}{2}\)]

(xii) sec² (tan^-1 2) + cosec² (cot^-1 3) = ________. (16, 14, 15)
Solution:
15

Question 2.
Write whether the following statements are true or false.
(i) sin^-1 \(\frac{1}{x}\) cosec^-1 x = 1
Solution:
False

(ii) cos^-1 \(\frac{4}{5}\) + tan^-1 \(\frac{2}{3}\) = tan^-1 \(\frac{17}{6}\)
Solution:
True

(iii) tan^-1 \(\frac{4}{3}\) + cot^-1 (\(\frac{-3}{4}\)) = π
Solution:
True

(iv) sec^-1 \(\frac{1}{2}\) + cosec^-1 \(\frac{1}{2}\) = \(\frac{\pi}{2}\)
Solution:
False

(v) sec^-1 (-\(\frac{7}{5}\)) = π – cos^-1 \(\frac{5}{7}\)
Solution:
True

(vi) tan^-1 (tan 3) = 3
Solution:
False

(vii) The principal value of tan^-1 (tan \(\frac{3 \pi}{4}\)) is \(\frac{3 \pi}{4}\)
Solution:
False

(viii) cot^-1 (-√3) is in the second quadrant.
Solution:
True

(ix) 3 tan^-1 3 = tan^-1 \(\frac{9}{13}\)
Solution:
False

(x) tan^-1 2 + tan^-1 3 = – \(\frac{\pi}{4}\)
Solution:
False

(xi) 2 sin^-1 \(\frac{4}{5}\) = sin^-1 \(\frac{24}{25}\)
Solution:
False

(xii) The equation tan^-1 (cotx) = 2x has exactly two real solutions.
Solution:
True

Question 3.
Express the value of the foilowing in simplest form.
(i) sin (2 sin^-1 0.6)
Solution:
sin (2 sin^-1 0.6)

(ii) tan (\(\frac{\pi}{4}\) + 2 cot^-1 3)
Solution:

(iii) cos (2 sin^-1 x)
Solution:

(iv) tan (cos^-1 x)
Solution:

(v) tan^-1 (\(\frac{x}{y}\)) – tan^-1 \(\frac{x-y}{x+y}\)
Solution:

(vi) cosec (cos^-1 \(\frac{3}{5}\) + cos^-1 \(\frac{4}{5}\))
Solution:

(vii) sin^-1 \(\frac{1}{\sqrt{5}}\) + cos^-1 \(\frac{3}{\sqrt{10}}\)
Solution:

(viii) sin cos^-1 tan sec √2
Solution:
sin cos^-1 tan sec √2
= sin cos^-1 tan sec \(\frac{\pi}{4}\)
= sin cos^-1 1 = sin 0 = 0

(ix) sin (2 tan^-1 \(\sqrt{\frac{1-x}{1+x}}\))
Solution:

(x) tan \(\left\{\frac{1}{2} \sin ^{-1} \frac{2 x}{1+x^2}+\frac{1}{2} \cos ^{-1} \frac{1-y^2}{1+y^2}\right\}\)
Solution:

(xi) sin cot^-1 cos tan^-1 x.
Solution:

(xii) tan^-1 \(\left(x+\sqrt{1+x^2}\right)\)
Solution:

Question 4.
Prove the following statements:
(i) sin^-1 \(\frac{3}{5}\) + sin^-1 \(\frac{8}{17}\) = cos^-1 \(\frac{36}{85}\)
Solution:

(ii) sin^-1 \(\frac{3}{5}\) + cos^-1 \(\frac{12}{13}\) = cos^-1 \(\frac{33}{65}\)
Solution:

(iii) tan^-1 \(\frac{1}{7}\) + tan^-1 \(\frac{1}{13}\) = tan^-1 \(\frac{2}{9}\)
Solution:
L.H.S = tan^-1 \(\frac{1}{7}\) + tan^-1 \(\frac{1}{13}\)

(iv) tan^-1 \(\frac{1}{2}\) + tan^-1 \(\frac{1}{5}\) + tan^-1 \(\frac{1}{8}\) = \(\frac{\pi}{4}\)
Solution:

(v) tan ( 2tan^-1 \(\frac{1}{5}\) – \(\frac{\pi}{4}\) ) + \(\frac{7}{17}\) = 0
Solution:

Question 5.
Prove the following statements:
(i) cot^-1 9 + cosec^-1 \(\frac{\sqrt{41}}{4}\) = \(\frac{\pi}{4}\)
Solution:

(ii) sin^-1 \(\frac{4}{5}\) + 2 tan^-1 \(\frac{1}{3}\) = \(\frac{\pi}{2}\)
Solution:

(iii) 4 tan^-1 \(\frac{1}{5}\) – tan^-1 \(\frac{1}{70}\) + tan^-1 \(\frac{1}{99}\) = \(\frac{\pi}{4}\)
Solution:

(iv) 2 tan^-1 \(\frac{1}{5}\) + sec^-1 \(\frac{5 \sqrt{2}}{7}\) + 2 tan^-1 \(\frac{1}{8}\) = \(\frac{\pi}{4}\)
Solution:

(v) cos^-1 \(\frac{12}{13}\) + 2 cos^-1 \(\sqrt{\frac{64}{65}}\) + cos^-1 \(\sqrt{\frac{49}{50}}\) = cos^-1 \(\frac{1}{\sqrt{2}}\)
Solution:


(vi) tan² cos^-1 \(\frac{1}{\sqrt{3}}\) + cot² sin^-1 \(\frac{1}{\sqrt{5}}\) = 6
Solution:
tan² cos^-1 \(\frac{1}{\sqrt{3}}\) + cot² sin^-1 \(\frac{1}{\sqrt{5}}\)
= tan² tan^-1 √2 + cot² cot^-1 (2)
= 2 + 4 = 6

(vii) cos tan^-1 cot sin^-1 x = x.
Solution.

Question 6.
Prove the following statements:
(i) cot^-1 (tan 2x) + cot^-1 (- tan 2x) = π
Solution:

(ii) tan^-1 x + cot^-1 (x + 1) = tan^-1 (x² + x + 1)
Solution:

(iii) tan^-1 (\(\frac{a-b}{1+a b}\)) + tan^-1 (\(\frac{b-c}{1+b c}\)) = tan^-1 a – tan^-1 c.
Solution:
tan^-1 (\(\frac{a-b}{1+a b}\)) + tan^-1 (\(\frac{b-c}{1+b c}\))
= tan^-1 a – tan^-1 b + tan^-1 b – tan^-1 c
= tan^-1 a – tan^-1 c.

(iv) cot^-1 \(\frac{p q+1}{p-q}\) + cot^-1 \(\frac{q r+1}{q-r}\) + cot^-1 \(\frac{r p+1}{r-p}\) = 0
Solution:

(v)

Solution:

Question 7.
Prove the following statements:
(i) tan^-1 \(\frac{2 a-b}{b \sqrt{3}}\) + tan^-1 \(\frac{2 b-a}{a \sqrt{3}}\) = \(\frac{\pi}{3}\)
Solution:

(ii) tan^-1 \(\frac{1}{x+y}\) + tan^-1 \(\frac{y}{x^2+x y+1}\) = tan^-1 \(\frac{1}{x}\)
Solution:

(iii) sin^-1 \(\sqrt{\frac{x-q}{p-q}}\) = cos^-1 \(\sqrt{\frac{p-x}{p-q}}\) = cot^-1 \(\sqrt{\frac{p-x}{x-q}}\)
Solution:

(iv) sin² (sin^-1 x + sin^-1 y + sin^-1 z) = cos² (cos^-1 x + cos^-1 y + cos^-1 z)
Solution:

(v) tan (tan^-1 x + tan^-1 y + tan^-1 z) = cot (cot^-1 x + cot^-1 y + cot^-1 z)
Solution:

Question 8.
(i) If sin^-1 x + sin^-1 y + sin^-1 z = π, show that x\(\sqrt{1-x^2}\) + x\(\sqrt{1-y^2}\) + x\(\sqrt{1-z^2}\) = 2xyz
Solution:
Let sin^-1 x = α, sin^-1 y = β, sin^-1 z = γ
∴ α + β + γ = π
∴ x = sin α, y = sin β, z = sin γ
or, α + β = π – γ
or, sin(α + β) = sin(π – γ) = sin γ
and cos(α + β) = cos(π – γ) = – cos γ

(ii) tan^-1 x + tan^-1 y + tan^-1 z = π show that x + y + z = xyz.
Solution:

(iii) tan^-1 x + tan^-1 y + tan^-1 z = \(\frac{\pi}{2}\). Show that xy + yz + zx = 1
Solution:

or, 1 – xy – yz – zx = 0
⇒ xy + yz + zx = 1

(iv) If r² = x² +y² + z², Prove that tan^-1 \(\frac{y z}{x r}\) + tan^-1 \(\frac{z x}{y r}\) + tan^-1 \(\frac{x y}{z r}\) = \(\frac{\pi}{2}\)
Solution:

(v) In a triangle ABC if m∠A = 90°, prove that tan^-1 \(\frac{b}{a+c}\) + tan^-1 \(\frac{c}{a+b}\) = \(\frac{\pi}{4}\). where a, b, and c are sides of the triangle.
Solution:
L.H.S. tan^-1 \(\frac{b}{a+c}\) + tan^-1 \(\frac{c}{a+b}\)

Question 9.
Solve
(i) cos (2 sin^-1 x) = 1/9
Solution:

(ii) sin^-1 x + sin^-1 (1 – x) = \(\frac{\pi}{2}\)
Solution:
sin^-1 x + sin^-1 (1 – x) = \(\frac{\pi}{2}\)
or, sin^-1 (1 – x) = \(\frac{\pi}{2}\) – sin^-1 x = cos^-1 x
or, sin^-1 (1 – x) = sin^-1 \(\sqrt{1-x^2}\)
or, 1 – x = \(\sqrt{1-x^2}\)
or, 1 + x² – 2x = 1 – x²
or, 2x² – 2x  = 0
or, 2x (x – 1) = 0
∴ x = 0 or, 1

(iii) sin^-1 (1 – x) – 2 sin^-1 x = \(\frac{\pi}{2}\)
Solution:
sin^-1 (1 – x) – 2 sin^-1 x = \(\frac{\pi}{2}\)
⇒ – 2 sin^-1 x = \(\frac{\pi}{2}\) – sin^-1 (1 – x)
⇒ cos^-1 (1 – x)
⇒ cos (– 2 sin^-1 x) = 1 – x      ….. (1)
Let sin^-1 Θ ⇒ sin Θ
Now cos (– 2 sin^-1 x) = cos (-2Θ)
= cos 2Θ = 1 – 2 sin² Θ = 1 – 2x²
Using in (1) we get
1 – 2x² = 1 – x
⇒ 2x² – x = 0 ⇒ x (2x – 1) = 0
⇒ x = 0, ½, But x = ½ does not
Satisfy the given equation, Thus x = 0.

(iv) cos^-1 x + sin^-1 \(\frac{x}{2}\) = \(\frac{\pi}{6}\)
Solution:

(v) tan^-1 \(\frac{x-1}{x-2}\) + tan^-1 \(\frac{x+1}{x+2}\) = \(\frac{\pi}{4}\)
Solution:

(vi) tan^-1 \(\frac{1}{2 x+1}\) + tan^-1 \(\frac{1}{4 x+1}\) = tan^-1 \(\frac{2}{x^2}\)
Solution:

(vii) 3 sin^-1 \(\frac{2 x}{1+x^2}\) – 4 cos^-1 \(\frac{1-x^2}{1+x^2}\) + 2 tan^-1 \(\frac{2 x}{1-x^2}\) = \(\frac{\pi}{3}\)
Solution:

(viii) cot^-1 \(\frac{1}{x-1}\) + cot^-1 \(\frac{1}{x}\) + cot^-1 \(\frac{1}{x+1}\) = cot^-1 \(\frac{1}{3x}\)
Solution:

(ix) cot^-1 \(\frac{1-x^2}{2 x}\) =  cosec^-1 \(\frac{1+a^2}{2 a}\) – sec^-1 \(\frac{1+b^2}{1-b^2}\)
Solution:

(x) sin^-1 \(\left(\frac{2 a}{1+a^2}\right)\) + sin^-1 \(\left(\frac{2 b}{1+b^2}\right)\) = 2 tan^-1 x
Solution:

(xi) sin^-1 y – cos^-1 x = cos^-1 \(\frac{\sqrt{3}}{2}\)
Solution:

(xii) sin^-1 2x + sin^-1 x = \(\frac{\pi}{3}\)
Solution:

Question 10.
Rectify the error ifany in the following:
sin^-1 \(\frac{4}{5}\) + sin^-1 \(\frac{12}{13}\) + sin^-1 \(\frac{33}{65}\)
Solution:

Question 11.
Prove that:
(i) cos^-1 \(\left(\frac{b+a \cos x}{a+b \cos x}\right)\) = 2 tan^-1 \(\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{x}{2}\right)\)
Solution:

(ii) tan \(\left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} \frac{a}{b}\right)\) + tan \(\left(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} \frac{a}{b}\right)\)
Solution:

(iii) tan^-1 \(\sqrt{\frac{x r}{y z}}\) + tan^-1 \(\sqrt{\frac{y r}{y x}}\) + tan^-1 \(\sqrt{\frac{z r}{x y}}\) = π where r = x + y +z.
Solution:

Question 12.
(i) If cos^-1 (\(\frac{x}{a}\)) + cos^-1 (\(\frac{y}{b}\)) = Θ, prove that \(\frac{x^2}{a^2}\) – \(\frac{2 x}{a b}\) cos Θ + \(\frac{y^2}{b^2}\) = sin² Θ.
Solution:

(ii) If cos^-1 (\(\frac{x}{y}\)) + cos^-1 (\(\frac{y}{3}\)) = Θ, prove that 9x² – 12xy cos Θ + 4y² = 36 sin² Θ.
Solution:

(iii) If sin^-1 (\(\frac{x}{a}\)) + sin^-1 (\(\frac{y}{b}\)) = sin^-1 (\(\frac{c^2}{a b}\)) prove that b²x² + 2xy \(\sqrt{a^2 b^2-c^4}\) a²y² = c²
Solution:

(iv) If sin^-1 (\(\frac{x}{a}\)) + sin^-1 (\(\frac{y}{b}\)) = α prove that \(\frac{x^2}{a^2}\) + \(\frac{2 x y}{a b}\) cos α + \(\frac{y^2}{b^2}\) = sin² α
Solution:

(v) If sin^-1 x + sin^-1 y + sin^-1 z = π prove that x² + y² + z² + 4x²y²z² = 2 ( x²y² + y²z² + z²x² )
Solution:

Question 13.
Solve the following equations:
(i) tan^-1 \(\frac{x-1}{x+1}\) + tan^-1 \(\frac{2 x-1}{2 x+1}\) = tan^-1 \(\frac{23}{36}\)
Solution:

(ii) tan^-1 \(\frac{1}{3}\) + tan^-1 \(\frac{1}{5}\) + tan^-1 \(\frac{1}{7}\) + tan^-1 x = \(\frac{\pi}{4}\)
Solution:

(iii) cos^-1 \(\left(x+\frac{1}{2}\right)\) + cos^-1 x+ cos^-1 \(\left(x-\frac{1}{2}\right)\) = \(\frac{3 \pi}{2}\)
Solution:

(iv) 3tan^-1 \(\frac{1}{2+\sqrt{3}}\) – tan^-1 \(\frac{1}{x}\) = tan^-1 \(\frac{1}{3}\)
Solution:

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 4 Matrices Ex 4(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Exercise 4(b)

Question 1.
State which of the following matrices are symmetric, skew-symmetric, both or not either:

Solution:
(i) Symmetric
(ii) Neither Symmetric nor skew-symmetric
(iii) Symmetric
(iv) Skew symmetric
(v) Both
(vi) Neither symmetric nor skew-symmetric
(vii) Skew symmetric

Question 2.
State ‘True’ or ‘False’:
(i) If A and B are symmetric matrices of the same order and AB – BA ≠ 0, then AB is not symmetric.
Solution:
True

(ii) For any square matrix A, AA’ is symmetric.
Solution:
True

(iii) If A is any skew-symmetric matrix, then A² is also skew-symmetric.
Solution:
False

(iv) If A is symmetric, then A², A³, …, Aⁿ are all symmetric.
Solution:
True

(v) If A is symmetric then A – A¹ is both symmetric and skew-symmetric.
Solution:
False

(vi) For any square matrix (A – A¹)² is skew-symmetric.
Solution:
True

(vii) A matrix which is not symmetric is skew-symmetric.
Solution:
False

Question 3.
(i) If A and B are symmetric matrices of the same order with AB ≠ BA, final whether AB – BA is symmetric or skew symmetric.
Solution:
A and B are symmetric matrices;
Thus A’ = A and B’ = B
Now (AB – BA)’ = (AB)’ – (BA)’
= B’A’ – A’B’
= BA – AB = – (AB – BA)
∴ AB – BA is skew symmetric.

(ii) If a symmetric/skew-symmetric matrix is expressed as a sum of a symmetric and a skew-symmetric matrix then prove that one of the matrices in the sum must be zero matrix.
Solution:
We know that zero matrix is both symmetric as well as skew-symmetric.
Let A is symmetric.
∴ A = A + O where A is symmetric and O is treated as skew-symmetric. If B is skew-symmetric then we can write B = O + B where O is symmetric and B is skew-symmetric.

Question 4.
A and B are square matrices of the same order, prove that
(i) If A, B and AB are all symmetric, then AB – BA = 0
Solution:
Let A, B and AB are all symmetric.
∴A’ = A, B’ = B and (AB)’ = AB
⇒ B’A’ = AB
⇒ BA = AB
⇒ AB – BA = 0

(ii) If A, B and AB are all skew symmetric then AB + BA = 0
Solution:
Let A, B and AB are all skew symmetric matrices
∴ A’ = -A, B’ = -B and (AB)’ = -AB
Now (AB)’ = -AB
⇒ B’A’ = -AB
⇒ (-B) (-A) = -AB
⇒ BA = -AB
⇒ AB + BA = 0

Question 5.
If A = \(\left[\begin{array}{rrr}
1 & 2 & 0 \\
0 & 1 & 3 \\
-2 & 5 & 3
\end{array}\right]\), then verify that A’ = \(\left[\begin{array}{ccc}
1 & 0 & -2 \\
2 & 1 & 5 \\
0 & 3 & 3
\end{array}\right]\)

(i) A+A’ is symmetric

(ii) A-A’ is skew-symmetric

Question 6.
Prove that a unit matrix is its own inverse. Is the converse true?
IfA = \(\left[\begin{array}{rrr}
0 & 1 & -1 \\
4 & -3 & 4 \\
3 & -3 & 4
\end{array}\right]\) show that A² = I and hence A= A^-1.
Solution:
No the converse is not true for example:

Question 7.
Here A is an involuntary matrix, recall the definition given earlier.
Solution:

Question 8.
Show that \(\left[\begin{array}{ll}
\mathbf{0} & \mathbf{1} \\
\mathbf{1} & \mathbf{0}
\end{array}\right]\) is its own inverse.
Solution:

Question 9.
Express as a sum of a symmetric and a skew symmetric matrix.

Solutions:

Question 10.
What is the inverse of

Question 11.
Find inverse of the following matrices by elementary row/column operation (transformations):
(i) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 5
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{cc}
4 & -2 \\
3 & 1
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{cc}
1 & 0 \\
2 & -3
\end{array}\right]\)
Solution:

(vi) \(\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\)
Solution:

Question 12.
Find the inverse of the following matrices using elementary transformation:
(i) \(\left[\begin{array}{lll}
\mathbf{0} & \mathbf{0} & 2 \\
\mathbf{0} & \mathbf{2} & \mathbf{0} \\
\mathbf{2} & \mathbf{0} & \mathbf{0}
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
3 & 1 & 1
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{ccc}
3 & -2 & 3 \\
2 & 1 & -1 \\
4 & -3 & 2
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{lll}
1 & 1 & 2 \\
0 & 1 & 2 \\
1 & 2 & 1
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 1 & 4 \\
1 & 0 & 2
\end{array}\right]\)
Solution:

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 5 Determinants Ex 5(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 5 Determinants Exercise 5(b)

Question 1.
Write the number of solutions of the following system of equations.
(i) x – 2y = 0
Solution:
No solution

(ii) x – y = 0 and 2x – 2y = 1
Solution:
Infinite

(iii) 2x + y = 2 and -x – 1/2y = 3
Solution:
No solution

(iv) 3x + 2y = 1 and x + 5y = 6
Solution:
One

(v) 2x + 3y + 1 = 0 and x – 3y – 4 = 0
Solution:
One

(vi) x + y + z = 1
x + y + z = 2
2x + 3y + z = 0
Solution:
No solution

(vii) x + 4y – z = 0
3x – 4y – z = 0
x – 3y + z = 0
Solution:
One

(viii) x + y – z = 0
3x – y + z = 0
x – 3y + z = 0
Solution:
One

(ix) a₁x + b₁y + c₁z = 0
a₂x + b₂y + c₂z = 0
a₃x + b₃y + c₃z = 0
and \(\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|\) = 0
Solution:
Infinite solutions as Δ = Δ₁ = Δ₂ = Δ₃ = 0

Question 2.
Show that the following system is inconsistent.
(a – b)x + (b – c)y + (c – a)z = 0
(b – c)x + (c – a)y + (a – b)z = 0
(c – a)x + (a – b)y + (b – c)z =1
Solution:

Question 3.
(i) The system of equations
x + 2y + 3z = 4
2x + 3y + 4z = 5
3x + 4y + 5z = 6 has
(a) infinitely many solutions
(b) no solution
(c) a unique solution
(d) none of the three
Solution:
(a) infinitely many solutions

(ii) If the system of equations
2x + 5y + 8z = 0
x + 4y + 7z = 0
6x + 9y – z = 0
has a nontrivial solution, then is equal to
(a) 12
(b) -12
(c) 0
(d) none of the three
Solution:
(b) -12

(iii) The system of linear equations
x + y + z = 2
2x + y – z = 3
3x +2y + kz = 4
has a unique solution if
(a) k ≠ 0
(b) -1 < k < 1
(c) -2 < k < 2
(d) k = 0
Solution:
(a) k ≠ 0

(iv) The equations
x + y + z = 6
x + 2y + 3z = 10
x + 2y + mz = n
give infinite number of values of the triplet (x, y, z) if
(a) m = 3, n ∈ R
(b) m = 3, n ≠ 10
(c) m = 3, n = 10
(d) none of the three
Solution:
(c) m = 3, n = 10

(v) The system of equations
2x – y + z = 0
x – 2y + z = 0
x – y + 2z = 0
has infinite number of nontrivial solutions for
(a) = 1
(b) = 5
(c) = -5
(d) no real value of
Solution:
(c) = -5

(vi) The system of equations
a₁x + b₁y + c₁z = 0
a₂x + b₂y + c₂z = 0
a₃x + b₃y + c₃z =0
with has
(a) more than two solutions
(b) one trivial and one nontrivial solutions
(c) No solution
(d) only trivial solutions
Solution:
(a) more than two solutions

Question 4.
Can the inverses of the following matrices be found?
(i) \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
Solution:
|A| = 0
∴ A^-1 can not be found.

(ii) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Solution:
∴ |A| = 4 – 6 = -2 ≠ 0
∴ A^-1 exists.

(iii) \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)
Solution:
|A| = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\) = 1 – 1 = 0
∴ A^-1 does not exist.

(iv) \(\left[\begin{array}{ll}
1 & 2 \\
2 & 4
\end{array}\right]\)
Solution:
|A| = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 4
\end{array}\right]\) = 4 – 4 = 0
∴ A^-1 does not exist.

(v) \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
Solution:
|A| = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) = 1 ≠ 0
∴ A^-1 exists.

Question 5.
Find the inverse of the following:
(i) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{cc}
2 & -1 \\
1 & 3
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{cc}
4 & -2 \\
3 & 1
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{cc}
1 & 0 \\
2 & -3
\end{array}\right]\)
Solution:

(vi) \(\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\)
Solution:

(vii) \(\left[\begin{array}{cc}
i & -i \\
i & i
\end{array}\right]\)
Solution:

(viii) \(\left[\begin{array}{ll}
x & -x \\
x & x^2
\end{array}\right]\), x ≠ 0, x ≠ -1
Solution:

Question 6.
Find the adjoint of the following matrices.
(i) \(\left[\begin{array}{ccc}
1 & 1 & -1 \\
2 & -1 & 2 \\
1 & 3 & -2
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ccc}
-2 & 2 & 3 \\
1 & 4 & 2 \\
-2 & -3 & 1
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{lll}
2 & 1 & 2 \\
2 & 2 & 1 \\
1 & 2 & 2
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ccc}
1 & 3 & 0 \\
2 & -1 & 6 \\
5 & -3 & 1
\end{array}\right]\)
Solution:

Question 7.
Which of the following matrices are invertible?
(i) \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
1 & 1 & 1 \\
2 & -1 & 1
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ccc}
2 & 1 & -2 \\
1 & 2 & 1 \\
3 & 6 & 4
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{ccc}
-1 & -2 & 3 \\
2 & 1 & -4 \\
-1 & 0 & 2
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ccc}
1 & 0 & 1 \\
2 & -2 & 1 \\
3 & 2 & 4
\end{array}\right]\)
Solution:

Question 8.
Examining consistency and solvability, solve the following equations by matrix method.
(i) x – y + z = 4
2x + y – 3z = 0
x + y + z = 2
Solution:

(ii) x + 2y – 3z = 4
2x + 4y – 5z = 12
3x – y + z = 3
Solution:
Let

(iii) 2x – y + z = 4
x + 3y + 2z = 12
3x + 2y + 3z = 16
Solution:

(iv) x + y + z = 4
2x + 5y – 2x = 3
x + 7y – 7z = 5
Solution:

(v) x + y + z = 4
2x – y + 3z = 1
3x + 2y – z = 1
Solution:

(vi) x + y – z = 6
2x – 3y + z = 1
2x – 4y + 2z = 1
Solution:

(vii) x – 2y = 3
3x + 4y – z = -2
5x – 3z = -1
Solution:

(viii) x + 2y + 3z = 14
2x – y + 5z = 15
2y + 4z – 3x = 13
Solution:

(ix) 2x + 3y +z = 11
x + y + z = 6
5x – y + 10z = 34
Solution:

Question 9.
Given the matrices
A = \(\left[\begin{array}{ccc}
1 & 2 & 3 \\
3 & -2 & 1 \\
4 & 2 & 1
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and C = \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)
write down the linear equations given by AX = C and solve it for x, y, z by matrix method.
Solution:

Question 10.
Find X, if \(\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1 & -1 \\
2 & 1 & -1
\end{array}\right]\) X = \(\left[\begin{array}{l}
6 \\
0 \\
1
\end{array}\right]\) where X = \(\left[\begin{array}{l}
x_1 \\
x_2 \\
x_3
\end{array}\right]\)
Solution:

Question 11.
Answer the following:
(i) If every element of a third order matrix is multiplied by 5, then how many times its determinant value becomes?
Solution:
125

(ii) What is the value of x if \(\left|\begin{array}{ll}
4 & 1 \\
2 & 1
\end{array}\right|^2=,\left|\begin{array}{ll}
3 & 2 \\
1 & x
\end{array}\right|-\left|\begin{array}{cc}
x & 3 \\
-2 & 1
\end{array}\right|\) ?
Solution:
x = 6

(iii) What are the values of x and y if \(\left|\begin{array}{ll}
x & y \\
1 & 1
\end{array}\right|=2,\left|\begin{array}{ll}
x & 3 \\
y & 2
\end{array}\right|=1\) ?
Solution:
x = 5, y = 3

(iv) What is the value of x if \(\left|\begin{array}{ccc}
x+1 & 1 & 1 \\
1 & 1 & -1 \\
-1 & 1 & 1
\end{array}\right|\) = 4?
Solution:
x = 0

(v) What is the value of \(\left|\begin{array}{ccc}
\mathbf{o} & -\mathbf{h} & -\mathbf{g} \\
\mathbf{h} & \mathbf{0} & -\mathbf{f} \\
\mathbf{g} & \mathbf{f} & \mathbf{0}
\end{array}\right|\)?
Solution:
0

(vi) What is the value of \(\left|\begin{array}{l}
\frac{1}{a} 1 \mathrm{bc} \\
\frac{1}{b} 1 c a \\
\frac{1}{c} 1 a b
\end{array}\right|\)
Solution:
0

(vii) What is the co-factor of 4 in the determinant \(\left|\begin{array}{rrr}
1 & 2 & -3 \\
4 & 5 & 0 \\
2 & 0 & 1
\end{array}\right|\)
Solution:
-2

(viii)In which interval does the determinant \(\left|\begin{array}{ccc}
1 & \sin \theta & 1 \\
-\sin \theta & 1 & \sin \theta \\
-1 & -\sin \theta & 1
\end{array}\right|\) lie?
Solution:
[2, 4]

(ix) Ifx + y + z = n, what is the value of Δ = \(\left|\begin{array}{ccc}
\sin (x+y+z) & \sin B & \cos C \\
-\sin B & 0 & \tan A \\
\cos (A+B) & -\tan A & 0
\end{array}\right|\) Where A, B, C are the angles of triangle.
Solution:
0

Question 12.
Evaluate the following determinants:
(i) \(\left|\begin{array}{ccc}
14 & 3 & 28 \\
17 & 9 & 34 \\
25 & 9 & 50
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
14 & 3 & 28 \\
17 & 9 & 34 \\
25 & 9 & 50
\end{array}\right|\)
= 2\(\left|\begin{array}{ccc}
14 & 3 & 28 \\
17 & 9 & 34 \\
25 & 9 & 50
\end{array}\right|\) = 0
(∵ C₁ = C₃)

(ii) \(\left|\begin{array}{ccc}
16 & 19 & 13 \\
15 & 18 & 12 \\
14 & 17 & 11
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
16 & 19 & 13 \\
15 & 18 & 12 \\
14 & 17 & 11
\end{array}\right|\) = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1 & 1 \\
14 & 17 & 11
\end{array}\right|\)
( R₁ = R₁ – R₂, R₂ = R₂ – R₃)
= 0 (∵ R₁ = R₂)

(iii) \(\left|\begin{array}{ccc}
224 & 777 & 32 \\
735 & 888 & 105 \\
812 & 999 & 116
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
224 & 777 & 32 \\
735 & 888 & 105 \\
812 & 999 & 116
\end{array}\right|\)
= 7\(\left|\begin{array}{ccc}
32 & 777 & 32 \\
105 & 888 & 105 \\
116 & 999 & 116
\end{array}\right|\) = 0
(∵ C₁ = C₂)

(iv) \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 3 & 4 \\
3 & 4 & 6
\end{array}\right|\)
Solution:

(v) \(\left|\begin{array}{ccc}
1 & 2 & 3 \\
3 & 5 & 7 \\
8 & 14 & 20
\end{array}\right|\)
Solution:

(vi) \(\left|\begin{array}{ccc}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2
\end{array}\right|\)
Solution:

= 225 – 256 – 4(100 – 144) + 9(64 – 81)
= -31 – 4(-44) + 9(-17)
= -31 + 176 – 153 = -184 + 176
= -8

(vii) \(\left|\begin{array}{ccc}
1 & 0 & -5863 \\
-7361 & 2 & 7361 \\
1 & 0 & 4137
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
1 & 0 & -5863 \\
-7361 & 2 & 7361 \\
1 & 0 & 4137
\end{array}\right|\)
= 2\(\left|\begin{array}{cc}
1 & -5863 \\
1 & 4137
\end{array}\right|\)
(expanding along 2nd column)
= 2(4137 + 5863)
= 2 × 10000 = 20000

(viii) \(\left|\begin{array}{lll}
265 & 240 & 219 \\
240 & 225 & 198 \\
219 & 198 & 181
\end{array}\right|\)
Solution:

(ix) \(\left|\begin{array}{ccc}
0 & a^2 & b \\
b^2 & 0 & a^2 \\
a & b^2 & 0
\end{array}\right|\)
Solution:

= -a² (0 –  a²) + b (b⁴ –  0) = a⁵ + b⁵

(x) \(\left|\begin{array}{ccc}
a-b & b-c & c-a \\
\boldsymbol{x}-\boldsymbol{y} & \boldsymbol{y}-\boldsymbol{z} & z-\boldsymbol{x} \\
\boldsymbol{p}-\boldsymbol{q} & \boldsymbol{q}-\boldsymbol{r} & \boldsymbol{r}-\boldsymbol{p}
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
a-b & b-c & c-a \\
x-y & y-z & z-x \\
p-q & q-r & r-p
\end{array}\right|\)
= \(\left|\begin{array}{lll}
0 & b-c & c-a \\
0 & y-z & z-x \\
0 & q-r & r-p
\end{array}\right|\) (C₁ = C₁ + C₂ + C₃)
= 0 ( ∵ C₁ = 0)

(xi) \(\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\)
= \(\left|\begin{array}{lll}
0 & b-c & c-a \\
0 & c-a & a-b \\
0 & a-b & b-c
\end{array}\right|\) (C₁ = C₁ + C₂ + C₃)
= 0

(xii) \(\left|\begin{array}{ccc}
-\cos ^2 \theta & \sec ^2 \theta & -0.2 \\
\cot ^2 \theta & -\tan ^2 \theta & 1.2 \\
-1 & 1 & 1
\end{array}\right|\)
Solution:

(Expanding along 3rd row)
= (-cos² θ + sec² θ) (-tan² θ – 1.2) – (sec² θ + 0.2) (cot² θ – tan² θ)
= sin² θ – 1.2 cos² θ – sec² θ tan² θ – 1.2 sec² θ – cosec² θ +  sec² θ tan² θ – 0.2 cot² θ + 0.2 tan² θ
= sin² θ – cosec² θ + 1.2 (cos² θ – sec² θ) + 0.2 (tan² θ – cot² θ) ≠ 0
The question seems to be wrong.

Question 13.
If \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+x & 1 \\
1 & 1 & 1+y
\end{array}\right|\) = 0 what are x and y?
Solution:

or, xy – 0 = 0 ⇒ xy = 0, ⇒ x = 0, or y = 0

Question 14.
For what value of x \(\left|\begin{array}{ccc}
2 x & 0 & 0 \\
0 & 1 & 2 \\
-1 & 2 & 0
\end{array}\right|\) = \(\left|\begin{array}{lll}
1 & 0 & 0 \\
2 & 3 & 4 \\
0 & 3 & 5
\end{array}\right|\)?
Solution:

Question 15.
Solve \(\left|\begin{array}{ccc}
x+a & 0 & 0 \\
a & x+b & 0 \\
a & 0 & x+c
\end{array}\right|\) = 0
Solution:
\(\left|\begin{array}{ccc}
x+a & 0 & 0 \\
a & x+b & 0 \\
a & 0 & x+c
\end{array}\right|\) = 0
or, (x – a) \(\left|\begin{array}{cc}
x+b & 0 \\
0 & x+c
\end{array}\right|\) = 0
or, (x + a) (x + b) (x + c) = 0
x = -a, x = -b, x = -c

Question 16.
Solve \(\left|\begin{array}{lll}
a+x & a-x & a-x \\
a-x & a+x & a-x \\
a-x & a-x & a+x
\end{array}\right|\) = 0
Solution:

Question 17.
Solve \(\left|\begin{array}{ccc}
x+a & b & c \\
a & x+b & c \\
a & b & x+c
\end{array}\right|\) = 0
Solution:

Question 18.
Show that x = 2 is a root of \(\left|\begin{array}{ccc}x & -6 & -1 \\ 2 & -3 x & x-3 \\ -3 & 2 x & x+2\end{array}\right|\) = 0 Solve this completely.
Solution:
Putting x = 2,

= (x – 1) (-15x + 30 – 5x² + 10x)
= (x – 1) (-5x² – 5x + 30)
= -5(x – 1) (x² + x – 6)
= -5(x – 1) (x + 3) (x – 2) = 0
⇒ x = 1 or, -3 or 2.

Question 19.
Evaluate \(\left|\begin{array}{ccc}
1 & a & b c \\
1 & b & c a \\
1 & c & a b
\end{array}\right|\) – \(\left|\begin{array}{lll}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2
\end{array}\right|\)
Solution:

= (a – b) (b – c) [(-a + c) – (b + c – a – b)]
= (a – b) (b – c) (-a + c – c + a) = 0

Question 20.
\(\left|\begin{array}{lll}
a & a^2-b c & 1 \\
b & b^2-a c & 1 \\
c & c^2-a b & 1
\end{array}\right|\)
Solution:

Question21.
For what value of X the system of equations
x + y + z = 6, 4x + λy – λz = 0, 3x + 2y – 4z = -5 does not possess a solution?
Solution:

= 24 – 6λ – 2λ = 24 – 8λ
when Δ = 0
We have 24 – 8λ, = 0 or, λ = 3
The system of equations does not posses solution for λ = 3.

Question 22.
If A is a 3 × 3 matrix and |A| = 2, then which matrix is represented by A × adj A?
Solution:

Question 23.
If A = \(\left[\begin{array}{cc}
0 & -\tan \frac{\alpha}{2} \\
\tan \frac{\alpha}{2} & 0
\end{array}\right]\)
show that (I + A) (I – A)^-1 = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\) where I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:

Question 24.
Prove the following:
(i) \(\left|\begin{array}{ccc}
a^2+1 & a b & a c \\
a b & b^2+1 & b c \\
a c & b c & c^2+1
\end{array}\right|\) = 1 + a² + b² + c²
Solution:

(ii) \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
a^3 & b^3 & c^3
\end{array}\right|\) = (b – c) (c – a) (a – b) (a + b + c)
Solution:

= (a – b) (b – c) (b² + bc + c² – a² – ab – b²)
= (a – b) (b- c) (c² – a² + bc – ab)
= (a – b) (b – c) {(c – a) (c + a) + b(c – a)}
= (a – b) (b – c) (c – a) (a + b + c) = R.H.S.
(Proved)

(iii) \(\left|\begin{array}{lll}
\boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c} \\
\boldsymbol{b} & \boldsymbol{c} & \boldsymbol{a} \\
\boldsymbol{c} & \boldsymbol{a} & \boldsymbol{b}
\end{array}\right|\) = 3abc – a³ – b³ – c³
Solution:

= (a + b + c) {(b – c) (a – b) – (c – a)²}
= (a + b + c) (a + b + c) (ab – b² – ca + bc – c² – a² + 2ca)
= (a + b + c) (-a² – b² – c² + ab + bc + ca)
= -(a + b + c) (a² + b² + c² – ab – bc – ca)
=- (a³ + b³ + c³ – 3abc)
= 3abc – a³ – b³ – c³

(iv) \(\left|\begin{array}{lll}
b^2-a b & b-c & b c-a c \\
a b-a^2 & a-b & b^2-a b \\
b c-a c & c-a & a b-a^2
\end{array}\right|\) = 0
Solution:

= (b² – a² + bc – ac) (a – b) {(-a + b) (c – a) – (bc – ac – ab + a²)}
= (b² – a² + bc – ac) (a – b) (- ca + a² + bc – ab – bc + ac + ab – a²)
= (b² – a² + bc – ac) (a – b) × 0 = 0
= R.H.S.
(Proved)

(v) \(\left|\begin{array}{ccc}
-a^2 & a b & a c \\
a b & -b^2 & b c \\
a c & b c & -c^2
\end{array}\right|\) = 4a²b²c²
Solution:

(vi) \(\left|\begin{array}{lll}
(b+c)^2 & a^2 & b c \\
(c+a)^2 & b^2 & c a \\
(a+b)^2 & c^2 & a b
\end{array}\right|\) = (a² + b² + c² ) (a + b + c) (b – c) (c – a) (a – b)
Solution:

= (a – b) (b – c) (a² + b² + c²) (-a² – ab + bc + c²)
= (a – b) (b – c) (a² + b² + c²) {(c² – a²) + b(c – a)}
= (a² + b² + c²) (a – b) (b – c) (c – a) (c + a + b)

(vii) \(\left|\begin{array}{lll}
b+c & a+b & a \\
c+a & b+c & b \\
a+b & c+a & c
\end{array}\right|\) = a³ + b³ + c³ – 3abc
Solution:

= (a + b +c) {(a – b) (a – c) – (c – b) (b – c)}
= (a + b + c) (a² – ac – ab + bc – bc + c² + b² – bc)
= (a + b + c) (a² + b² + c² – ab – bc – ca)
= (a³ + b³ + c³ – 3abc)

(viii) \(\left|\begin{array}{ccc}
a+b+c & -c & -b \\
-c & a+b+c & -a \\
-b & -a & a+b+c
\end{array}\right|\) = 2(b + c) (c + a) (a + b)
Solution:

= -2(a + b) (b + c) (-a – b – c + b)
= 2(a + b) (b + c) (c + a)

(ix) \(\left|\begin{array}{ccc}
a x-b y-c z & a y+b x & a z+c x \\
b x+a y & b y-c z-a x & b z+c y \\
c x+a z & a y+b z & c z-a x-b y
\end{array}\right|\) = (a² + b² + c²) (ax + by + cz) (x² + y² + z²)
Solution:

Question 25.
If 2s = a + b + c show that \(\left|\begin{array}{ccc}
a^2 & (s-a)^2 & (s-a)^2 \\
(s-b)^2 & b^2 & (s-b)^2 \\
(s-c)^2 & (s-c)^2 & c^2
\end{array}\right|\) = 2s³ (s – a) (s – b) (s – c)
Solution:
Let s – a = A, s – b = B, s – c = C
A + B + C = 3s – (a + b + c)
= 3s – 2s = s
Also B + C = s – b + s – c = 2s – (b + c)
= (a + b + c) – b + c = a
Similarly C + A = b, A + B = c

= 2 ABC (A + B + C)²
[Refer Q.No.9 (xii) of Exercise 5(a)]
= 2(s – a) (s – b)(s – c) s³

Question 26.
if \(\left|\begin{array}{ccc}
x & x^2 & x^3-1 \\
y & y^2 & y^3-1 \\
z & z^2 & z^3-1
\end{array}\right|\) = 0 then prove that xyz =1 when x, y, z are non zero and unequal.
Solution:

= (x – y) (y – z) (z – x) (xyz – 1)
It is given that
(x – y) (y – z) (z – x) (xyz – 1) = 0
⇒ xyz – 1 (as x ≠ y ≠ z)

Question 27.
Without expanding show that the following determinant is equal to Ax + B where A and B are determinants of order 3 not involving x.
\(\left|\begin{array}{ccc}
x^2+x & x+1 & x-2 \\
2 x^2+3 x-1 & 3 x & 3 x-3 \\
x^2+2 x+3 & 2 x-1 & 2 x-1
\end{array}\right|\)
Solution:

Question 28.
If x, y, z are positive and are the pth, qth and rth terms of a G.P. then prove that \(\left|\begin{array}{lll}
\log x & p & 1 \\
\log y & q & 1 \\
\log z & r & 1
\end{array}\right|\) = 0
Solution:
Let the G.P. be
a, aR, aR², aR³ …..aR^n-1
p th term = aR^p-1
q th term = aR^q-1
r th term = aR^r-1
x = aR^p-1, y= aR^q-1, z = aR^r-1
log x = log a + (p – 1) log R,
log y = log a + (q – 1) log R,
log z = log a + (r – 1) log R

Question 29.
If D_j = \(\left|\begin{array}{ccc}
j & a & n(n+2) / 2 \\
j^2 & b & n(n+1)(2 n+1) / 6 \\
j^3 & c & n^2(n+1)^2 / 4
\end{array}\right|\) then prove that \(\sum_{j=1}^n\)D_j = 0.
Solution:

Question 30.
Ifa₁, a₂,……a_n are in G.P. and a_i > 0 for every i, then find the value of
\(\left|\begin{array}{ccc}
\log a_n & \log a_{n+1} & \log a_{n+2} \\
\log a_{n+1} & \log a_{n+2} & \log a_{n+3} \\
\log a_{n+2} & \log a_{n+3} & \log a_{n+4}
\end{array}\right|\)
Solution:

Question 31.
If f(x)= \(\left|\begin{array}{ccc}
1+\sin ^2 x & \cos ^2 x & 4 \sin ^2 x \\
\sin ^2 x & 1+\cos ^2 x & 4 \sin 2 x \\
\sin ^2 x & \cos ^2 x & 1+4 \sin ^2 x
\end{array}\right|\) what is the least value of f(x)?
Solution:

As minimum value of sin 2x is 0. So the minimum value of above function f(x) is 2.

Question 32.
If f_r(x), g_r(x), h_r(x), r = 1, 2, 3 are polynomials in x such that f_r(a) = g_r(a) = h_r(a) and
F(x) = \(\left[\begin{array}{lll}
f_1(x) & f_2(x) & f_3(x) \\
g_1(x) & g_2(x) & g_3(x) \\
h_1(x) & h_2(x) & h_3(x)
\end{array}\right]\) find F'(x) at x = a.
Solution:
We have

[Since f₁a) = g₁(a) = h₁(a), f₂(a) = g₂(a) = h₂(a) and f₃(a) = g₃(a) = h₃(a) So that each determinant is zero due to presence of two identical rows.]

Question 33.
If f(x) = \(\left[\begin{array}{ccc}
\cos x & \sin x & \cos x \\
\cos 2 x & \sin 2 x & 2 \cos 2 x \\
\cos 3 x & \sin 3 x & 3 \cos 3 x
\end{array}\right]\) find f'(\(\frac{\pi}{2}\)).
Solution:

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 4 Matrices Ex 4(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Exercise 4(a)

Question 1.
State the order of the following matrices.
(i) [abc]
(ii) \(\left[\begin{array}{l}
1 \\
2
\end{array}\right]\)
(iii) \(\left[\begin{array}{ll}
x & y \\
y & z \\
z & x
\end{array}\right]\)
(iv) \(\left[\begin{array}{cccc}
1 & 0 & 1 & 4 \\
2 & 1 & 3 & 0 \\
-3 & 2 & 1 & 3
\end{array}\right]\)
Solution:
(i) (1 x 3)
(ii) (2 x 1)
(iii) (3 x 2)
(iv) (3 x 4)

Question 2.
How many entries are there in a
(i) 3 x 3 matrix
(ii) 3 x 4 matrix
(iii) p x q matrix
(iv) a sqare matrix of order p?
Solution:
(i) 9
(ii) 12
(iii) pq
(iv) p²

Question 3.
Give an example of
(i) 3 x 1 matrix
(ii) 2 x 2 matrix
(iii) 4 x 2 matrix
(iv) 1 x 3 matrix
Solution:
(i) \(\left(\begin{array}{l}
a \\
b \\
c
\end{array}\right)\)
(ii) \(\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right)\)
(iii) \(\left(\begin{array}{ll}
a & b \\
c & d \\
e & f \\
g & h
\end{array}\right)\)
(iv) (1, 2, 3)

Question 4.
Let A = \(\left[\begin{array}{lllll}
1 & 2 & 3 & 4 & 1 \\
4 & 5 & 6 & 1 & 2 \\
3 & 9 & 1 & 1 & 6
\end{array}\right]\)
(i) What is the order of A?
(ii) Write down the entries a₃₁, a₂₅, a₂₃
(iii) Write down A^T.
(iv) What is the order of A^T?
Solution:
A = \(\left[\begin{array}{lllll}
1 & 2 & 3 & 4 & 1 \\
4 & 5 & 6 & 1 & 2 \\
3 & 9 & 1 & 1 & 6
\end{array}\right]\)
(i) Order of A is (3 x 5)
(ii) a₃₁ = 3, a₂₅= 2, a₂₃ = 6
(iii) A^T = \(\left[\begin{array}{lll}
1 & 4 & 3 \\
2 & 5 & 9 \\
3 & 6 & 1 \\
4 & 1 & 1 \\
1 & 2 & 6
\end{array}\right]\)
(iv) Order of A^T is (5 x 3).

Question 5.
Matrices A and B are given below. Find A + B, B + A, A – B and B – A. Verify that A + B = B + A and B – A = -(A – B)
(i) A = \(\left[\begin{array}{l}
7 \\
1
\end{array}\right]\), B = \(\left[\begin{array}{c}
-6 \\
9
\end{array}\right]\)
Solution:

(ii) A = \(\left[\begin{array}{cc}
1 & 2 \\
3 & -1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
4 & 1 \\
-3 & 2
\end{array}\right]\)
Solution:

(iii) A = \(\left[\begin{array}{ll}
\frac{1}{2} & \frac{1}{4} \\
\frac{1}{3} & \frac{1}{5}
\end{array}\right]\), B = \(\left[\begin{array}{ll}
\frac{1}{3} & \frac{1}{2} \\
\frac{1}{2} & \frac{4}{5}
\end{array}\right]\)
Solution:

(iv) A = \(\left[\begin{array}{cc}
1 & a-b \\
a+b & -3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & b \\
-a & 5
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{rrr}
1 & -2 & 5 \\
-1 & 4 & 3 \\
1 & 2 & -3
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
-1 & 2 & -5 \\
1 & -3 & -3 \\
1 & 2 & 4
\end{array}\right]\)
Solution:

Question 6.
(i) Find the 2×2 matrix X
if X + \(\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\) = \(\left[\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right]\)
Solution:

(ii) Given
[x y z] – [-4 3 1] = [-5 1 0] derermine x, y, z.
Solution:
[x y z] – [-4 3 1] = [-5 1 0]
∴ (x y z) = (-4 3 1) + (-5 1 0) = (-9 4 1)
∴ x = -9, y = 4, z = 1

(iii) If \(\left[\begin{array}{ll}
x_1 & x_2 \\
y_1 & y_2
\end{array}\right]\) – \(\left[\begin{array}{ll}
2 & 3 \\
0 & 1
\end{array}\right]\) = \(\left[\begin{array}{ll}
3 & 5 \\
1 & 2
\end{array}\right]\) determine x₁, x₂, y₁, y₂.
Solution:

(iv) Find a matrix which when added to \(\left[\begin{array}{cc}
2 & -3 \\
-4 & 7
\end{array}\right]\) gives \(\left[\begin{array}{ll}
4 & 1 \\
3 & 2
\end{array}\right]\)
Solution:

Question 7.
Calculate whenever possible, the following products.
(i) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\left[\begin{array}{l}
2 \\
3
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{l}
2 \\
3
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
4 & 3
\end{array}\right]\)
Solution:
\(\left[\begin{array}{l}
2 \\
3
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
4 & 3
\end{array}\right]\) is impossible because number of columns of 1st ≠ number of rows of second.

(iii) \(\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\left[\begin{array}{ll}
3 & 1 \\
1 & 1
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{cc}
1 & -2 \\
-2 & 3
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4
\end{array}\right]\)
Solution:

Question 8.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]\), C = \(\left[\begin{array}{ll}
2 & 2 \\
1 & 3
\end{array}\right]\)
Calculate (i) AB (ii) BA (iii) BC (iv) CB (v) AC (vi) CA
Solution:

Question 9.
Find the following products.
(i) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & 3 \\
1 & 4
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ll}
1 & 3 \\
1 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{cc}
1 & i \\
i & -1
\end{array}\right]^2\) where i = √-1
Solution:

(vi) \(\left[\begin{array}{ll}
\mathbf{0} & \mathbf{1} \\
\mathbf{1} & \mathbf{0}
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Solution:

(vii) \(\left[\begin{array}{ll}
0 & k \\
1 & 0
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Solution:

(viii) \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\)
Solution:

(ix) \(\left[\begin{array}{ll}
1 & 0 \\
0 & k
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Solution:

(x) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)
Solution:
\(\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\) = \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)

Question 10.
Write true or false in the following cases:
(i) The sum of a 3 x 4 matrix with a 3 x 4 matrix is a 3 x 3 matrix.
Solution:
False

(ii) k[0] = 0, k ∈ R
Solution:
False

(iii) A – B = B – A, if one of A and B is zero and A and B are of the same order.
Solution:
False

(iv) A + B = B + A, if A and B are matrices of the same order.
Solution:
True

(v) \(\left[\begin{array}{cc}
1 & 0 \\
-2 & 0
\end{array}\right]\) + \(\left[\begin{array}{cc}
-1 & 0 \\
2 & 0
\end{array}\right]\) = 0
Solution:
True

(vi) \(\left[\begin{array}{ll}
3 & 1 \\
6 & 2
\end{array}\right]\) = 3 \(\left[\begin{array}{ll}
1 & 1 \\
2 & 2
\end{array}\right]\)
Solution:
False

(vii) With five elements a matrix can not be constructed.
Solution:
False

(viii)The unit matrix is its own transpose.
Solution:
True

Question 11.
If A = \(\left[\begin{array}{cc}
2 & 4 \\
3 & 13
\end{array}\right]\) and I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) find A – α I, α ∈ R.
Solution:

Question 12.
Find x and y in the following.
(i) \(\left[\begin{array}{cc}
x & -2 y \\
0 & -2
\end{array}\right]=\left[\begin{array}{cc}
1 & -8 \\
0 & -2
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{c}
x+3 \\
2-y
\end{array}\right]=\left[\begin{array}{c}
1 \\
-3
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{c}
2 x-y \\
x+y
\end{array}\right]=\left[\begin{array}{c}
3 \\
-9
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{l}
x \\
y
\end{array}\right]+\left[\begin{array}{l}
3 \\
4
\end{array}\right]=\left[\begin{array}{c}
2 \\
-1
\end{array}\right]\)
Solution:

(v) [2x -y] + [y 3x] = 5 [1 0]
Solution:

Question 13.
The element of ith row and ith column of the following matrix is i +j. Complete the matrix.

Question 14.
Write down the matrix

Question 15.
Construct a 2 x 3 matrix having elements given by
(i) a_ij = i + j
(ii) a_ij = i – j
(iii) a_ij = i × j
(iv) a_ij = i / j
Solution:

Question 16.
If \(\left[\begin{array}{cc}
2 x & y \\
1 & 3
\end{array}\right]+\left[\begin{array}{cc}
4 & 2 \\
0 & -1
\end{array}\right]=\left[\begin{array}{ll}
8 & 3 \\
1 & 2
\end{array}\right]\)
Solution:

Question 17.
Find A such that
\(\left[\begin{array}{ccc}
2 & 3 & 4 \\
1 & 0 & -2 \\
3 & 1 & -1
\end{array}\right]+A=\left[\begin{array}{ccc}
1 & 2 & -1 \\
2 & -1 & 0 \\
1 & 3 & 2
\end{array}\right]\)
Solution:

Question 18.
If

Question 19.
What is the order of the matrix B if [3 4 2] B = [2 1 0 3 6]
Solution:
(3 4 2) B = (2 1 0 3 6)
Let A = (3 4 2), C = (2 1 0 3 6)
∴ Order of A = (1 x 3)
Order of C = (1 x 5)
∴ Order of B = (3 x 5)

Question 20.
Find A if \(\left[\begin{array}{l}
4 \\
1 \\
3
\end{array}\right]\) A = \(\left[\begin{array}{rrr}
-4 & 8 & 4 \\
-1 & 2 & 1 \\
-3 & 6 & 3
\end{array}\right]\)
Solution:

Question 21.
Find B if B² = \(\left[\begin{array}{cc}
17 & 8 \\
8 & 17
\end{array}\right]\)
Solution:

∴ a² + bc = 17, ab + bd= 8
ca + cd = 8, bc + d² = 17
∴ a² + bc = bc + d²
or, a² + d² or, a = d
or, ca + cd = ab + bd
or, cd + cd – bd + bd
or, 2cd = 2bd = 8
or, b = c and bd = 4 = cd
∴ ab + bd= 8
or, ab + 4 = 8
or, ab = 4
Again, a² + bc = 17
or, a² + b . b = 17 (∵ b = c)
or, a² + b² = 17
Also (a + b)² = a² + b² + 2ab
∴ (a + b)² = 17 + 8 = 25
or, a + b = 5
And (a – b)² = 17 – 8 = 9
or, a – b = 3
∴ a = 4, b = 1, So d = 4, c = 1
∴ B = \(\left[\begin{array}{ll}
4 & 1 \\
1 & 4
\end{array}\right]\)

Question 22.
Find x and y when

Question 23.
Find AB and BA given that:

Question 24.
Evaluate

Question 25.
If

Show that AB = AC though B ≠ C. Verify that
(i) A + (B + C) = (A + B) + C
(ii) A(B + C) = AB + AC
(iii) A(BC) = (AB)C
Solution:

Question 26.
Find A and B where

Question 27.
If A = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]\) and I be the 2 × 2 unit matrix find (A – 2I) (A – 3I)
Solution:

Question 28.
Verify that [AB]^T = B^TA^T where

Question 29.
Verify that A = \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\) satisfies the equation x² – (a + d)x + (ad – bc)I = 0 where I is the 2 x 2 matrix.
Solution:

Question 30.
If A = \(\left[\begin{array}{rrr}
1 & 2 & 3 \\
3 & -2 & 1 \\
4 & 2 & 1
\end{array}\right]\), show that A³ – 23 A – 40 I = 0
Solution:

Question 31.

Question 32.
If A and B are matrices of the same order and AB = BA, then prove that
(i) A² – B² = (A – B) (A + B)
(ii) A² + 2AB + B² = (A + B)²
(iii) A² – 2AB + B² = (A – B)²
Solution:
(i) (A – B) (A + B)
= A² + AB – BA – B²
= A² + AB – AB- B²(∵ AB = BA)
= A² – B²
(ii) (A + B)² = (A + B) (A + B)
= A² + AB + BA + B²
= A² + AB + AB + B² (∵ AB = BA)
= A² + 2AB + B²
(iii) (A – B)² = (A – B) (A – B)
= A² – AB – BA + B²
= A² – AB – AB + B² (∵ AB = BA)
= A² – 2AB + B²

Question 33.
If α and β are scalars and A is a square matrix then prove that
(A – αI) . (A – βI) = A² – (α + β) A + αβI, where I is a unit matrix of same order as A.
Solution:
(A – αI) (A – βI)
= A² – AβI – αIA + αβI²
= A² – βAI – αA + αβI
(∵ IA = A, I² = I)
= A² – βA – αA + αβI) (∵ AI = A)
= A² – (α + β) A + αβI

Question 34.
If α and β are scalars such that A = αβ + βI, where A, B and the unit matrix I are of the same order, then prove that AB = BA.
Solution:
We have A = αβ + βI
AB (αβ + βI) B
= α βB + βI B
= α βB + βB = (α + I) βB
= βB (α + 1)
(∵ Scalar mltiβlication is associative)
= Bβ (α + 1)
= Bβα + Bβ = Bαβ + BIβ
(∵ BI = B)
= B (αβ + βi) = BA
AB = BA
(proved)

Question 35.

Question 36.

Question 37.

Question 38.

Question 39.

Question 40.

Question 41.

Question 42.

Question 43.

	Men	Women	Children
Family A →	4	6	2
Family B →	2	2	4

	Family B
	Calory	Proteins
Men	2400	45
Women	1900	55
Children	1800	33

Solution:
The given informations can be written in matrix form as

∴ Calory requirements for families A and B are 24600 and 15800 respectively and protein requirements are 576 gm and 332 gm respectively.

Question 44.
Let the investment in first fund = ₹x and in the second fund is ₹(50000-x)
Investment matrix A=[x  50000-x]

⇒ 300000 – x = 278000
⇒ x = 22000
∴ He invests ₹22000 in first bond and ₹28000 in the second bond.

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 7 Continuity and Differentiability Ex 7(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 7 Continuity and Differentiability Exercise 7(b)

Question 1.
Differentiate from definition
(i) e^3x
Solution:

(ii) 2^x2
Solution:

(iii) In (3x + 1)
Solution:

(iv) logx⁵ (Hint : logx⁵ = \(\frac{\ln 5}{\ln x}\))
Solution:

(v) In sin x
Solution:

(vi) x² a^2x
Solution: