Class 12

Odisha State Board CHSE Odisha Class 12 Invitation to English 1 Solutions Chapter 1 My Greatest Olympic Prize Textbook Exercise Questions and Answers.

CHSE Odisha 12th Class English Solutions Chapter 1 My Greatest Olympic Prize

CHSE Odisha Class 12 English My Greatest Olympic Prize Text Book Questions and Answers

Unit-wise Gist and Glossary:

UNIT – I:
Gist:
Jesse Owens takes us back to the 1936 Summer Olympics held in Berlin where nationalistic feelings were running high because of Hitler’s reference to his country’s participants who belonged to a ‘master race’. His words produced no effect on Owens. Everyone looked forward to winning his long jump event, because a year before, he as a university student had set a world record in that field. But his surprise knew no bounds at the sight of a German called Luz Long touching the pit at almost 26 feet on his practice. Owens learned that Hitler hoped to win the jump. In his view, Luz Long’s victory would cement the Nazi’s Aryan superiority theory. He was a Negro and was bent on showing his superiority.
ସାରମର୍ମ :
ଜେସି ଓୟେସ ଆମ୍ଭମାନଙ୍କୁ ୧୯୩୬ ମସିହାରେ ବର୍ଲିନ୍‌ଠାରେ ଅନୁଷ୍ଠିତ ଗ୍ରୀଷ୍ମକାଳୀନ ଅଲିମ୍ପିକ୍ କ୍ରୀଡ଼ାର ପୃଷ୍ଠଭୂମିକୁ ନେଇ ଯାଇଛନ୍ତି ଯେଉଁଠାରେ କି ନିଜ ଦେଶର ଖେଳାଳିମାନେ ଶ୍ରେଷ୍ଠ ଜାତିର ଅନ୍ତର୍ଭୁକ୍ତ ବୋଲି ହିଲର୍‌ଙ୍କ ମନ୍ତବ୍ୟ କାରଣରୁ ପ୍ରବଳ ଜାତୀୟତା ଭାବନା ସୃଷ୍ଟି ହୋଇଥିଲା । ତାଙ୍କ କଥାର କୌଣସି ପ୍ରଭାବ ଓୟେସଙ୍କ ଉପରେ ପଡ଼ିନଥିଲା । ଲମ୍ବଡ଼ିଆରେ ତାଙ୍କର ବିଜୟକୁ ସମସ୍ତେ ଆଗ୍ରହର ସହିତ ଅପେକ୍ଷା କରିଥିଲେ, କାରଣ ବର୍ଷକ ପୂର୍ବରୁ ଜଣେ ବିଶ୍ବବିଦ୍ୟାଳୟ ଛାତ୍ର ଭାବରେ ସେ ଏହି କ୍ଷେତ୍ରରେ ବିଶ୍ଵରେକର୍ଡ ପ୍ରତିଷ୍ଠା କରିଥିଲେ । ମାତ୍ର ଲୁଜ୍ ଲଙ୍ଗ୍ ନାମକ ଜଣେ ଜର୍ମାନ୍ ଅଭ୍ୟାସ ପର୍ଯ୍ୟାୟରେ ପ୍ରାୟ ୨୬ ଫୁଟ୍ ଡେଇଁବାର ଦେଖୁ ତାଙ୍କ ବିସ୍ମୟର ସୀମା ରହିଲା ନାହିଁ । ହିଟ୍‌ଲର୍ ତାଙ୍କୁ ଡିଆଁରେ ବିଜୟୀ ହେବାର ଆଶା ରଖୁଛନ୍ତି ବୋଲି ଓୟେସ ଜାଣିବାକୁ ପାଇଲେ । ତାଙ୍କ ମତରେ, ଲୁଜ୍ ଲଙ୍ଗ୍‌ଙ୍କ ବିଜୟ ନାଜୀମାନଙ୍କର ‘ଆର୍ଯ୍ୟ-ଶ୍ରେଷ୍ଠତ୍ୱ’ ସିଦ୍ଧାନ୍ତକୁ ନିଶ୍ଚିତରୂପେ ଦୃଢ଼ୀଭୂତ କରିବ । ସେ ଜଣେ ନିଗ୍ରୋ ଥିଲେ ଏବଂ ନିଜର ଶ୍ରେଷ୍ଠତ୍ଵ ପ୍ରତିପାଦନ ପାଇଁ ବଦ୍ଧପରିକର ହେଲେ ।

Glossary:
Olympic Games: a modern revival of the greatest of games or festivals of ancient Greece. The Olympic Games are held every four years, each time in a different country. (ଅଲିମ୍ପିକ୍ କ୍ରୀଡ଼ା)
Adolf Hitler : (1889-1945) Nazi dictator of Germany (ଜର୍ମାନୀର ନାଜୀ ଶାସକ)
childishly : ପିଲାଳିଆ ଭାବରେ
performers : competitors (ପ୍ରତିଯୋଗୀମାନେ)
master race: superior to all other races (ଶ୍ରେଷ୍ଠ ଜାତି)
Hitler held that Germans were superior to all other races.
nationalistic: promoting nationalism (especially, a narrow kind of nationalism) (ଜାତୀୟତା)
all-time high: the highest ever (ସର୍ବାଧ୍ବକ)
I …. six years: Owens had tried hard for six years.
set : established (ପ୍ରତିଷ୍ଠିତ)
26 feet 8 inches: 8.13 metres (୮.୧୩ ମିଟର)
hands down: very easily (ଅତି ସହଜରେ)
I surprise : Owens’ surprise knew no bounds (ଓବେନ୍ସଙ୍କ ଆଶ୍ଚର୍ୟ୍ୟର ସୀମା ନ ଥିଲା)
startled : greatly shocked and surprised (ଆଶ୍ଚର୍ଯ୍ୟ ହୋଇଗଲେ |)
hitting : touching (ଛୁଇଁବା)
leaps : jumps
evidently : clearly
under wraps : secret
Nazis: members of Hitler’s National Socialist German Workers’ Party (NSDAP)
Aryan-superiority: The Aryans are superior to all other races. (ଆର୍ଯ୍ୟ-ଶ୍ରେଷ୍ଠତ୍ୱ)
After all : ମୋଟାମୋଟି ଭାବେ
hot under the collar: very angry (ରାଗୀ)
determined: firmly decided
Der Fuhrer: the leader in German (Used with special reference to Hitler) (ଜର୍ମାନ୍ ନେତା ହିଟଲର୍ )

Think it out:
Question 1.
Why were nationalistic feelings running high during the 1936 Summer Olympics in Berlin?
Answer:
Nationalistic feelings were running high during the 1936 Summer Olympics in Berlin because of Hitler’s Nazi theory that Germans were superior to all other races.

Question 2.
‘I wasn’t too worried about all this’. What does “this” refer to – Hitler’s beliefs or winning a gold medal?
Answer:
‘This’ refers to Hitler’s beliefs.

Question 3.
Why wasn’t Owens worried?
Answer:
Owens was not worried, because he had shed his blood, sweat, and tears for six years, with the Games in his mind.

Question 4.
Why did everyone expect Owens to win the long jump easily?
Answer:
Everyone expected Owens to win the long jump easily, because, a year before the advent of the Berlin Olympic Games, he, as a university student, had established the world record of 26 feet 8] inches (8.13 meters).

Question 5.
What was the surprise that awaited Jesse Owens in Berlin?
Answer:
The surprise that awaited Jesse Owens was a tall German boy, Luz Long’s amazing performance of hitting the pit at almost 26 feet on his practice jumps.

Question 6.
What did he learn from people about Luz Long?
Answer:
He learned from people about Luz Long that Hitler had kept him secretly hoping he would be the jump winner.

Question 7.
Do you think Nazis’ Aryan-superiority theory meant that Germans were superior to Negroes? How did Owens feel about it – angry or bothered?
Answer:
I don’t think Nazis’ Aryan-superior theory meant that Germans were superior to Negroes. Owens felt angry about it.

Question 8.
What made Owens determined to beat Luz Long?
Answer:
The fact that made Owens determined to beat Luz Long was that he was a Negro and against this backdrop, he would disprove Hitler’s Aryan superiority theory.

UNIT – II

Gist:
In the writer’s view, anger is the worst enemy of an athlete, because this base passion leads him or her to commit mistakes. The results of the first two qualifying jumps for Owens were dismal. He was utterly disgusted. His failure in the two qualifying jumps made him kick the pit. In the meantime, to his stunned disbelief, he found Luz Long, the tall German long jumper, offered him a firm handshake. He wore a nice look. Owens tried to conceal his nervousness, but Long understood his anger.

In spite of being trained in the Nazi youth movement, he was a glorious exception. He did not believe in the concept of Aryan supremacy. The blue-eyed and remarkably handsome Long eventually noticed that his anger had abated and advised Owens to draw a line a few inches at the back of the board and focus on making his take-off from there. He said to Owens that to come first in the trials was of no use and the next day was crucial. Luz Long’s words worked wonders. Owens’ tension vanished and he qualified for the jump with great confidence.
ସାରମର୍ମ :
ଲେଖକଙ୍କର ମତରେ, କ୍ରୋଧ ଖେଳାଳିର ସବୁଠାରୁ ବଡ଼ ଶତ୍ରୁ । କାରଣ ଏହି ଘୃଣ୍ୟ ପ୍ରବୃତ୍ତି ଯୋଗୁଁ ସେ ଭୁଲ୍ କରି ବସେ । ଓୟେସ୍‌ଙ୍କର ଯୋଗ୍ୟତା ପର୍ଯ୍ୟାୟରେ ପ୍ରଥମ ଦୁଇଟି ଲମ୍ଫ ନୌରାଶ୍ୟଜନକ ଥିଲା । ସେ ଭୀଷଣ ଭାବରେ ବିରକ୍ତ ହୋଇଗଲେ । ଯୋଗ୍ୟତା ପର୍ଯ୍ୟାୟର ପ୍ରଥମ ଦୁଇଟି ଡିଆଁରେ ଅସଫଳ ହୋଇ ସେ ଭୂଇଁକୁ ଗୋଇଠା ମାରିଥିଲେ । ଏହି ସମୟରେ ସେ ଜର୍ମାନ୍‌ର ଡେଙ୍ଗା ଲମ୍ବଡିଆଁ ପ୍ରତିଯୋଗୀ ଲୁଜ୍ ଲଙ୍ଗ୍ ତାଙ୍କ ସହିତ କରମର୍ଦ୍ଦନ କରିବାକୁ ହାତ ବଢ଼ାଇଥବା ଦେଖୁ ବିସ୍ମିତ ହେଲେ । ସେ ବନ୍ଧୁତ୍ଵପୂର୍ଣ୍ଣ ଦୃଷ୍ଟିରେ ଚାହିଁ ରହିଥିଲେ । ଓୟେନ୍ସ ନିଜର କ୍ରୋଧକୁ ଲୁଚାଇବାକୁ ଚାହୁଁଥିଲେ ହେଁ ଲଙ୍ଗ୍ ତାହା ବୁଝିପାରିଥିଲେ । ନାଜି ଯୁବ ଆନ୍ଦୋଳନରେ ପ୍ରଶିକ୍ଷିତ ହୋଇଥିଲେ ହେଁ ସେ ଏକ ଚମତ୍କାର ବ୍ୟତିକ୍ରମ ଥିଲେ । ସେ ଆର୍ଯ୍ୟ-ଶ୍ରେଷ୍ଠତ୍ୱରେ ବିଶ୍ଵାସ କରୁନଥିଲେ । ନୀଳାଭ ନୟନ ଓ ସୁଗଠିତ ଶରୀରଧାରୀ ଲଙ୍ଗ୍ ଦେଖ‌ିଲେ ଯେ ଓୟେନ୍‌ସ୍‌ଙ୍କ ରାଗ ପ୍ରଶମିତ ହୋଇଗଲାଣି । ସେ କାଠପଟା କିଛି ଇଞ୍ଚ ପଛରୁ ଏକ ଗାର ଟାଣି ଓ ସେହି ଗାରକୁ ନଜରରେ ରଖି ସେହିଠାରୁ ଡିଆଁ ଆରମ୍ଭ କରିବାକୁ ଓୟେସ୍‌ଙ୍କୁ ଉପଦେଶ ଦେଲେ । ଯୋଗ୍ୟତା ପର୍ଯ୍ୟାୟରେ ପ୍ରଥମ ହେବାର କିଛି ଆବଶ୍ୟକତା ନାହିଁ ଏବଂ ବାସ୍ତବରେ ପରବର୍ତ୍ତୀ ଦିନ ହିଁ ଗୁରୁତ୍ଵପୂର୍ଣ୍ଣ ବୋଲି ସେ ଓୟେସ୍‌ଙ୍କୁ କହିଥିଲେ । ଲୁଜ୍ ଲଙ୍ଗ୍‌ଙ୍କର ପରାମର୍ଶ ଯାଦୁ ଭଳି କାମ କଲା । ଓୟେସ୍‌ଙ୍କ ଚିନ୍ତା ଉଭେଇଗଲା ଏବଂ ସେ ଦୃଢ଼ ଆତ୍ମବିଶ୍ଵାସ ସହ ଶେଷ ଡିଆଁ ପାଇଁ ଯୋଗ୍ୟତା ହାସଲ କରିଥିଲେ ।

Glossary:
athlete : କ୍ରୀଡ଼ାବିତ୍
exception : ବ୍ୟତିକ୍ରମ
leapt : jumped (ଡେଇଁଲେ)
beyond : ବାହାରେ
bitterly : with hatred ଭାବରେ )
kicked : ଗୋଇଠା ମାରିଲେ
disgustedly : ବିରକ୍ତିପୂର୍ଣ୍ଣ ଭାବେ
firm handshake : ଦୃଢ଼ ହ୍ୟାଣ୍ଡସେକ
twist : (here) speech accent (ଉଚ୍ଚାରଣ ଭଙ୍ଗୀ )
hide : ଲୁଚାଇବା
mastered : acquired complete knowledge or skill (ଦକ୍ଷତା ହାସଲ କରିବା)
a bit : a little (ସ୍ଵଳ୍ପ/ଅଳ୍ପ)
slang : words used informally; words used in talk by a group or class of people (ଅନୌପଚାରିକ ଭାଷା)
must be eating you : must be agitating your mind
anger : କ୍ରୋଧ
took pain : took trouble (ଅସୁବିଧାରେ ପକାଇଲା)
reassure : to say something to make somebody less frightened (ପୁନଃ ଆଶ୍ୱାସନା ଦେବା )
schooled : trained (ପ୍ରଶିକ୍ଷିତ)
movement : ଚଳନ
strikingly : impressively
handsome : ସୁନ୍ଦର
calmed : cooled (ଶାନ୍ତ ହେଲା )
counts : matters (ଆବଶ୍ୟକ କରେ)

Think it out:
Question 1.
What does a coach say about an angry athlete?
Answer:
A coach says that an angry athlete will commit mistakes. In other words, he says that anger is an athlete’s worst enemy.

Question 2.
What were the results of the first two qualifying jumps for Owens?
Answer:
The results of the first two qualifying jumps for Owens were miserable. He jumped from several inches outside the take-off board for a no-jump.

Question 3.
Why did Owens kick the pit?
Answer:
Owens kicked the pit because he failed during the trials. He was disgusted.

Question 4.
Who offered Owens a firm handshake? Was he friendly or hostile?
Answer:
Luz Long, a German long jumper offered him a firm handshake. He was friendly.

Question 5.
Why did Long speak to Owens during the trials? Did he mean to make friends with Owens or to find out what was troubling him?
Answer:
Long spoke to Owens during the trials to help him. He wanted to find out what was troubling Owens.

Question 6.
“he really looked the part” – What does this mean? Does it mean Long was trying to play the part of an Aryan or he looked as if he belonged to a superior race?
Answer:
‘He really looked the part’ means Luz Long was trying to play the part of an Aryan.

Question 7.
How did Luz Long help Jesse Owens in qualifying for the final jumps?
Answer:
Luz Long helped Jesse Owens in qualifying for the final jumps by advising him to draw a line a few inches at the back of the take-off board and focussing on his start from there.

Question 8.
“Tomorrow is what counts.” – What did Long mean by this? Does he mean that Owens would win the next day or that their performance the next day would matter much?
Answer:
Long means that Owens would win the next day.

Question 9.
Did Owens qualify for the final jump? How did he do that?
Answer:
Thanks to Long’s friendly advice, Owens qualified for the final jump. Brimming with confidence, he drew a line a full foot behind the board and advanced to jump from there and qualified for the final jump.

UNIT – III

Gist:
A real friendship sprang up between Jesse Owens and Luz Long when the former went to the latter’s room and dwelt on varied topics for two hours. The moment they had been waiting for had arrived at last. Luz smashed his own past record and encouraged Owens to give his best performance. Jesse Owens won the event, setting the Olympic record of 26 feet 5 4 inches. Luz congratulated him and shook his hand warmly in spite of Hitler’s angry look at them. Owens felt genuine friendship for Luz at that moment. The most fabulous Olympic prize for him was the friendship he formed with. Long, but not the gold medal he won in the long jump. In Owens’ view, Long epitomized the philosophy of Pierre de Coubertin, the founder of modem Olympic Games – the essence of the Olympic Games lies not in winning but in participating. Good fight, but not conquest is the hallmark of life.
ସାରମର୍ମ :
ଯେତେବେଳେ ଓୟେସ୍ ଲୁଜ୍ ଲଙ୍ଗ୍‌ଙ୍କ କୋଠରିକୁ ଯାଇ ଦୁଇ ଘଣ୍ଟା ଧରି ବିଭିନ୍ନ ବିଷୟରେ ଆଲୋଚନା କଲେ, ସେତେବେଳେ ଦୁଇଜଣଙ୍କ ମଧ୍ୟରେ ପ୍ରକୃତ ବନ୍ଧୁତା ଗଢ଼ି ଉଠିଲା । ସେମାନଙ୍କର ଅପେକ୍ଷା କରିବାର ମୁହୂର୍ତ୍ତ ଆସିଗଲା । ଲୁଜ୍ ନିଜର ପୂର୍ବ ରେକର୍ଡ ଭାଙ୍ଗିଲେ ଏବଂ ଶ୍ରେଷ୍ଠତ୍ୱ ପ୍ରତିପାଦନ କରିବାକୁ ଓୟେସ୍‌ଙ୍କୁ ଉତ୍ସାହିତ କଲେ । ଓୟେସ୍ ପ୍ରତିଯୋଗିତାରେ ଜିତିଲେ ଏବଂ ୨୬ ଫୁଟ ୫୪ ଇଞ୍ଚ ଡେଇଁ ଅଲିମ୍ପିକ୍ ରେକର୍ଡ ସ୍ଥାପନ କଲେ । ଲୁଜ୍ ଲଙ୍ଗ୍ ହିଟ୍‌ଲର୍‌ଙ୍କ କ୍ରୋଧପୂର୍ଣ୍ଣ ଚାହାଣି ସତ୍ତ୍ଵେ ତାଙ୍କୁ ଅଭିନନ୍ଦନ ଜଣାଇଲେ ଏବଂ ଖୁସିରେ କରମର୍ଦ୍ଦନ କଲେ । ସେହି ମୁହୂର୍ତ୍ତରେ ଓୟେନ୍ସ ଲୁଜ୍‌ଙ୍କ ପ୍ରତି ଅନାବିଳ ବନ୍ଧୁତ୍ଵଭାବ ଅନୁଭବ କଲେ । ଲମ୍ବଡ଼ିଆରେ ସ୍ଵର୍ଣ୍ଣପଦକ ଜିତିବା ଅପେକ୍ଷା ଲୁଜ୍‌ଙ୍କ ସହ ସ୍ଥାପିତ ସମ୍ପର୍କ ତାଙ୍କ ପାଇଁ ସର୍ବଶ୍ରେଷ୍ଠ ଅଲିମ୍ପିକ୍ ପୁରସ୍କାର ଥିଲା । ଆଧୁନିକ ଅଲିମ୍ପିକ୍ କ୍ରୀଡ଼ାର ପ୍ରତିଷ୍ଠାତା ପେରୀ ଡି କୁବରଟିନ୍‌ଙ୍କ ଦର୍ଶନ ଯାହାକି ଅଲିମ୍ପିକ୍ କ୍ରୀଡ଼ାର ମହତ୍ତ୍ବ ବିଜୟୀ ହେବାରେ ନୁହେଁ ଅଂଶଗ୍ରହଣ କରିବାରେ ରହିଛି, ଲଙ୍ଗ ତାହାର ଜ୍ଵଳନ୍ତ ଉଦାହରଣ ଥିଲେ । ଜୀବନର ମହତ୍ତ୍ବ ବିଜୟପ୍ରାପ୍ତ କରିବା ନୁହେଁ ଉତ୍ତମରୂପେ ସଂଘର୍ଷ କରିବା ଉପରେ ପର୍ଯ୍ୟବସିତ ।

Glossary:
real : genuine (ବାସ୍ତବ)
beat : defeat (ହରେଇବା )
peak performance : best ever performance ( ସର୍ବୋତ୍କୃଷ୍ଟ କୃତିତ୍ଵ)
at the instant: at once (ସଙ୍ଗେ ସଙ୍ଗେ)
congratulating : ଅଭିନନ୍ଦନ
26 feet 5 1/4 inches: 8.6 metres (୮.୬ ମିଟର)
despite : in spite of (ସତ୍ତ୍ୱେ)
glared : looked with anger (କ୍ରୋଧରେ ଚାହିଁଲେ)
fake : false (କୃତ୍ରିମ)
24-carat friendship : genuine friendship (ପ୍ରକୃତ ବନ୍ଧୁତା)
epitome : (here) a typical representation of the ideal (ପ୍ରକୃଷ୍ଟ ଉଦାହରଣ )
taking part : participating (ଭାଗ ନେବା)
conquering : winning (ଜିତିବା)

Think about it:
Question 1.
When did Owens and Long realize that they had become friends?
Answer:
Owens and Long realized that they had become friends after the former went to the latter’s room and talked for two hours concerning track and field, themselves, the global scenario, and a dozen other topics.

Question 2.
Who was Coubertin? What was his ideal?
Answer:
Coubertin was the founder of the modem Olympic Games. His idea was that in life not winning but fighting in the right spirit was very important.

Question 3.
Why has Luz Long been called a fine example of Coubertin’s ideal?
Answer:
Luz Long has been called a fine example of Coubertin’s ideal because the former took a leaf out of the latter’s book, ‘The important thing in the Olympic Games is not winning but taking part. The essential thing in life is not conquering but fighting well.”

Question 4.
What do you think was the greatest Olympic Prize for Jesse Owens – the gold medal he won in the long jump, or the friendship he formed with Luz Long?
Answer:
I think the greatest Olympic Prize for Jesse Owens was the friendship he formed with Luz Long.

Post-Reading Activities:
Doing with words :
(a) ‘Childish’ is an adjective. We can make it an adverb by adding ‘ly’ – ‘childishly’. Now add ‘ly’ to make the following adjectives adverbs: easy, real, bitter, disgusted, clear, physical, friend, final, certain, sudden
Answer:
easy – easily
real – really
bitter – bitterly
disgusted – disgustedly
clear – clearly
physical – physically
friend – friendly
final – finally
certain – certainly
sudden – suddenly

(b) Replace the italicized words in each of the following sentences with idiomatic expressions given in brackets :
(an all-time high, hands down, under wraps, hot under the collar, look the part)
(i) The plan was carefully kept secret.
(ii) Tendulkar’s double century is the highest-ever individual score in a one-day cricket match.
(iii) You’d never guess he was a security guard; he doesn’t appear to be suited to the job.
(iv) Delhi daredevils won the IPL cup very easily.
(v) The policeman was very angry because the criminal escaped.
Answer:
(i) The plan was carefully kept under wraps.
(ii) Tendulkar’s double century is an all-time high individual score in a one-day cricket match
(iii) You’d never guess he was a security guard; he doesn’t look the part.
(iv) Delhi daredevils won the IPL cup hands down.
(v) The policeman was hot under the collar because the criminal escaped.

(c) Make sentences of your own using the following expressions :
(i) Make a fool of oneself
(ii) have one’s eye on
(iii) (to be) in for a surprise
(iv) ebb out
(v) no exception
Answer:
(i) Make a fool of oneself – He made a fool of himself by turning up drunk to a TV chat show.
(ii) have one’s eye on – I have got my eye on a new DVD player.
(iii) (to be) in for a surprise – The players could be in for a surprise if they expect an easy victory.
(iv) ebb out – Enthusiasm for reform ebbed out.
(v) no exception – Climbers are brave people, and Sharat is no exception.

CHSE Odisha Class 12 English My Greatest Olympic Prize Important Questions and Answers

I. Multiple-Choice Questions (MCQs) with Answers:

Question 1.
Who is the writer of “My Greatest Olympic Prize”?
(A) Jessie Owens
(B) Luz Long
(C) Adolf Hitler
(D) Churchill
Answer:
(A) Jessie Owens

Question 2.
Jessie Owens belongs to which country?
(A) America
(B) England
(C) Germany
(D) Italy
Answer:
(A) America

Question 3.
Why had Jessie Owens come to Germany?
(A) to play football
(B) to play cricket
(C) to participate in the Commonwealth Games
(D) to participate in the Olympic event
Answer:
(D) to participate in the Olympic event

Question 4.
In which year this Olympic event was organized?
(A) 1935
(B) 1937
(C) 1936
(D) 1938
Answer:
(C) 1936

Question 5.
In which season this Olympic event was organized?
(A) Winter
(B) Summer
(C) Spring
(D) Rainy
Answer:
(B) Summer

Question 6.
What did Adolf Hitler childishly insist?
(A) His performers were members of a ‘master race’
(B) His performers were members of Nordic races
(C) His performers were members of Aryan races
(D) All the above
Answer:
(D) All the above

Question 7.
Why was not Jessie Owens worried about Hitler’s attitude?
(A) because he had known him
(B) because he had not full confidence in himself
(C) He had trained himself for six years
(D) None of the above
Answer:
(C) He had trained himself for six years

Question 8.
What was he thinking when he was coming over the boat?
(A) to fight well
(B) was confused about what to do
(C) to take the gold medal
(D) to play whatever may be
Answer:
(C) to take the gold medal

Question 9.
On which event had he decided to participate?
(A) high jump
(B) running
(C) long jump
(D) swimming
Answer:
(C) long jump

Question 10.
What was the record he had created a year before as a university student?
(A) by jumping 26 feet 8 1/4 inches
(B) by jumping 26 feet 7 1/4 inches
(C) by jumping 26 feet 8 1/2 inches
(D) by jumping 26 feet 8 1/3 inches
Answer:
(A) by jumping 26 feet 8 1/4 inches

Question 11.
Why was he surprised when the time came for the long jump trials?
(A) he saw Hitler there inspiring his performers
(B) he saw a tall boy hitting the pit at almost 26 feet on his practice leaps
(C) he saw a tall boy hitting the pit at almost 25 feet
(D) he saw how Hitler was encouraging them to win the gold medal
Answer:
(B) he saw a tall boy hitting the pit at almost 26 feet on his practice leaps

Question 12.
What was the name of Jessie Owen’s rival?
(A) Hitler
(B) Churchill
(C) Luz Long
(D) None of the above
Answer:
(C) Luz Long

Question 13.
Why had Hitler kept him under secret?
(A) Hoping Luz Long would not be known to others.
(B) Hoping Luz Long should not talk to others.
(C) Hoping Luz Long would win the jump.
(D) All the above
Answer:
(C) Hoping Luz Long would win the jump.

Question 14.
Why did Jessie Owens think if Long won, it would add some new support to the Nazis’ Aryan Superiority Theory?
(A) because Hitler was a great leader
(B) because Hitler had organized the Olympic event in Berlin
(C) because Hitler had told his performers were members of a ‘master race’
(D) All the above
Answer:
(C) because Hitler had told his performers were members of a ‘master race’

Question 15.
What did Jessie Owens determine?
(A) to respect Hitler’s thoughts
(B) became nervous to know Hitler’s attitude
(C) promised to show the leader and his master race who was superior and who wasn’t
(D) None of the above
Answer:
(C) promised to show the leader and his master race who was superior and who wasn’t

Question 16.
What does an angry athlete do?
(A) An angry athlete easily wins the match
(B) An angry athlete becomes a looser
(C) An angry athlete makes mistakes
(D) All the above
Answer:
(C) An angry athlete makes mistakes

Question 17.
Why was Jessie Owens disqualified in his first two trials?
(A) He was nervous.
(B) He was afraid of Hitler.
(C) He jumped from several inches beyond the take-off board for a no-jump.
(D) He could not understand the rule.
Answer:
(C) He jumped from several inches beyond the take-off board for a no-jump.

Question 18.
Jessie Owens could not clear two of the three long jump trials because he
(A) was nervous
(B) was over-confident
(C) was angry over the ‘master race’ theory of Hitler
(D) feared that Luz Long might defeat him
Answer:
(C) was angry over the ‘master race’ theory of Hitler

Question 19.
The important thing in Olympics is
(A) taking part
(B) playing tricks
(C) giving trials
(D) All the above
Answer:
(A) taking part

Question 20.
The essential thing in life is
(A) conquering
(B) earning money
(C) fighting well
(D) winning prize
Answer:
(C) fighting well

Question 21.
Who is referred as Der Fuhrer?
(A) Luz Long
(B) Jessie Owens
(C) Hitler
(D) None of the above
Answer:
(C) Hitler

Question 22.
Jessie Owens was
(A) an American Negro
(B) an Australian
(C) a German
(D) a swimmer
Answer:
(A) an American Negro

Question 23.
The motto of the Olympics is
(A) Slow and steady wins the race
(B) Participation is more important than winning
(C) Faster, Higher, Stronger
(D) Winning is more important than participation
Answer:
(B) Participation is more important than winning

Question 24.
Luz Long, the German athlete had
(A) a dull face
(B) a strikingly handsome, chiseled face
(C) a tanned face
(D) a dusky complexion
Answer:
(B) a strikingly handsome, chiseled face

Question 25.
Luz Long suggested Owens to
(A) draw a line a few inches in the back of the board and then take off
(B) run fast
(C) not to participate in the finals
(D) foul in the last attempt
Answer:
(A) draw a line a few inches in the back of the board and then take off

Question 26.
Jessie Owens considers his friendship with Luz Long as a
(A) 18-carat friendship
(B) 22-carat friendship
(C) 24-carat friendship
(D) 25-carat friendship
Answer:
(C) 24-carat friendship

Question 27.
The founder the Modem Olympic Games is
(A) Bill Gates
(B) MalalaYousafzae
(C) Pierre de Coubertin
(D) Mahatma Gandhi
Answer:
(C) Pierre de Coubertin

Question 28.
Luz Long was schooled in
(A) an International English medium school
(B) Nazi Youth Movement
(C) an urban school in Germany
(D) none of the above
Answer:
(B) Nazi Youth Movement

Question 29.
The two friends talked for two hours on
(A) the political situation of Germany
(B) about Hitler’s behavior
(C) about track and field, the world situation, and a dozen other things
(D) all the above
Answer:
(C) about track and field, the world situation, and a dozen other things

Question 30.
What helped Owens qualifying for the finals?
(A) Long’s true and comforting words
(B) His anger for Hitler
(C) His determination
(D) Long qualifying for the finals easily
Answer:
(A) Long’s true and comforting words

Question 31.
Where did Owens walk over to that night?
(A) To the Olympic ground
(B) To the Olympic village
(C) Luz Long’s room
(D) To his coach’s quarters
Answer:
(C) Luz Long’s room

Question 32.
How long did Owens and Long talk?
(A) For an hour
(B) For two hours
(C) For few hours
(D) Till morning
Answer:
(B) For two hours

Question 33.
Owens and Luz Long didn’t talk about _____________.
(A) track and fields
(B) themselves
(C) the world situation
(D) other athletes
Answer:
(D) other athletes

Question 34.
What did Owens know Luz wanted him to do?
(A) Give his best
(B) Let him win
(C) Try to beat him
(D) Participate in the games
Answer:
(A) Give his best

Question 35.
Luz long wanted Owens to give his best, even if that meant _____________.
(A) Owen’s win
(B) Proving the Aryan supremacy theory wrong
(C) Owen’s defeat
(D) Hitler getting angry
Answer:
(A) Owen’s win

Question 36.
Who broke his own past record?
(A) Luz Long
(B) Jesse Owens
(C) Both Long and Owens
(D) None of them
Answer:
(A) Luz Long

Question 37.
Luz Long breaking his own past record pushed Owens on to _____________.
(A) difficult situation
(B) peak performance
(C) annoying situation
(D) breaking his own record
Answer:
(B) peak performance

Question 38.
What was the Olympic record set by Owens?
(A) 26 feet 8 1/4 inches
(B) 28 feet 61/4 inches
(C) 26 feet 5 1/4 inches
(D) 28 feet 8 1/4 inches
Answer:
(C) 26 feet 5 1/4 inches

Question 39.
How far were the stands where Hitler was glaring at the two athletes?
(A) Less than a hundred yards
(B) A hundred meters
(C) Less than a hundred meters
(D) A hundred inches
Answer:
(A) Less than a hundred yards

Question 40.
Who was/were by the narrator’s side congratulating him for the win?
(A) Jesse Owens
(B) Adolf Hitler
(C) Luz Long
(D) Other American athletes
Answer:
(C) Luz Long

Question 41.
What was the greatest Olympic prize for Jesse Owens?
(A) Setting the Olympic record
(B) Proving Hitler wrong
(C) Beating Hitler’s best athlete
(D) The friendship of Luz Long
Answer:
(D) The friendship of Luz Long

Question 42.
Who is the father of the modem Olympic games?
(A) Jesse Owens
(B) Pierre de Coubertin
(C) Luz Long
(D) Adolf Hitler
Answer:
(B) Pierre de Coubertin

Question 43.
What according to Coubertin is the most important thing in the Olympic Games?
(A) Winning
(B) Participating
(C) Making friends
(D) Setting world records
Answer:
(B) Participating

Question 44.
Coubertin said that the most important thing in life is not conquering but _____________.
(A) participating
(B) playing with a friendly spirit
(C) helping each other in need
(D) fighting well
Answer:
(D) fighting well

Question 45.
Who was/were the epitome of Coubertin’s ideal?
(A) Jesse Owens
(B) Luz Long
(C) The Olympic participants
(D) German athletes
Answer:
(B) Luz Long

Question 46.
Which of the following is not an adverb?
(A) Easily
(B) Bitterly
(C) Physically
(D) Silly
Answer:
(D) Silly

Question 47.
He had kept his plans _____________.
(A) hands down
(B) hot under collars
(C) under secret
(D) under wraps
Answer:
(D) under wraps

Question 48.
Tendulkar’s double century is the _____________ individual score in a one-day cricket match.
(A) all-time highest
(B) all-time high
(C) all-time best
(D) all-time record
Answer:
(B) all-time high

Question 49.
You’d not believe he was a security guard, he doesn’t _____________.
(A) appear like that
(B) seem like that
(C) look that part
(D) look the part
Answer:
(D) look the part

Question 50.
He was expected to win the match very easily. (Replace the itallic portion with a suitable idiomatic expression).
(A) under hands
(B) hands down
(C) hands up
(D) under wraps
Answer:
(B) hands down

Question 51.
He has always been very angry with the ways of his neighbor. [Replace the bold word with a suitable idiomatic expression]
(A) on guards
(B) hot-headed
(C) red under the hands
(D) hot under the collar
Answer:
(D) hot under the collar

Question 52.
Which of the following means “to behave in a very silly way”?
(A) Have one’s eyes on
(B) To be in for a surprise
(C) Hot under the collar
(D) Make a fool of oneself
Answer:
(D) Make a fool of oneself

Question 53.
Owens’ had his _____________the long jump.
(A) hands down
(B) eyes on
(C) wraps under
(D) eyes at
Answer:
(B) eyes on

Question 54.
He doesn’t know that he is _____________when he reaches home.
(A) making fool of himself
(B) no exception
(C) little hot under the collar
(D) in for a surprise
Answer:
(D) in for a surprise

Question 55.
All his tension seemed to _____________.
(A) get out
(B) take out
(C) go out
(D) ebb out
Answer:
(A) get out

II. Short Type Questions with Answers:

Question 1.
Why were nationalistic feelings running high during the 1936 Summer Olympics in Berlin?
Answer:
Nationalistic feelings were running high during the 1936 Summer Olympics in Berlin because of Hitler’s Nazi theory that Germans were superior to all other races.

Question 2.
How did Luz Long push the narrator on to setting the Olympic record?
Answer:
Luz Long went out to the field the next day trying to beat Owens if he could. But Owens knew that Luz Long wanted him to do his best even if that meant his winning. As it turned out, Luz broke his own past record. In doing so he pushed the narrator on to setting the Olympic record, the peak of performance.

Question 3.
Why did everyone expect Owens to win the long jump easily?
Answer:
Everyone expected Owens to win the long jump easily, because, a year before the advent of the Berlin Olympic Games, he, as a university student, had established the world record of 26 feet 8\ inches (8.13 meters).

Question 4.
What was the surprise that awaited Jesse Owens in Berlin?
Answer:
The surprise that awaited Jesse Owens was a tall German boy, Luz Long’s amazing performance of hitting the pit at almost 26 feet on his practice jumps.

Question 5.
What made Owens determined to beat Luz Long?
Answer:
The fact that made Owens determined to beat Luz Long was that he was a Negro and against this backdrop, he would disprove Hitler’s Aryan superiority theory.

Question 6.
What does a coach say about an angry athlete?
Answer:
A coach says that an angry athlete will commit mistakes. In other words, he says that anger is an athlete’s worst enemy.

Question 7.
What were the results of the first two qualifying jumps for Owens?
Answer:
The results of the first two qualifying jumps for Owens were miserable. He jumped from several inches outside the take-off board for a no-jump.

Question 8.
How did Luz Long help Jesse Owens in qualifying for the final jumps?
Answer:
Luz Long helped Jesse Owens in qualifying for the final jumps by advising him to draw a line a few inches at the back of the take-off board and focussing on his start from there.

Question 9.
Did Owens qualify for the final jump? How did he do that?
Answer:
Thanks to Long’s friendly advice, Owens qualified for the final jump. Brimming with confidence, he drew a line a full foot behind the board and advanced to jump from there and qualified for the final jump.

Question 10.
When did Owens and Long realize that they had become friends?
Answer:
Owens and Long realized that they had become friends after the former went to the latter’s room and talked for two hours concerning track and field, themselves, the global scenario, and a dozen other topics.

Question 11.
What did they discuss in Luz Long’s room in the Olympic village?
Answer:
They discussed in Luz Long’s room in the Olympic village for two hours about track and field, themselves, the world situation, and a dozen of other things.

Question 12.
When did Owens and Long realize that they had become friends?
Answer:
After discussing a lot of things like the track, and field, the world situation, etc. in Luz Long’s room in the Olympic village, Owens finally got up to leave, and they both knew that a real friendship had been formed.

Question 13.
Who was Coubertin? What was his ideal?
Answer:
Coubertin was the founder of the Modem Olympic Games. His ideal was ‘The important thing in the Olympic Games is not winning but taking part.

Question 14.
Why has Luz Long been called a fine example of Coubertin’s ideal?
Answer:
Luz Long has been called a fine example of Coubertin’s ideal because he proved this by supporting Owens who is his immediate rival in the games when he was disturbed. He was a real hero.

Question 15.
Throw light on Hitler’s theory of the ‘master race’.
Answer:
Hilter’s theory of ‘master race’ states that the Germans belonged to the Aryan race that cut other peoples to size. There was a tinge of arrogance about his tone.

Question 16.
“I wasn’t too worried about all this.” What did ‘this’ signify here?
Answer:
‘This’ signified the fact that Owens was not bothered about Hitler’s slogan of Aryan superiority which gave rise to unprecedented nationalistic feelings.

Introducing the Author:
James Cleveland “Jesse” Owens (1913-1980), an American track and field athlete, is an icon in the world of sports. In 1936, Owens arrived in Berlin to compete for the United States in the Summer Olympics. Adolf Hitler was using the games to show the world a resurgent Nazi Germany. He and other government officials had high hopes that German athletes would dominate the games with victories (the German athletes achieved a “top of the table” medal haul). Meanwhile, Nazi propaganda promoted concepts of “Aryan racial Superiority” and depicted ethnic Africans as inferior.

Owens’ surprised many by winning four- gold medals: On August 3, 1936, he won the 100 m sprint, defeating Ralph Metcalfe; on August 4, the long jump (later crediting friendly and helpful advice from Luz Long, the German competitor he ultimately defeated), on August 5, the 200 m sprint; and after he was added to the 4 x 100 m relay team, following a request by the Germans to replace a Jewish-American sprinter, he won his fourth on August 9, a performance not equaled until Carl Lewis, won gold medals in the same events at the 1984 Summer Olympics. These four gold, medals made Jesse Owens globally famous. In 1955, President Dwight D. Eisenhower honored Owens by naming him ‘an Ambassador of sports’.

About the Topic:
In this essay, Jesse Owens gives vent to his experiences of the 1936 Summer Olympics held in Berlin. Nationalistic feelings were running high in Germany. However, Owens was not worried at all. He was endowed with an unflinching faith in his abilities. Owens set a world record in the long jump defeating the famous German Athlete Luz Long. This essay also deals with Owens’ lasting friendship with him and the spirit of the Olympic Games.

Summary:
The writer takes us back to the summer of 1936 when the Olympic Games took place in Berlin. Adolf Hitler’s slogan of ‘Aryan racial superiority’ sparked intense patriotic feelings. However, Owens was unmoved. He had shed blood, sweat, and tears for the last six years for this moment. He was keen on winning the gold medal, especially in the long jump. Everyone expected him to come out successful in that final event quite easily. A great surprise was in store for Owens.

He noticed a tall German boy named Luz Long perform an amazing performance on his practice leaps. He learned from people that Hitler had kept him secret. The Nazi leader hoped Luz Long to win the jump. Owens was a Negro. Hitler’s theory that Germans were superior to Negroes filled him with anger. Owens was determined to cut Hitler’s vanity to size. Anger had an adverse effect on Owens. The first two of his three qualifying jumps were a dismal failure. His setback in the trial disgusted him. Bitterness gripped him.

To – his stunned disbelief, Luz Long came to Owens and talked to him in a cordial manner. He understood that the American athlete was angry. He frankly said that he did not believe in Aryan supremacy. Luz Long had a lean, muscular frame, clear blue eyes, fair hair, and an impressively handsome face. He saw that Owens’ anger had abated. Lung advised him to draw a line a few inches at the back of the board and focus on his start from there. His advice worked wonders. Owens qualified for the final jump.

That night Owens met Luz Long in his room in the Olympic village to thank him for his timely advice. Their two-hour talk embraced so many things. They were bound by a genuine friendship. The moment everyone had waited for came at last. Luz surpassed his own record. His spectacular feat compelled Owens to give his best performance. And he set the world record of 26 feet 5 inches in his final jump. Despite Hitler’s angry look at them, Luz congratulated Owens and warmly shook his hand with a sweet smile.

‘Owens’ feeling for Luz was indefinable at that moment. In short, the greatest Olympic v prize for Jesse Owens was not the gold medal he won in the long jump, but the friendship > he established with Luz Long. Owens states that Luz Long was a perfect example of an athlete as epitomized by the philosophy of Pierre de Coubertin, founder of the modern Olympic Games. To Coubertin, participation in the Olympics Games is more important than victory. Besides, the essence of life lies not in conquering but in fighting in the right spirit.

ସାରାଂଶ:
ଲେଖକ ଆମ୍ଭମାନଙ୍କୁ ୧୯୩୬ ମସିହା ଗ୍ରୀଷ୍ମଋତୁରେ ବର୍ଲିନ୍‌ଠାରେ ଅନୁଷ୍ଠିତ ହୋଇଥ‌ିବା ଅଲିମ୍ପିକ୍ କ୍ରୀଡ଼ାର ପୃଷ୍ଠଭୂମିକୁ ନେଇ ଯାଇଛନ୍ତି । ଆଡ଼ଲଫ୍ ହିଟ୍‌ଲର୍‌ଙ୍କ ଆର୍ଯ୍ୟ-ଶ୍ରେଷ୍ଠତ୍ଵ ପ୍ରଚାରବାଣୀ ପ୍ରବଳ ଦେଶପ୍ରେମ ଭାବନା ସୃଷ୍ଟି କରିଥିଲା । ମାତ୍ର ଏହା ଓୟେସଙ୍କୁ ପ୍ରଭାବିତ କରିନଥିଲା । ଏହି ମୁହୂର୍ତ୍ତ ପାଇଁ ସେ ସ୍ବେଦ, ଅଶ୍ରୁ, ରକ୍ତକଣିକା ଦେଇ ବିଗତ ୬ ବର୍ଷ ଧରି ନିଜକୁ ପ୍ରସ୍ତୁତ କରିଥିଲେ । ବିଶେଷତଃ ଲମ୍ବଡ଼ିଆଁରେ ସ୍ବର୍ଣ୍ଣପଦକ ଜିଣିବାପାଇଁ ସେ ନିଶ୍ଚିତ ଥିଲେ । ସେ ଅନ୍ତିମ ପର୍ଯ୍ୟାୟରେ ସହଜରେ ସଫଳ ହେବେ ବୋଲି ସମସ୍ତେ ଆଶା କରିଥିଲେ । ଗୋଟିଏ ବିରାଟ ବିସ୍ମୟ ଓୟେସ୍‌ଙ୍କ ପାଇଁ ଅପେକ୍ଷା କରି ରହିଥିଲା । ଜଣେ ଡେଙ୍ଗା ଜର୍ମାନ୍ ବାଳକର ଅଭ୍ୟାସ ଡିଆଁରେ ବିସ୍ମୟକର କୃତିତ୍ଵ ସେ ଦେଖିବାକୁ ପାଇଲେ । ହିଟ୍‌ଲର୍ ତାଙ୍କୁ ଗୋପନୀୟଭାବେ ରଖିଥ‌ିବାର ସେ ଲୋକମାନଙ୍କଠାରୁ ଜାଣିବାକୁ ପାଇଲେ ।

ନାଜି ନେତାଜଣକ ଲୁଜ୍ ଲଙ୍ଗ ଲମ୍ବଡିଆରେ ପଦକ ଜିତିବେ ବୋଲି ଆଶା କରୁଥିଲେ । ଓୟେସ୍ ଜଣେ ନିଗ୍ରୋ ଥିଲେ । ‘ଜର୍ମାନ୍‌ମାନେ ନିଗ୍ରୋମାନଙ୍କଠାରୁ ଉତ୍କୃଷ୍ଟ’ – ହିଟଲର୍‌ଙ୍କର ଏହି ସିଦ୍ଧାନ୍ତ ତାଙ୍କ ମନରେ କ୍ରୋଧ ସୃଷ୍ଟି କରିଥିଲା । ସେ ହିଲର୍‌ଙ୍କର ବୃଥା ଗର୍ବକୁ ଖର୍ଚ କରିବାକୁ ନିଶ୍ଚୟ କଲେ । କ୍ରୋଧ ଓୟେସଙ୍କ ଉପରେ ପ୍ରତିକୂଳ ପ୍ରଭାବ ପକାଇଲା । ତାଙ୍କର ଯୋଗ୍ୟତା ପର୍ଯ୍ୟାୟ ପ୍ରଥମ ତିନୋଟି ଡିଆଁ ମଧ୍ୟରୁ ପ୍ରଥମ ଦୁଇଟି ଦୟନୀୟ ଭାବେ ବିଫଳ ହେଲା । ଯୋଗ୍ୟତା ପର୍ଯ୍ୟାୟର ବିଫଳତା ତାଙ୍କୁ ଅସନ୍ତୁଷ୍ଟ କଲା । ତିକ୍ତତା ତାଙ୍କୁ ଜାବୁଡ଼ି ଧରିଲା । ତାଙ୍କ ପାଇଁ ଅବିଶ୍ଵାସ୍ୟ ମନେ ହେଉଥିଲେ ହେଁ ଲୁଜ ଲଙ୍ଗ ଓୟେସଙ୍କ ନିକଟକୁ ଆସି ଆନ୍ତରିକ ଭାବେ କଥାବାର୍ତ୍ତା କଲେ । ଆମେରିକାନ୍‌ କ୍ରୀଡ଼ାବିତ୍ ଜଣକ ରାଗି ଯାଇଛନ୍ତି ବୋଲି ସେ ବୁଝିପାରିଥିଲେ । ସେ ‘ଆର୍ଯ୍ୟ- ଶ୍ରେଷ୍ଠତ୍ୱ’ ସିଦ୍ଧାନ୍ତକୁ ବିଶ୍ଵାସ କରନ୍ତି ନାହିଁ ବୋଲି ସ୍ପଷ୍ଟଭାବେ ପ୍ରକାଶ କଲେ ।

ଲୁଜ୍ ଲଙ୍ଗଙ୍କର ପତଳା, ମାଂସପେଶୀ ବହୁଳ ଚେହେରା, ଉଜ୍ଜ୍ଵଳ ନୀଳ ଆଖୁ, ସୁନ୍ଦର କେଶ ଓ ଅନ୍ୟମାନଙ୍କୁ ପ୍ରଭାବିତ କରୁଥ‌ିବା ଭଳି କମନୀୟ ମୁଖମଣ୍ଡଳ ଥିଲା । ଓୟେସଙ୍କ କ୍ରୋଧ ପ୍ରଶମିତ ହେଉଥବାର ସେ ଦେଖ‌ିଲେ । ବୋର୍ଡର କିଛି ଇଞ୍ଚ୍ ପଛରେ ଏକ ଗାର ଟାଣି ସେଠାରେ ଧ୍ୟାନ କେନ୍ଦ୍ରୀଭୂତ କରି ଡେଇଁବାକୁ ଲୁଜ୍ ଲଙ୍ଗ୍ ତାଙ୍କୁ ଉପଦେଶ ଦେଲେ । ତାଙ୍କର ଉପଦେଶ ବିସ୍ମୟକରଭାବେ କାମ କଲା । ଶେଷ ଡିଆଁ ପାଇଁ ଓୟେ ଯୋଗ୍ୟ ବିବେଚିତ ହେଲେ । ସେହି ରାତିରେ ଓୟେନ୍ସ ଲୁଜ୍ ଲଙ୍ଗ୍ଙ୍କୁ ତାଙ୍କ ଅଲିମ୍ପିକ୍ ଗ୍ରାମସ୍ଥିତ ପ୍ରକୋଷ୍ଠରେ ସାକ୍ଷାତ୍ କରି ସମେୟାପଯୋଗୀ ଉପଦେଶ ନିମନ୍ତେ ଧନ୍ୟବାଦ ଜ୍ଞାପନ କଲେ । ସେମାନଙ୍କ ଦୁଇଘଣ୍ଟାର କଥୋପକଥନ ଅନେକ ବିଷୟ ଉପରେ ପର୍ଯ୍ୟବସିତ ଥିଲା । ସେମାନେ ମିତ୍ରତା ବନ୍ଧନରେ ଆବଦ୍ଧ ହୋଇଗଲେ । ସମସ୍ତଙ୍କ ଅପେକ୍ଷିତ ମୁହୂର୍ତ୍ତ ଶେଷରେ ଉପସ୍ଥିତ ହେଲା। ଲୁଜ୍ ତାଙ୍କ ପୂର୍ବ ରେକର୍ଡ ଭଙ୍ଗ କଲେ ।

ତାଙ୍କର ଦର୍ଶନୀୟ ଲମ୍ଫ ଓୟେସଙ୍କୁ ତାଙ୍କ ସର୍ବଶ୍ରେଷ୍ଠ କୃତିତ୍ୱ ପାଇଁ ବାଧ୍ୟ କଲା । ସେ ଶେଷ ଡିଆଁରେ ୨୬ ଫୁଟ୍ ୫% ଇଞ୍ଚ ଡେଇଁ ବିଶ୍ଵ ରେକର୍ଡ ପ୍ରତିଷ୍ଠା କଲେ । ହିଟ୍‌ଲର୍‌ଙ୍କ କ୍ରୋଧପୂର୍ଣ୍ଣ ଚାହାଣି ସତ୍ତ୍ବେ ଲୁଜ୍ ଲଙ୍ଗ ମିଠା ହସ ହସି ତାଙ୍କ କରମର୍ଦ୍ଦନ କଲେ । ସେହି ସମୟରେ ଲୁଜ୍‌ଙ୍କ ପ୍ରତି ତାଙ୍କ ମନର ଭାବନା ଅବର୍ଣ୍ଣନୀୟ ଥିଲା । ସଂକ୍ଷେପରେ କହିବାକୁ ଗଲେ, ଲମ୍ବଡିଆଁରେ ଜିତିଥିବା ସ୍ଵର୍ଣ୍ଣପଦକ ନୁହେଁ, ବରଂ ଲୁଜ୍ ଲଙ୍ଗଙ୍କ ସହ ସ୍ଥାପିତ ବନ୍ଧୁତ୍ବ ତାଙ୍କ ଶ୍ରେଷ୍ଠ ଅଲିମ୍ପିକ୍ ପୁରସ୍କାର ଥିଲା । ଓୟେନ୍ସ କହିଛନ୍ତି ଯେ ଆଧୁନିକ ଅଲିମ୍ପିକ କ୍ରୀଡ଼ାର ପ୍ରତିଷ୍ଠାତା ପେରୀ ଡି କୁବରଟିନ୍ ଚିନ୍ତାଧାରା ଅନୁସାରେ ଲୁଜ୍ ଲଙ୍ଗ ଜଣେ କ୍ରୀଡ଼ାବିତ୍‌ ପ୍ରକୃଷ୍ଟ ଉଦାହରଣ ଥିଲେ । କୁବରଟିନ୍‌ଙ୍କ ମତରେ, ଅଲିମ୍ପିକ୍ କ୍ରୀଡ଼ାରେ ବିଜୟୀ ହେବା ଅପେକ୍ଷା ଅଂଶଗ୍ରହ କରିବା ଅଧୂକ ଗୁରୁତ୍ଵପୂର୍ଣ୍ଣ । ଏତଦ୍‌ବ୍ୟତୀତ ଜୀବନର ମହତ୍ତ୍ଵ କେବଳ ବିଜୟପ୍ରାପ୍ତି ଉପରେ ନୁହେଁ, ବରଂ ଉତ୍ତମ ମନୋଭାବ ନେଇ ସଂଘର୍ଷ କରିବା ଉପରେ ପର୍ଯ୍ୟବସିତ ।

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 8 Application of Derivatives Ex 8(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 8 Application of Derivatives Exercise 8(a)

Question 1.
Find the velocity and acceleration at the end of 2 seconds of the particle moving according to the following rules.
(i) s = 2t² + 3t + 1
Solution:
s = 2t² + 3t + 1

∴ Velocity is 11 units/sec and acceleration is 4 units/sec².

(ii) s = √t +1
Solution:
s = √t +1

(iii) s = \(\frac{3}{2 t+1}\)
Solution:

(iv) s = t³ – 6t² + 15t + 12
Solution:
s = t³ – 6t² + 15t + 12

∴ Velocity is 3 units/sec and acceleration is 0.

Question 2.
The sides of an equilateral triangle are increasing at the rate of √3 cm/sec. Find the rate at which the area of the triangle is increasing when the side is 4 cm long.
Solution:
Let x be the lenght of each side of an equilateral triangle.

∴ Area of the triangle is increasing at the rate of 6 cm²/sec

Question 3.
Find the rate at which the volume of a spherical balloon will increase when its radius is 2 metres if the rate of increase of its radius is 0.3 m/min.
Solution:
Let r be the radius of a spherical balloon.

∴ The volume increase at the rate of 4.8π m³/min.

Question 4.
The surface area of a cube is decreasing at the rate of 15 sq. cm/sec. Find the rate at which its edge is decreasing when the length of the edge is 5 cm.
Solution:
Let s be the surface area of a cube.
Let x be the length of each side of the cube.

∴ The edge is decreasing at the rate of 0.25 cm/sec

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

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

Question 1.
Verify Rolle’s theorem for the function
f(x) = x (x – 2)², 0 ≤ x ≤ 2.
Solution:
f(x) = x (x – 2)², 0 ≤ x ≤ 2
Here a = 0, b = 2
f(x) is a polynomial function hence it is continuous and also differentiable.
∴ f is continuous on [0. 2]
f is differentiable on (0, 2)
f(0) = 0 = f(2)
Thus conditions of Rolle’s theorem are satisfied.
f'(x) = (x – 2)² + 2x (x – 2)
= (x – 2) (x – 2 + 2x)
= (x – 2) (3x – 2)
f'(x)= 0 ⇒ x = 2, x = \(\frac{2}{3}\)
But x = 2 ∉ (0, 2). Thus c = \(\frac{2}{3}\) such that f'(c) = 0
Thus Rolle’s theorem is verified.

Question 2.
Examine if Rolle’s theorem is applicable to the following functions:
(i) f(x) = |x| on [-1, 1]
Solution:
f(x) = |x| on [-1, 1]
As f(x) = |x| is not differentiable at x = 0 ∈ (-1, 1)
We have Rolle’s theorem is not applicable.

(ii) f(x) = [x] on [-1, 1]
Solution:
f(x) = [x] on [-1, 1]
f(x) = [x] is not continuous at 0 ∈ [-1, 1]
Rolle’s theorem is not applicable.

(iii) f(x) = sin x on [0, π]
Solution:
f(x) = sin x on [0, π]
f is a trigonometric function hence continuous and differentiable on its domain.
∴ f is continuous on [0, π]
f is differentiable on (0, π]
f(0) = f(π)
Thus Rolle’s theorem is applicable for f(x) = sin x on [0, π]

(iv) f(x) = cot x on [0, π]
Solution:
f(x) = cot x on [0, π]
Clearly cot (0) and cot (π] are not defined hence Rolle’s theorem is not applicable.

Question 3.
Verify Lagrange’s Mean-Value theorem for
F(x) = x³ – 2x² – x + 3 on [1, 2]
Solution:
f(x) = x³ – 2x² – x + 3 on [1, 2]
f is a polynomial function hence continuous as well as differentiable.
∴ f is continuous on [1, 2]
f is differentiable on (1, 2)
Thus Largange’s mean value theorem is applicable.
Now f(1) = 1 – 2 – 1 + 3 = 1
f(2) = 8 – 8 – 2 + 3 = 1
∴ f(x) = 2x² – 4x – 1

Thus Lagrange’s mean value theorem is verified.

Question 4.
Test if Lagrange’s mean value theorem holds for the functions given in question no. 2.
Solution:
(i) f(x) = |x| is not differentiable at x = 0 ∈ (-1, 1)
Thus Lagrange’s mean value theorem, does not hold.

(ii) f(x) = [x] is discontinuous at 0 ∈ [-1, 1]
Thus Lagrange’s mean value theorem is not applicable.

(iii) f(x) = sin x is a trigonometric function, which is continuous as well as differentiable in its domain.
∴ f is continuous on [0, π]
f is differentiable on (0, π)
Thus conditions of Lagrange’s mean value theorem are satisfied.
Hence mean value theorem is applicable.

(iv) f(x) = cot x
Which is undefined x = 0 and x = π
Thus Lagrange’s mean value theorem is not applicable.

Question 5.
(Not for examination) Verify Cauchy’s mean value theorem for the functions x² and x³
in [1, 2].
Solution:
Let f(x) = x², and g(x) = x³ on [1, 2]
Both f and g are polynomial functions, hence continuous and differentiable.
∴ f and g are continuous on [1, 2]
f and g are differentiable on (1, 2)
g'(x) = 3x² ≠ 0 ∀ x ∈ (1, 2)
Thus conditions of Cauchy’s mean value theorem are satisfied.
Now f(1) = 1, f(2) = 4, g(1) = 1, g(2) = 8
f'(x) = 2, and g'(x) = 3x²

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

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

Differentiate.
Question 1.
√x w.r.t x².
Solution:

Question 2.
sin x. w.r.t. cot x.
Solution:
Let y = sin x and z = cot x

Question 3.
\(\frac{1-\cos x}{1+\cos x}\) w.r.t \(\frac{1-\sin x}{1+\sin x}\)
Solution:

Question 4.
tan-1 x w.r.t. tan^-1 \( \sqrt{1+x^2} \)
Solution:

Question 5.
sin^-1 \(\left(\frac{2 x}{1+x^2}\right)\) w.r.t cos^-1 \(\left(\frac{1-x^2}{1+x^2}\right)\)
Solution:
Let y = sin^-1 \(\frac{2 x}{1+x^2}\) and z = cos^-1 \(\frac{1-x^2}{1+x^2}\)
Then y = 2 tan^-1 x and z = 2 tan^-1 x
So y = z
∴ \(\frac{d y}{d z}\) = 1.

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 6 Probability Ex 6(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 6 Probability Exercise 6(c)

Question 1.
There are 3 bags B₁, B₂ and B₃ having respectively 4 white, 5 black; 3 white, 5 black and 5 white, 2 black balls. A bag is chosen at random and a ball is drawn from it. Find the probability that the ball is white.
Solution:
Let
E₁ = The selected bag is B₁,
E₂ = The selected bag is B₂,
E₃ = The selected bag is B₃.
A = The ball drawn is white.

Question 2.
There are 25 girls and 15 boys in class XI and 30 boys and 20 girls in class XII. If a student chosen from a class, selected at random, happens to be a boy, find the probability that he has been chosen from class XII.
Solution:
Let
E₁ = The student is choosen from class XI.
E₂ = The student is choosen from class XII.
A = The student is a boy.

Question 3.
Out of the adult population in a village 50% are farmers, 30% do business and 20% are service holders. It is known that 10% of the farmers, 20% of the business holders and 50% of service holders are above poverty line. What is the probability that a member chosen from any one of the adult population, selected at random, is above poverty line?
Solution:
Let
E₁ = The person is a farmer.
E₂ = The person is a businessman.
E₃ = The person is a service holder.
A = The person is above poverty line.

Question 4.
Take the data of question number 3. If a member from any one of the adult population of the village, chosen at random, happens to be above poverty line, then estimate the probability that he is a farmer.
Solution:
P (a farmer / he is above poverty line)
= P (E₁ | A)

Question 5.
From a survey conducted in a cancer hospital it is found that 10% of the patients were alcoholics, 30% chew gutka and 40% have no specific carcinogenic habits. If cancer strikes 80% of the smokers, 70% of alcoholics, 50% of gutka chewers and 10% of the nonspecific, then estimates the probability that a cancer patient chosen from any one of the above types, selected at random,
(i) is a smoker
(ii) is alcoholic
(iii) chews gutka
(iv) has no specific carcinogenic habits.
Solution:
The question should be modified as from a survey conducted in a cancer hospital, 10% patients are smokers, 20% alcoholics, 30% chew gutka and 40% have no specific carcinogenic habits. If cancer strikes 80% of the smokers, 70% of alcoholic, 50% of gutka chewers and 10% of non specific, then estimate the probability that a cancer patient chosen from any one of the following types selected at random
(i) is a smoker
(ii) is alcoholic
(iii) chews gutka
(iv) has no specific carcinogenic habits.
Let
E₁ = The person is a smoker.
E₂ = The person is alcoholic.
E₃ = The person chew Gutka.
E₄ = The person have no specific habits.
A = The person is a cancer patient.
According to the question

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 6 Probability Ex 6(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 6 Probability Exercise 6(b)

Question 1.
A bag contains 5 white and 3 black marbles and a second bag contains 3 white and 4 black marbles. A bag is selected at random and a marble is drawn from it. Find the probability that it is white. Assume that either bag can be chosen with the same probability.
Solution:
A bag contains 5 white and 3 black marbles and a second bag contains 3 white and 4 black marbles. A bag is selected at random and a marble is drawn from it.
Let W₁ be the event that 1st bag is choosen and a white marble is drawn and let W₂ be the event that 2nd bag is choosen and a white marble is drawn and these two events are mutually exclusive.

Question 2.
A bag contains 5 white and 3 black balls; a second bag contains 4 white and 5 black balls; a third bag contains 3 white and 6 black balls. A bag is selected at random and a ball is drawn. Find the probability that the ball is black.
(i) Do the problem assuming that the probability of choosing each bag is same.
(ii) Do the problem assuming that the probability of choosing the first bag is twice as much as choosing the second bag, which is twice as much as choosing the third bag.
Solution:
A bag contains 5 white and 3 black balls, a 2nd bag contains 4 white and 5 black balls, a 3rd bag contains 3 white and 6 black balls. A bag is selected at random and a ball is drawn.

(i) Let B₁, B₂, B₃ be the events that 1st bag is choosen and a black ball is draw. 2nd bag is choosen and a black ball is drawn, 3rd bag is drawn and a black ball is drawn. These events are mutually exclusive.
Probability of drawing a black ball
= P(B₁) + P(B₂) + P(B₃)
= \(\frac{1}{3}\) × \(\frac{3}{8}\) + \(\frac{1}{3}\) × \(\frac{5}{9}\) + \(\frac{1}{3}\) × \(\frac{6}{9}\) = \(\frac{115}{216}\)

(ii) Let the probability of choosing 1st bag be 4x. The probability of choosing the 2nd bag is 2x and that of 3rd bag is x.
It is obvious that probability of choosing 3 bags = 1
4x + 2x + x = 1 or, 7x = 1.
x = \(\frac{1}{7}\)
Probability of drawing a black ball
= \(\frac{4}{7}\) × \(\frac{3}{8}\) + \(\frac{2}{7}\) × \(\frac{5}{9}\) + \(\frac{1}{7}\) × \(\frac{6}{9}\)
= \(\frac{108+80+48}{8 \times 9 \times 7}\) = \(\frac{236}{8 \times 9 \times 7}\) = \(\frac{59}{126}\)

Question 3.
A and B play a game by alternately throwing a pair of dice. One who throws 8 wins the game. If A starts the game, find their chances of winning.
Solution:
A and B play a game by alternately throwing a pair of dice. One who throws 8 wins the game. A starts the game.
We can obtain 8 as follows:
{(6, 2), (5, 3), (4, 4) (3, 5), (6, 2)}.
∴ |S| = 6² = 36
∴ P(B) = \(\frac{5}{36}\)
⇒ P(not 8) = 1 – \(\frac{5}{36}\) = \(\frac{31}{36}\)
Since A starts the game, A can win the following situations.
(i) A throws 8
(ii) A does not throws, B does not throw 8, A throws 8,
(iii) A does not throw 8, B does not throw 8, A does not throw 8, B does not throw 8, A throws 8, etc.

Question 4.
A, B, C play a game by throwing a pair of dice in that order. One who gets 8 wins the game. If A starts the game, find their chances of winning.
Solution:
A, B, C play a game by throwing a pair of dice in that order. One who gets 8 wins the game and A starts the game.
P(B) = \(\frac{5}{36}\), P(not 8) = \(\frac{31}{36}\)

If A starts the game, then
(i) A throws 8.
(ii) A does not throw 8, B does not throw 8, C does not throw 8, A throws 8.
(iii) A does not throw 8, B does not throw 8, C does not throw 8, A does not throw 8, B does not throw 8, C does not throw 8, A throws 8, etc.

Similarly, If B wins the game, then
(i) A does not throw 8, B throw 8.
(ii) A does not throw 8, B does not throw 8, C does not throw 8, A does not throw 8, B throws 8.
(iii) A does not throw 8, B does not throw 8, C does not throw 8, A does not throw 8, B does not throw 8, C does not throw 8, A does not throw 8, B throws 8, etc.

Question 5.
There are 6 white and 4 black balls in a bag. If four are drawn successively (and not replaced), find the probability that they are alternately of different colour.
Solution:
There are 6 white and 4 black balls in a bag. Four balls are drawn without replacement. Let W and B denotes the white and black ball. There are two mutually exclusive cases WBWB and BWBW.

Question 6.
Five boys and four girls randomly stand in a line. Find the probability that no two girls come together.
Solution:
Five boys and 4 girls randomly stand in a line such that no two girls come together.
|S| = 9!

The 4 girls can stand in 6 positions in 6P4 ways. Further 5 boys can stand in 5! ways.
Probability that they will stand in a line such that no two girls come together.
= \(\frac{5 ! \times{ }^6 P_4}{9 !}\) = \(\frac{5}{42}\)

Question 7.
If you throw a pair of dice n times, find the probability of getting at least one doublet. [When you get identical members you call it a doublet. You can get a double in six ways: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6); thus the probability of getting a doublet is \(\frac{6}{36}\) = \(\frac{1}{6}\), so that the probability of not getting a doublet in one throw is \(\frac{5}{6}\)].
Solution:
A pair of dice is thrown n times. We get
the doublet as (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).
Probability of getting a doublet in one throw
= \(\frac{6}{36}\) = \(\frac{1}{6}\)
Probability of not getting a doublet
= 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
If a pair of dice is thrown n-times, the probability of not getting a doublet
= \(\left(\frac{5}{6}\right)^n\)
Probability of getting atleast one doublet
= 1 – \(\left(\frac{5}{6}\right)^n\)

Question 8.
Suppose that the probability that your alarm goes off in the morning is 0.9. If the alarm goes off, the probability is 0.8 that you attend your 8 a.m. class. If the alarm does not go off, the probability that you make your 8 a.m. class is 0.5. Find the probability that you make your 8 a.m. class.
Solution:
Let A be the event that my alarm goes off and let B be the event that I make my 8 a. m. class.
Since S = a ∪ A’, B = (B ∩ A) ∪ (B ∩ A’)
Where B ∩ A and B ∩ A’ are mutually
exclusive events.
P(B) = P (B ∩ A) + P (B ∩ A’)
= P(A). P(\(\frac{B}{A}\)) + P(A’). P(\(\frac{\mathrm{B}}{\mathrm{A}^{\prime}}\))
= 0.9 × 0.8 + 0.1 × 0.5 = 0.77

Question 9.
If a fair coin is tossed 6 times, find the probability that you get just one head.
Solution:
A fair coin is tossed 6 times.
∴ |S| = 2⁶
The six mutually exclusive events are
HTTTTT, THTTTT, TTHTTT, TTTHTT, TTTTHT, TTTTTH.
Probability of getting just one head = \(\frac{6}{2^6}\)

Question 10.
Can you generalize this situation? If a fair coin is tossed six times, find the probability of getting exactly 2 heads.
Solution:
A fair coin is tossed 6 times. Let A be the event of getting exactly 2 heads.
∴ |A| = ⁶C₂ = 15
∴ P(A) = \(\frac{15}{2^6}\)
Yes we can generalize the situation, i.e., if a fair coin is tossed n-times, then probability of getting exactly 2 heads
= \(\frac{{ }^n \mathrm{C}_2}{2^n}\) = \(\frac{{ }^6 C_2}{2^6}\)

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

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

Question 1.
Examine the continuity of the following functions at indicated points.
(i) f(x) = \(\left\{\begin{array}{cl}
\frac{x^2-a^2}{x-a} & \text { if } x \neq a \\
a & \text { if } x=a
\end{array}\right.\) at x = a
Solution:

(ii) f(x) = \(\left\{\begin{aligned}
\frac{\sin 2 x}{x} & \text { if } x \neq 0 \\
2 & \text { if } x=0
\end{aligned}\right.\) at x = 0
Solution:

(iii) f(x) = \(\begin{cases}(1+2 x)^{\frac{1}{x}} & \text { if } x \neq 0 \\ e^2 & \text { if } x=0\end{cases}\) at x = 0
Solution:

(iv) f(x) = \(\left\{\begin{array}{l}
x \sin \frac{1}{x} \text { if } x \neq 0 \\
0
\end{array}\right.\) at x = 0
Solution:

(v) f(x) = \(\left\{\begin{array}{l}
\frac{x^2-1}{x-1} \text { if } x \neq 1 \\
2
\end{array}\right.\) at x = 1
Solution:

(vi) f(x) = \(\begin{cases}\sin \frac{1}{x} & \text { if } x \neq a \\ 0 & \text { if } x=0\end{cases}\) at x = 0
Solution:

(vii) f(x) = [3x + 11] at x = –\(\frac{11}{3}\)
Solution:

(viii) f(x) = \(\left\{\begin{array}{l}
\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1} \text { if } x \neq 0 \\
0
\end{array}\right.\) at x= 0
Solution:

(ix) f(x) = \(\left\{\begin{array}{l}
\frac{1}{x+[x]} \text { if } x<0 \\
-1 \quad \text { if } x \geq 0
\end{array}\right.\) at x = 0
Solution:

because [- h]is the greatest integer not exceeding – h
and so [- h ] = – 1
As L.H.L. = R.H.L. = f(0)
f(x) is cntinuous at x = 0.

(x) f(x) = \(\begin{cases}\frac{|x|}{x} & \text { if } x \neq 0 \\ 0 & \text { if } x=0\end{cases}\) at x = 0
Solution:

(xi) f(x) = \(\left\{\begin{array}{l}
2 x+1 \text { if } x \leq 0 \\
x \quad \text { if } 0<x<1 \\
2 x-1 \text { if } x \geq 1
\end{array}\right.\) at x = 0, 1
Solution:

(xii) f(x) = \(\left\{\begin{array}{l}
\frac{1}{e^{\frac{1}{x}}-1} \text { if } x>0 \\
0
\end{array} \text { if } x \leq 0\right.\) at x = 0
Solution:

(xiii) f(x) = sin\(\frac{\pi[x]}{2}\) at x = 0
Solution:

(xiv) f(x) = \(\frac{g(x)-g(1)}{x-1}\) at x = 1
Solution:
g(x) = |x – 1|
Then g(1) = |1 – 11| = 0
Now f(1) = \(\frac{g(1)-f(1)}{1-1}\) = 0/0
which we cannot determine.
Hence f(x) is discontinuous at x = 1.

Question 2.
If a function is continuous at x = a, then find
(i) \(\lim _{h \rightarrow 0}+\frac{1}{2}\{f(a+h)+f(a-h)\}\)
(ii) \(\lim _{h \rightarrow 0}+\frac{1}{2}\{f(a+h)-f(a-h)\}\)
Solution:

Question 3.
Find the value ofa such that the function f defined by \(\begin{cases}\frac{\sin a x}{\sin x} & \text { if } x \neq 0 \\ \frac{1}{a} & \text { if } x=0\end{cases}\)
is continuous at x = 0.
Solution:

Question 4.
If f(x) = \(\left\{\begin{array}{l}
a x^2+b \text { if } x<1 \\
1 \quad \text { if } x=1 \\
2 a x-b \text { if } x>1
\end{array}\right.\)
is continuous at x = 1, then find a and b.
Solution:
Let f(x) be continuous at x = 1
Then L.H.L. = R.H.L. = f(1)

Question 5.
Show that sin x is continuous for every real x.
Solution:
Let f(x) = sin x
Consider the point x = a, where ‘a’ is any real number.
Then f(a) = sin a

Thus L.H.L. = R.H.L. = f(a)
Hence f(x) = sin x is continuous for every real x.
(Proved)

Question 6.
Show that the function f defined by \(\left\{\begin{array}{l}
1 \text { if } x \text { is rational } \\
0 \text { if } x \text { is irrational }
\end{array}\right.\) is discontinuous ∀ ≠ 0 ∈ R.
Solution:
Consider any real point x = a
If a is rational then f(a) = 1.
Again lim_x→a+f(x) = lim_h→0f(a + h)
which does not exist because a + h may be rational or irrational
Similarly lim_x→a-f(x) does not exist.
Thus f(x) is discontinuous at any rational point. Similarly we can show that f(x) is discontinuous at any irrational point.
Hence f(x) is discontinuous for all x ∈ R
(Proved)

Question 7.
Show that the function f defined by f(x) = \(f(x)=\left\{\begin{array}{l}
x \text { if } x \text { is rational } \\
-x \text { if } x \text { is irrational }
\end{array}\right.\)
is continuous at x = 0 and discontinuous ∀ x ≠ 0 ∈ R.
Solution:
f(0) = 0
L.H.L. = lim_x→0f(x) = lim_h→0f(-h)
= \(\lim _{h \rightarrow 0} \begin{cases}-h & \text { if } h \text { is rational } \\ h & \text { if } h \text { is irrational }\end{cases}\) = 0
Similarly R.H.L. = 0
Thus L.H.L. = R.H.L. = f(0)
Hence f(x) is continuous at x = 0.
We can easily show that f(x) is discontinuous at all real points x ≠ 0.

Question 8.
Show that the function f defined by
f(x) = \(\left\{\begin{array}{l}
x \text { if } x \text { is rational } \\
0 \text { if } x \text { is irrational }
\end{array}\right.\)
is discontinuous everywhere except at x = 0.
Solution:
f(0) = 0
L.H.L. = lim_h→0f(-h)
= lim_h→0\(\begin{cases}-h & \text { if }-h \text { is rational } \\ 0 & \text { if }-h \text { is irrational }\end{cases}\)
Similarly R.H.L. = 0
Thus L.H.L. = R.H.L = f(0)
Hence f(x) is continuous at x = 0.
Let a be any real number except 0.
If a is rational then f(a) = a.
L.H.L. = lim_h→0f(a – h) which does not exist because a – h may be rational or may be irrational.
Similarly R.H.L. does not exist.
Thus f(x) is discontinuous at any rational point x – a ≠ 0.
Similarly f(x) is discontinuous at any irrational point.
Hence f(x) is discontinuous everywhere except at x = 0.
(Proved)

Question 9.
Show that f(x) = \(\begin{cases}x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0\end{cases}\) is continuous at x = 0.
Solution:
Refer to No. 1(iv) of Exercise – 7(a).

Question 10.
Prove that e^x – 2 = 0 has a solution between 0 and 1. [Hints: Use continuity of e^x– 2 and fact – 2]
Solution:

∴ f(x) is continuous in [0, 1]
f(0). f(1) = (-1) (e – 2) < 0
∴ f(x) has a zero between 0 and 1
i.e. e^x – 2 = 0 has a solution between 0 and 1

Question 11.
So that x⁵ + x +1 = 0 for some value of x between -1 and 0.
Solution:
Let f(x) = x⁵ + x + 1 and any a ∈ (-1, 0)
f(a) = a⁵ + a + 1

= -1 = f(-1)
∴ f is continuous on [-1, 0]
But f(-1) f(0) = 1 × -1 < 0
∴ f has a zero between -1 and 0
⇒ x5 + x + 1 = 0 for some value of x between -1 and 0.

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 6 Probability Additional Exercise Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 6 Probability Additional Exercise

(A) Multiple Choice Questions (Mcqs) With Answers

Question 1.
If \(\left|\begin{array}{ccc}
1+x & x & x^2 \\
x & 1+x & x^2 \\
x^2 & x & 1+x
\end{array}\right|\) = a + bx + cx² + dx³ + ex⁴ + fx⁵ then write the value of a.
(a) 0
(b) 2
(c) 1
(d) 3
Answer:
(c) 1

Question 2.
If every element of a third order determinant of value 8 is multiplied by 2, then write the value of the new determinant.
(a) 32
(b) 64
(c) 16
(d) 128
Answer:
(b) 64

Question 3.
If A is a 4 x 5 matrix and B is a matrix such that A^TB and BA^T both are defined, then write the order of B
(a) 4 x 5
(b) 1 x 5
(c) 5 x 4
(d) None of these
Answer:
(a) 4 x 5

Question 4.
If \(\left[\begin{array}{lll}
3 & 5 & 3 \\
2 & 4 & 2 \\
\lambda & 7 & 8
\end{array}\right]\) is a singular matrix, write die value of 1.
(a) λ = 2
(b) λ = 1
(c) λ = 4
(d) λ = 8
Answer:
(d) λ = 8

Question 5.
Determine the maximum value of \(\left|\begin{array}{rl}
\cos x & \sin x \\
-\sin x & \cos x-1
\end{array}\right|\)
(a) 1
(b) 2
(c) 3
(d) 0
Answer:
(b) 2

Question 6.
If \(\left[\begin{array}{cc}
x & y \\
x & \frac{x}{2}+t
\end{array}\right]\) + \(\left[\begin{array}{cc}
y & x+t \\
x+2 & \frac{x}{2}
\end{array}\right]\) = \(\left[\begin{array}{ll}
1 & 4 \\
2 & 3
\end{array}\right]\) then find x.
(a) x = 1
(b) x = 0
(c) x = 2
(d) x = -1
Answer:
(b) x = 0

Question 7.
If \(\left[\begin{array}{cc}
x & y \\
x & \frac{x}{2}+t
\end{array}\right]\) + \(\left[\begin{array}{cc}
y & x+t \\
x+2 & \frac{x}{2}
\end{array}\right]\) = \(\left[\begin{array}{ll}
1 & 4 \\
2 & 3
\end{array}\right]\) then find y.
(a) y = 1
(b) y = 3
(c) y = 2
(d) y = 0
Answer:
(a) y = 1

Question 8.
If \(\left[\begin{array}{cc}
x & y \\
x & \frac{x}{2}+t
\end{array}\right]\) + \(\left[\begin{array}{cc}
y & x+t \\
x+2 & \frac{x}{2}
\end{array}\right]\) = \(\left[\begin{array}{ll}
1 & 4 \\
2 & 3
\end{array}\right]\) then find t.
(a) t = 1
(b) t = 2
(c) t = 3
(d) t = 0
Answer:
(c) t = 3

Question 9.
Which matrix is a unit matrix?
(a) \(\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)\)
(b) \(\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)\)
(c) \(\left(\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right)\)
(d) \(\left(\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right)\)
Answer:
(b) \(\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)\)

Question 10.
If \(\left(\begin{array}{cc}
\mathbf{x}_1 & \mathbf{x}_2 \\
\mathbf{y}_1 & \mathbf{y}_2
\end{array}\right)\) – \(\left(\begin{array}{ll}
2 & 3 \\
0 & 1
\end{array}\right)\) = \(\left(\begin{array}{ll}
3 & 5 \\
1 & 2
\end{array}\right)\) then find x₁, x₂, y₁, y₂.
(a) x₁ = 8, x₂ = 5, y₁ = 3, y₂ = 1
(b) x₁ = 1, x₂ = 8, y₁ = 5, y₂ = 3
(c) x₁ = 5, x₂ = 8, y₁ = 1, y₂ = 3
(d) x₁ = 3, x₂ = 1, y₁ = 8, y₂ = 5
Answer:
(c) x₁ = 5, x₂ = 8, y₁ = 1, y₂ = 3

Question 11.
If \(\left|\begin{array}{ll}
2 & 4 \\
k & 6
\end{array}\right|\) = 0, what is the value of k?
(a) 3
(b) 4
(c) 2
(d) 6
Answer:
(a) 3

Question 12.
If \(\left|\begin{array}{ll}
\mathbf{a}_1 & \mathbf{b}_1 \\
\mathbf{c}_1 & \mathbf{d}_1
\end{array}\right|\) = k \(\left|\begin{array}{ll}
a_1 & c_1 \\
b_1 & d_1
\end{array}\right|\) hen what is the value of k?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 13.
If A = \(\left(\begin{array}{lll}
1 & 0 & 2 \\
5 & 1 & x \\
1 & 1 & 1
\end{array}\right)\) is a singular matrix then what is the value of x?
(a) 6
(b) 7
(c) 8
(d) 9
Answer:
(d) 9

Question 14.
Evaluate \(\left|\begin{array}{ccc}
-6 & 0 & 0 \\
3 & -5 & 7 \\
2 & 8 & 11
\end{array}\right|\)
(a) 66
(b) 666
(c) 6666
(d) 6
Answer:
(b) 666

Question 15.
Evaluate \(\left|\begin{array}{lll}
1 & 1 & b+c \\
1 & b & c+a \\
1 & c & a+b
\end{array}\right|\)
(a) 0
(b) 1
(c) 11
(d) 2
Answer:
(a) 0

Question 16.
Evaluate \(\left|\begin{array}{ccc}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2
\end{array}\right|\)
(a) 54
(b) 58
(c) -54
(d) 60
Answer:
(c) -54

Question 17.
If A and B are square matrices of order 3, such that |A| = -1, |B| = 3 then |3 AB| = –
(a) 1
(b) 11
(c) 9
(d) 81
Answer:
(d) 81

Question 18.
For what k
x + 2y – 3z = 2
(k + 3)z = 3
(2k + 1)y + z = 2 is inconsistent?
(a) -3
(b) -6
(c) 3
(d) 6
Answer:
(a) -3

Question 19.
The sum of two nonintegral roots of \(\left|\begin{array}{lll}
x & 2 & 5 \\
3 & x & 3 \\
5 & 4 & x
\end{array}\right|\) = 0 is ______.
(a) 5
(b) -5
(c) 3
(d) 15
Answer:
(b) -5

Question 20.
The value of \(\left|\begin{array}{ccc}
1 & 2 & 3 \\
3 & 5 & 2 \\
8 & 14 & 20
\end{array}\right|\) is ______.
(a) 1
(b) 2
(c) 0
(d) 3
Answer:
(c) 0

Question 21.
If [x 1] \(\left[\begin{array}{cc}
1 & 0 \\
-2 & 0
\end{array}\right]\) = 0, then x equals:
(a) 0
(b) -2
(c) -1
(d) 2
Answer:
(d) 2

Question 22.
The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:
(a) 27
(b) 18
(c) 81
(d) 512
Answer:
(d) 512

Question 23.
If A = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\) , and A + A’ = I, then the value of α is
(a) \(\frac{\pi}{6}\)
(b) \(\frac{\pi}{3}\)
(c) π
(d) \(\frac{3 \pi}{2}\)
Answer:
(b) \(\frac{\pi}{3}\)

Question 24.
Matrix A and B will be inverse of each other only if
(a) AB = BA
(b) AB = BA = 0
(c) AB = 0, BA = I
(d) AB = BA = I
Answer:
(d) AB = BA = I

Question 25.
The matrix P = \(\left[\begin{array}{lll}
0 & 0 & 4 \\
0 & 4 & 0 \\
4 & 0 & 0
\end{array}\right]\) is a
(a) square matrix
(b) diagonal matrix
(c) unit matrix
(d) None of these
Answer:
(a) square matrix

Question 26.
If A and B are symmetric matrices of same order, then AB – BA is a
(a) Skew-symmetric matrix
(b) Symmetric matrix
(c) Zero matrix
(d) Identity
Answer:
(a) Skew-symmetric matrix

Question 27.
If A is a square matrix of order 3, such that A(adj A) = 10I, then |adj A| is equal to
(a) 1
(b) 10
(c) 100
(d) 1000
Answer:
(c) 100

Question 28.
Let A be a square matrix of order 2 × 2, then |KA| is equal to
(a) K|A|
(b) K²|A|
(c) K³|A|
(d) 2K|A|
Answer:
(b) K²|A|

Question 29.
If A and B are invertible matrices then which of the following is not correct
(a) Adj A = |A|. A^-1
(b) det (A^-1) = (det A)^-1
(c) (AB)^-1 = B^-1A^-1
(d) (A + B)^-1 = A^-1 + B^-1
Answer:
(d) (A + B)^-1 = A^-1 + B^-1

Question 30.
If A is a skew-symmetric matrix of order 3, then the value of |A| is
(a) 3
(b) 0
(c) 9
(d) 27
Answer:
(b) 0

Question 31.
If A is a square matrix of order 3, such that A(adjA) = 10I, then ladj Al is equal to
(a) 1
(b) 10
(c) 100
(d) 1000
Answer:
(c) 100

Question 32.
Let A be a non-angular square matrix of order 3 x 3, then |A. adj Al is equal to
(a) |A|³
(b) |A|²
(c) |A|
(d) 3|A|
Answer:
(a) |A|³

Question 33.
Let A be a square matrix of order 3 × 3 and k a scalar, then |kA| is equal to
(a) k|A|
(b) |k||A|
(c) k³|A|
(d) none of these
Answer:
(c) k³|A|

Question 34.
If a, b, c are all distinct, and \(\left|\begin{array}{lll}
a & a^2 & 1+a^3 \\
b & b^2 & 1+b^3 \\
c & c^2 & 1+c^3
\end{array}\right|\) = 0 then the value of abc is
(a) 0
(b) -1
(c) 3
(d) -3
Answer:
(b) -1

Question 35.
If a, b, c are in AP, then the value of \(\left|\begin{array}{lll}
x+1 & x+2 & x+a \\
x+2 & x+3 & x+b \\
x+3 & x+4 & x+c
\end{array}\right|\) is:
(a) 4
(b) -3
(c) 0
(d) abc
Answer:
(c) 0

Question 36.
If A is a skew-symmetric matrix of order 3, then the value of |A| is
(a) 3
(b) 0
(c) 9
(d) 27
Answer:
(b) 0

Question 37.
A bag contains 3 white, 4 black and 2 red balls. If 2 balls are choosen at random (without replacement), then the probability that both the balls are white is:
(a) \(\frac{1}{18}\)
(b) \(\frac{2}{9}\)
(c) \(\frac{1}{12}\)
(d) \(\frac{1}{24}\)
Answer:
(c) \(\frac{1}{12}\)

Question 38.
Three diece are thrown simultaneously. The probability of obtaining a total score of 5 is:
(a) \(\frac{5}{216}\)
(b) \(\frac{1}{6}\)
(c) \(\frac{1}{36}\)
(d) \(\frac{1}{49}\)
Answer:
(c) \(\frac{1}{36}\)

Question 39.
An urn contains 6 balls of which two are red and four are black. Two balls are drawn at random. Probability that they are of the different colour is:
(a) \(\frac{2}{5}\)
(b) \(\frac{1}{15}\)
(c) \(\frac{8}{15}\)
(d) \(\frac{4}{15}\)
Answer:
(d) \(\frac{4}{15}\)

Question 40.
The probability of obtaining an even prime number on each die when a pair of dice is rolled is:
(a) 0
(b) \(\frac{1}{3}\)
(c) \(\frac{1}{12}\)
(d) \(\frac{1}{36}\)
Answer:
(d) \(\frac{1}{36}\)

Question 41.
Two events A and B are said to be independent if:
(a) A and B are mutually exclusive
(b) P (A’B’) = [1 – P(A)][1 – P(B)]
(c) P(A) = P(B)
(d) P(A) + P(B) = 1
Answer:
(b) P (A’B’) = [1 – P(A)][1 – P(B)]

Question 42.
A die is. thrown once, then the probability of getting number greater than 3 is:
(a) \(\frac{1}{2}\)
(b) \(\frac{2}{3}\)
(c) 6
(d) 0
Answer:
(a) \(\frac{1}{2}\)

Question 43.
If P(A) = \(\frac{6}{11}\), P(B) = \(\frac{5}{11}\) and P(A ∪ B) = \(\frac{7}{11}\), then P(A/B) is:
(a) \(\frac{2}{5}\)
(b) \(\frac{3}{5}\)
(c) \(\frac{4}{5}\)
(d) 1
Answer:
(c) \(\frac{4}{5}\)

Question 44.
Let the target be hit A and B: the target be hit by B and C: the target be hit by A and C. Then the probability that A, B and C all will hit, is:
(a) \(\frac{4}{5}\)
(b) \(\frac{3}{5}\)
(c) \(\frac{2}{5}\)
(d) \(\frac{1}{5}\)
Answer:
(c) \(\frac{2}{5}\)

Question 45.
What is the probability that ‘none of them will hit the target’?
(a) \(\frac{1}{30}\)
(b) \(\frac{1}{60}\)
(c) \(\frac{1}{15}\)
(d) \(\frac{2}{15}\)
Answer:
(b) \(\frac{1}{60}\)

(B) Very Short Type Questions With Answers

Question 1.
If \(\left|\begin{array}{ccc}
1+\mathbf{x} & \mathbf{x} & \mathbf{x}^2 \\
\mathbf{x} & 1+\mathbf{x} & \mathbf{x}^2 \\
\mathbf{x}^2 & \mathbf{x} & 1+\mathbf{x}
\end{array}\right|\) = a + bx + cx² + dx³ + ex⁴ + fx⁵ then write the value of a.
Solution:
\(\left|\begin{array}{ccc}
1+\mathbf{x} & \mathbf{x} & \mathbf{x}^2 \\
\mathbf{x} & 1+\mathbf{x} & \mathbf{x}^2 \\
\mathbf{x}^2 & \mathbf{x} & 1+\mathbf{x}
\end{array}\right|\)
= a + bx + cx² + dx³ + ex⁴ + fx⁵
which is an identity
Putting x = 0 we get
a = \(\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|\) = 1

Question 2.
If every element of a third order determinant of value 8 is multiplied by 2, then write the value of the new determinant.
Solution:
According to the question
|A| = 8
Now |KA| = Kⁿ|A|
⇒ |2A| = 2³|A| = 8 × 8 = 64
Value of the new determinant is 64.

Question 3.
If I is an identity matrix of order n, then k being a natural number, write the matrix I^k_n.
Solution:
If I is an identity matrix of order n, then I^k_n = I_n

Question 4.
If A is a 4 × 5 matrix and B is a matrix such that A^TB and BA^T both are defined, then write the order of B.
Solution:
Order of A = 4 × 5
Order of A^T = 5 × 4
Let order of B = m × n
A^TB is well defined ⇒ m = 4
BA^T is well defined ⇒ n = 5
Order of B = 4 × 5

Question 5.
Write the matrix which when added to the matrix \(\left[\begin{array}{cc}
2 & -3 \\
-4 & 7
\end{array}\right]\) gives the matrix \(\left[\begin{array}{ll}
4 & 1 \\
3 & 2
\end{array}\right]\)
Solution:
Let the required matrix is A.
\(\left(\begin{array}{cc}
2 & -3 \\
-4 & 7
\end{array}\right)\) + A = \(\left(\begin{array}{ll}
4 & 1 \\
3 & 2
\end{array}\right)\)
A = \(\left(\begin{array}{ll}
4 & 1 \\
3 & 2
\end{array}\right)\) – \(\left(\begin{array}{cc}
2 & -3 \\
-4 & 7
\end{array}\right)\) = \(\left(\begin{array}{cc}
2 & 4 \\
7 & -5
\end{array}\right)\)

Question 6.
Determine the maximum value of \(\left|\begin{array}{rl}
\cos x & \sin x \\
-\sin x & \cos x-1
\end{array}\right|\)
Solution:
Let f(x) = \(\left|\begin{array}{rl}
\cos x & \sin x \\
-\sin x & \cos x-1
\end{array}\right|\)
= cos²x – cos x + sin²x = 1 – cos x
As – 1 < cos x ≤ 1
⇒ 1 >- cos x ≥ – 1
⇒ 2 > 1 – cos x ≥ 0
The maximum value of f(x) = 2.

Question 7.
Write the value of k if:
\(\left|\begin{array}{lll}
\mathbf{a a _ { 1 }} & \mathbf{a a}_2 & \mathbf{a} \mathbf{a}_3 \\
\mathbf{a b _ { 1 }} & \mathbf{a b}_2 & \mathbf{a b} \\
\mathbf{a c _ { 2 }} & \mathbf{a c}_2 & \mathbf{a c _ { 3 }}
\end{array}\right|\) = k\(\left|\begin{array}{lll}
\mathbf{a}_1 & \mathbf{b}_1 & \mathbf{c}_1 \\
\mathbf{a}_2 & \mathbf{b}_2 & \mathbf{c}_2 \\
\mathbf{a}_3 & \mathbf{b}_3 & \mathbf{c}_3
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
\mathbf{a a _ { 1 }} & \mathbf{a a}_2 & \mathbf{a} \mathbf{a}_3 \\
\mathbf{a b _ { 1 }} & \mathbf{a b}_2 & \mathbf{a b} \\
\mathbf{a c _ { 2 }} & \mathbf{a c}_2 & \mathbf{a c _ { 3 }}
\end{array}\right|\) = k\(\left|\begin{array}{lll}
\mathbf{a}_1 & \mathbf{b}_1 & \mathbf{c}_1 \\
\mathbf{a}_2 & \mathbf{b}_2 & \mathbf{c}_2 \\
\mathbf{a}_3 & \mathbf{b}_3 & \mathbf{c}_3
\end{array}\right|\)
k = a³.

Question 8.
If A is a 3 × 3 matrix and |A| = 3, then write the matrix represented by A × adj A.
Solution:
|A| = 3 ⇒ A × Adj A = \(\left(\begin{array}{lll}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{array}\right)\)

Question 9.
If ω is a complex cube root of 1, then for what value of λ the determinant
\(\left|\begin{array}{ccc}
1 & \omega & \omega^2 \\
\omega & \lambda & 1 \\
\omega^2 & 1 & \omega
\end{array}\right|\) = 0?
Solution:

⇒ for any value of ‘A,’ the given determinant is ‘0’

Question 10.
If [1 2 3] A = [0], then what is the der of the matrix A?
Solution:
If [1 2 3] A = [0]
A is a 3 × 1 matrix

Question 11.
What is A + B if A = \(\left(\begin{array}{cc}
1 & 2 \\
3 & -1
\end{array}\right)\), B = \(\left(\begin{array}{cc}
0 & -1 \\
-2 & 1
\end{array}\right)\)?
Solution:
For A = \(\left(\begin{array}{cc}
1 & 2 \\
3 & -1
\end{array}\right)\), B = \(\left(\begin{array}{cc}
0 & -1 \\
-2 & 1
\end{array}\right)\)
A + B = \(\left(\begin{array}{ll}
1 & 1 \\
1 & 0
\end{array}\right)\)

Question 12.
Give an example of a unit matrix.
Solution:
\(\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)\) is a unit matrix of 2nd order.

Question 13.
Construct a 2 × 3 matrix having elements defined by a_ij = i – j.
Solution:
a_ij = i – j
a₁₁ =0, a₁₂ = 1 – 2 = – 1, a₁₃ = 1 – 3 =- 2
a₂₁ = 2 – 1 = 1, a₂₂ = 2 – 2 = 0, a₂₃ = 2 – 3 = -1
∴ The required matrix is \(\left(\begin{array}{ccc}
0 & -1 & -2 \\
0 & 0 & -1
\end{array}\right)\).

Question 14.
Find x, y if A = A’ where A = \(\left(\begin{array}{ll}
5 & \mathbf{x} \\
\mathbf{y} & 0
\end{array}\right)\)
Solution:
A = A’
⇒ \(\left(\begin{array}{ll}
5 & \mathbf{x} \\
\mathbf{y} & 0
\end{array}\right)\) = \(\left(\begin{array}{ll}
5 & \mathbf{y} \\
\mathbf{x} & 0
\end{array}\right)\) ⇒ x = y
∴ x and y are any real number where x = y

Question 15.
Cana matrix be constructed by taking 29 elements?
Solution:
Only two matrices can be formed by taking 29 elements. They are of order 1 × 29 and 29 × 1.

Question 16.
If \(\left|\begin{array}{ll}
2 & 4 \\
k & 6
\end{array}\right|\) = 0 , what is the value of k?
Solution:
\(\left|\begin{array}{ll}
2 & 4 \\
k & 6
\end{array}\right|\) = 0 ⇒ 12 – 4k = 0 ⇒ k = 3

Question 17.
If \(\left|\begin{array}{ll}
\mathbf{a}_1 & \mathbf{b}_1 \\
\mathbf{c}_{\mathbf{1}} & \mathbf{d}_1
\end{array}\right|\) = k = \(\left|\begin{array}{ll}
\mathbf{a}_1 & \mathbf{c}_1 \\
\mathbf{b}_1 & \mathbf{d}_1
\end{array}\right|\) then what is the value of k?
Solution:
k = 1

Question 18.
If A and B are square matrices of order 3, such that |A| = -1, |B| = 3 then |3 AB| = ______.
Solution:
|3 AB| = 27 |A| |B| = 81

Question 19.
Solve: \(\left|\begin{array}{ccc}
2 & 2 & x \\
-1 & x & 4 \\
1 & 1 & 1
\end{array}\right|\) = 0
Solution:
\(\left|\begin{array}{ccc}
2 & 2 & x \\
-1 & x & 4 \\
1 & 1 & 1
\end{array}\right|\) = 0 => \(\left|\begin{array}{ccc}
0 & 2 & x \\
-1-x & x & 4 \\
0 & 1 & 1
\end{array}\right|\) = 0
⇒ – (- 1 – x) (2 – x) = 0 ⇒ x = -1, x = 2.

(C) Short Type Questions With Answers

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

Question 2.
Solve: \(\left|\begin{array}{ccc}
\mathbf{x + 1} & \omega & \omega \\
\omega & x+\omega^2 & 1 \\
\omega^2 & 1 & x+\omega
\end{array}\right|\) = 0
Solution:

Question 3.
If A = \(\left[\begin{array}{ccc}
1 & 2 & 0 \\
0 & 1 & 3 \\
-2 & 5 & 3
\end{array}\right]\), then verify that A + A’ is symmetric and A – A’ is skew symmetric.
Solution:

Question 4.
If A, B, C are matrices of order 2 × 2 each and 2A + B + C = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 0
\end{array}\right]\), A + B + C = \(\left[\begin{array}{ll}
0 & 1 \\
2 & 1
\end{array}\right]\) and A + B – C = \(\left[\begin{array}{ll}
1 & 2 \\
1 & 0
\end{array}\right]\), then find A, B and C.
Solution:

Question 5.
Find the inverse of the following matrix: \(\left[\begin{array}{lll}
1 & 1 & 2 \\
0 & 1 & 2 \\
1 & 2 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 1 & 2 \\
0 & 1 & 2 \\
1 & 2 & 1
\end{array}\right]\)
Method – I
Let us find A^-1 by using elementary row transformation.
Let A = IA

Method – II
|A| = 1(1 – 4) – 1(0- 2) + 2(0- 1)
= 1(-3) – 1(-2) + 2(-1)
= -3 ≠ 0
∴ A^-1 exists.
A₁₁ = -3, A₁₂ = 2, A₁₃ = -1

Question 6.
Show that \(\left|\begin{array}{ccc}
a-b-c & 2 a & 2 a \\
2 b & b-c-a & 2 b \\
2 c & 2 c & c-a-b
\end{array}\right|\) = (a+b +c)³
Solution:

= (a + b + c)³ (1 – 0) = (a + b + c)³

Question 7.
Find the inverse of the following matrix: \(\left[\begin{array}{lll}
0 & 0 & 2 \\
0 & 2 & 0 \\
2 & 0 & 0
\end{array}\right]\)
Solution:
A = \(\left[\begin{array}{lll}
0 & 0 & 2 \\
0 & 2 & 0 \\
2 & 0 & 0
\end{array}\right]\)
|A| = 2(- 4) = – 8 ≠ 0
∴ A^-1 exists.
A₁₁ = 0, A₁₂ = 0, A₁₃ = – 4
A₂₁ = 0, A₂₂ = – 4, A₂₃ = 0
A₃₁ = 0,A₃₂ = 0, A₃₃ = – 4
∴ The matrix of cofactors

Question 8.
If the matrix A is such that \(\left[\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right]\)A = \(\left[\begin{array}{cc}
-4 & 1 \\
7 & 7
\end{array}\right]\), find A.
Solution:

Question 9.
Show that (a + 1) is a factor of \(\left|\begin{array}{ccc}
a+1 & 2 & 3 \\
1 & a+1 & 3 \\
3 & -6 & a+1
\end{array}\right|\).
Solution:

= (a + 1) (a² + 2a + 1 + 18) – 2(a + 1 – 9) + 3(-6 – 3a – 3)
= (a + 1)(a² + 2a + 19) – 2a + 16 – 27 – 9a
= (a + 1) (a² + 2a + 19) – 11a – 11
= (a + 1) (a² + 2a + 19) – 11(a + 1)
= (a + 1) (a² + 2a + 8)
⇒ (a + 1) is a factor of the given determinant.

Question 11.
If A = \(\left[\begin{array}{ll}
\alpha & 0 \\
1 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
1 & 0 \\
5 & 1
\end{array}\right]\) show that for no values of α, A² = B.
Solution:

⇒ α2 = 1 and α + 1 = 5
⇒ α = ± 1 and α = 4
Which is not possible.
There is no α for which A² = B

Question 12.
If A = \(\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]\), then show that A^k = \(\left[\begin{array}{cc}
1+2 \mathrm{k} & -4 \mathrm{k} \\
\mathrm{k} & 1-2 \mathrm{k}
\end{array}\right]\), k ∈ N.
Solution:

Question 13.
If A = \(\left[\begin{array}{ccc}
1 & -2 & 2 \\
3 & 1 & -1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & 4 \\
1 & 2 \\
3 & -1
\end{array}\right]\), verify that (AB)^T = B^TA^T.

Question 14.
Show that for each real value of λ the system of equations
(λ + 3) + λy = 0
x + (2λ + 5)y = 0 has a unique solution.
Solution:
Given system of equations is a
homogeneous system of linear
equations.
Now
Δ = \(\left|\begin{array}{cc}
\lambda+3 & \lambda \\
1 & 2 \lambda+5
\end{array}\right|\)
= (λ + 3)(2λ + 5) – λ
= 2λ² + 11λ + 15 – λ
= 2λ² + 10λ + 15
As for 2λ² + 10A + 15, D = 100 – 120 < 0
the polynomial 2λ² + 10λ + 15 has no roots i.e. Δ ≠ 0.
Thus the system has a unique trivial solution for every real value of λ.

Question 15.
If A and B are square matrices of same order then show by means of an example that AB ≠ BA in general.
Solution:

∴ AB ≠ BA we have AB ≠ BA in general.

Question 16.
If A = \(\left|\begin{array}{cc}
0 & -\tan \frac{\theta}{2} \\
\tan \frac{\theta}{2} & 0
\end{array}\right|\), then prove that det{(I + A)(I – A)-1} = 1
Solution:

Question 17.
Solve for x: \(\left|\begin{array}{ccc}
15-2 x & 11 & 10 \\
11-3 x & 17 & 16 \\
7-x & 14 & 13
\end{array}\right|\) = 0
Solution:

Question 18.
If A = \(\left[\begin{array}{ccc}
-1 & 3 & 5 \\
1 & -3 & -5 \\
-1 & 3 & 5
\end{array}\right]\) find A³ – A².
Solution:

A² = A ⇒ A²A = A²
⇒ A³ – A² = 0

Question 19.
Prove that: A = \(\left|\begin{array}{ccc}
2 & 3 & 4 \\
1 & -2 & -3 \\
3 & 1 & -8
\end{array}\right|\) ⇒ A² – 5A + 71 = 0.
Solution:

Question 20.
Test whether the following system of equations have non-zero solution.
Write the solution set:
2x + 3y + 4z = 0,
x – 2y – 3z = 0,
3x + y – 8z = 0.
Solution:
Given equations are
2x + 3y + 4z = 0
x – 2y – 3z = 0
3x + y – 8z = 0
Now \(\left|\begin{array}{ccc}
2 & 3 & 4 \\
1 & -2 & -3 \\
3 & 1 & -8
\end{array}\right|\)
= 2(19) – 3(1) + 4(7) 0
∴ The system has no non-zero solution.
The solution set is x = 0; y = 0, z = 0.

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 6 Probability Ex 6(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 6 Probability Exercise 6(a)

Question 1.
Two balls are drawn from a bag containing 5 white and 7 black balls. Find the probability of selecting 2 white balls if
Solution:
Two balls are drawn from a bag containing 5 white and 7 black balls.
∴ |S| = 12.

(i) the first ball is replaced before drawing the second.
Solution:
The 1st ball is replaced before the 2nd ball is drawn. We are to select 2 white balls. So in both the draws we will get white balls. Drawing a white ball in 1st draw and in 2nd draw are independent events.
Probability of getting 2 white balls = \(\frac{5}{12}\) × \(\frac{5}{12}\) = \(\frac{25}{144}\)

(ii) the first ball is not replaced before drawing the second.
Solution:
Here the 1st ball is not replaced before the 2nd ball is drawn. Since we are to get 2 white balls in each draw, we must get a white ball.
Now probability of getting a white ball in 1st draw = \(\frac{5}{12}\).
Probability of getting a white ball in 2nd = \(\frac{4}{11}\).
Since the two draws are independent, we have the probability of getting 2 white balls
= \(\frac{5}{12}\) × \(\frac{4}{11}\) = \(\frac{20}{132}\).

Question 2.
Two cards are drawn from a pack of 52 cards; find the probability that
(i) they are of different suits.
(ii) they are of different denominations.
Solution:
Two cards are drawn from a pack of 52 cards. The cards are drawn one after another. Each suit has 13 cards.
|S| = ⁵²C₂
(i) As the two cards are of different suits, their probability
= \(\frac{52}{52}\) × \(\frac{39}{51}\)
(ii) Each denomination contains 4 cards. As the two cards drawn are of different denominations, their probability
= \(\frac{52}{52}\) × \(\frac{48}{51}\)

Question 3.
Do both parts of problem 2 if 3 cards drawn at random.
Solution:
(i) 3 cards are drawn one after another. As they are of different suits, we have their probability
= \(\frac{52}{52}\) × \(\frac{39}{51}\) × \(\frac{26}{50}\).
(ii) As the 3 cards are of different denominations, we have their probability
= \(\frac{52}{52}\) × \(\frac{48}{51}\) × \(\frac{44}{50}\).

Question 4.
Do both parts of problem 2 if 4 cards are drawn at random.
Solution:
(i) 4 cards are drawn one after another. As they are of different suits, we have their probability
= \(\frac{52}{52}\) × \(\frac{39}{51}\) × \(\frac{26}{50}\) × \(\frac{13}{49}\).
(ii) As the cards are of different denominations, we have their probability
= \(\frac{52}{52}\) × \(\frac{48}{51}\) × \(\frac{44}{50}\) × \(\frac{40}{49}\).

Question 5.
A lot contains 15 items of which 5 are defective. If three items are drawn at random, find the probability that
(i) all three are defective
(ii) none of the three is defective.
Do this problem directly.
Solution:
(i) A lot contains 15 items of which 5 are defective. Three items are drawn at random. As the items are drawn one after another.
Their probability = \(\frac{5}{15}\) × \(\frac{4}{14}\) × \(\frac{3}{13}\)
(ii) As none of the 3 items are defective, we have to draw 3 non-defective items one after another.
Their probability = \(\frac{10}{15}\) × \(\frac{9}{14}\) × \(\frac{8}{13}\)

Question 6.
A pair of dice is thrown. Find the probability of getting a sum of at least 9 if 5 appears on at least one of the dice.
Solution:
A pair of dice is thrown. Let A be the event of getting at least 9 points and B, the event that 5 appears on at least one of the dice.
∴ B = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6)}
A = {(3, 6), (4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}
∴ A ∩ B = {(4, 5), (5, 4), (5, 5), (5, 6), (6, 5)}
∴ P (A | B) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}\) = \(\frac{\frac{5}{36}}{\frac{31}{36}}\) = \(\frac{5}{11}\)

Question 7.
A pair of dice is thrown. If the two numbers appearing are different, find the probability that
(i) the sum of points is 8.
(ii) the sum of points exceeds 8.
(iii) 6 appears on one die.
Solution:
A pair of dice is thrown as two numbers are different
We have |S| = 30
(i) Let A be the. event that the sum of points on the dice is 8, where the numbers on the dice are different.
A = {(2, 6), (3, 5), (5, 3), (6, 2)}
P(A) = \(\frac{|\mathrm{A}|}{|\mathrm{S}|}\) = \(\frac{4}{30}\)
(ii) Let B be the event that sum of the points exceeds 8.
B = {(3, 6), (4, 5), (5, 4), (6, 3), (5, 6), (6, 5), (4, 6), (6, 4)}
P(B) = \(\frac{|\mathrm{B}|}{|\mathrm{S}|}\) = \(\frac{8}{30}\)
(iii) Let C be the event that 6 appears on one die.
C = {(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
P(C) = \(\frac{|\mathrm{C}|}{|\mathrm{S}|}\) = \(\frac{10}{30}\) = \(\frac{1}{3}\)

Question 8.
In a class 30% of the students fail in Mathematics, 20% of the students fail in English and 10% fail in both. A student is selected at random.
Solution:
In a class 30% of the students fail in Mathematics, 20% of the students fail in English and 10% fail in both. Let A be the event that a student fails in Mathematics and B be the events that he fails in English.
P(A) = \(\frac{30}{100}\), P(B) = \(\frac{20}{100}\)
Where |S| = 100, P (A ∩ B) = \(\frac{10}{100}\)

(i) If he has failed in English, what is the probability that he has failed in Mathematics?
Solution:
If he has failed in English, then the probability that he has failed in Mathematics.
i.e., P\(\left(\frac{A}{B}\right)\) = \(\frac{P(A \cap B)}{P(B)}\) = \(\frac{\frac{10}{100}}{\frac{20}{100}}\) = \(\frac{1}{2}\)

(ii) If he has failed in Mathematics, what is the probability that he has failed in English?
Solution:
If he has failed in Mathematics, then the probability that he has failed in English
i.e., P\(\left(\frac{B}{A}\right)\) = \(\frac{P(A \cap B)}{P(A)}\) = \(\frac{\frac{10}{100}}{\frac{30}{100}}\) = \(\frac{1}{3}\)

(iii) What is the probability that he has failed in both?
Solution:
Probability that he has failed in both
i.e., P (A ∩ B) = \(\frac{10}{100}\) = \(\frac{1}{10}\)

Question 9.
IfA, B are two events such that
P(A) = 0.3, P(B) = 0.4, P (A ∪ B) = 0.6
Find
(i) P (A | B)
(ii) P (B | A)
(iii) P (A | B^c)
(iv) P (B | A^c)
Solution:
A and B are two set events such that
P(A) = 0.3, P(B) = 0.4, P (A ∪ B) = 0.6
We have
P (A ∪ B) = P(A) + P(B) – P (A ∩ B)
or, 0.6 = 0.3 + 0.4 – P (A ∩ B)
or, P (A ∩ B) = 0.7 – 0.6 = 0.1

Question 10.
If A, B are events such that P(A) = 0.6, P(B) = 0.4 and P (A ∩ B) = 0.2, then find
(i) P (A | B)
(ii) P (B | A)
(iii) P (A | B^c)
(iv) P (B | A^c)
Solution:
A and B are events such that
P(A) = 0.6, P(B) = 0.4, P (A ∩ B) = 0.2

Question 11.
If A and B are independent events, show that
(i) A^c and B^c are independent,
(ii) A and B^c are independent,
(iii) A^c and B are independent.
Solution:
A and B are independent events.
(i) We have P (A ∩ B) = P(A). P(B)
P (A’ ∩ B’) = P (A ∪ B)’ = 1 – P (A ∪ B)
= 1 – [P(A) + P(B) – P (A ∩ B)]
= 1 – P(A) – P(B) + P(A) P(B)
= 1 [1 – P(A)] – P(B) [1 – P(A)]
= [1 – P(A)] [1 – P(B)] = P(A’) P(B’)
∴ A’ and B’ are independent events.

(ii) P (A ∩ B^c) = P (A – B)
= P(A) – P (A ∩ B)
= P(A) – P(A) P(B)
= P(A) [1 – P(B)]
= P(A). P(B^c).
∴ A and B^c are independent events.

(iii) P (A^c ∪ B) = P (B – A)
= P(B) – P (A ∩ B)
= P(B) – P(A) P(B)
= P(B) [1 – P(A)] = P(B) P(A^c)
∴ A^c and B are independent events.

Question 12.
Two different digits are selected at random from the digits 1 through 9.
(i) If the sum is even, what is the probability that 3 is one of the digits selected?
(ii) If the sum is odd, what is the probability that 3 is one of the digits selected?
(iii) If 3 is one of the digits selected, what is the probability that the sum is odd?
(iv) If 3 is one of the digits selected, what is the probability that the sum is even?
Solution:
Two different digits are selected at random from the digits 1 through 9.
(i) Let A be the event that the sum is even and B be the event that 3 is one of the number selected.
We have to find P (B | A).
There are 4 even digits and 5 odd digits.
∴ The sum is even if both the numbers are odd or both are even.
∴ |A| = ⁴C₂ + ⁵C₂ = 6 + 10 = 16
∴ P(A) = \(\frac{16}{{ }^9 \mathrm{C}_2}\) = \(\frac{16}{36}\)
Also B = {(1, 3), (5, 3), (7, 3), (9, 3), (3, 2), (3, 4), (3, 8), (3, 6)}
∴ A ∩ B = {(1, 3), (5, 3), (7, 3), (9, 3)}
P(B) = \(\frac{8}{36}\) P (A ∩ B) = \(\frac{4}{36}\)
P(\(\frac{B}{A}\)) = \(\frac{P(A \cap B)}{P(A)}\) = \(\frac{\frac{4}{36}}{\frac{16}{36}}\) = \(\frac{1}{4}\)

(ii) Let A be the event that the sum is odd. The sum is odd if one of the numbers selected is odd and other is even.
∴ P(A) = \(\frac{5}{9}\) × \(\frac{4}{8}\) + \(\frac{4}{9}\) × \(\frac{5}{8}\) = \(\frac{20}{36}\)
Let B be the event that one of the numbers selected is 3.
∴ B = {(1, 3), (2, 3), (4, 3), (5, 3), (6, 3), (7, 3), (8, 3), (9, 3)}
∴ A ∩ B = {(2, 3), (4, 3), (6, 3), (8, 3)}

Question 13.
If P(A) = 0.4, P (B | A) = 0.3 and P (B^c | A^c) = 0.2. find
(i) P (A | B)
(ii) P (B | A^c)
(iii) P(B)
(iv) P(A^c)
(v) P (A ∪ B)
Solution:
P(A) = 0.4, P (B | A) = 0.3 and P (B^c | A^c) = 0.2.


(iii) P(B) = 0.6
= \( \frac{6}{10} \)
= \( \frac{3}{5} \)

(iv) P(A^c) = 1 – P(A)
= 1 – 0.4 = 0.6
= \( \frac{6}{10} \) = \( \frac{3}{5} \)

(v) P (A ∪ B) = P(A) + P(B) – P (A ∩ B)
= 0.4 + 0.6 – 0.12
= 1.0 – 0.12 = 0.88

Question 14.
If P(A) = 0.6, P (B | A) = 0.5, find P (A ∪ B) if A, B are independent.
Solution:
P(A) = 0.6, P (B | A) = 0.5
We have P (B | A) = \(\frac{\mathrm{P}(\mathrm{B} \cap \mathrm{A})}{\mathrm{P}(\mathrm{A})}\) = 0.5
or, P (B ∩ A) = 0.5 × P(A)
= 0.5 × 0.6 = 0.3
As A and B are independent events, we have
P (B ∩ A) = P(B) P(A) = 0.3
or, P(B) = \(\frac{0.3}{\mathrm{P}(\mathrm{A})}\)
= \(\frac{0.3}{0.6}\)
= \(\frac{1}{2}\) = 0.5
P (A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.6 + 0.5 – 0.3 = 0.8

Question 15.
Two cards are drawn in succession from a deck of 52 cards. What is the probability that both cards are of denomination greater than 2 and less than 9?
Solution:
Two cards are drawn in succession from a deck of 52 cards.
There are 6 denominations which are greater than 2 and less than 9. So there are 24 cards whose denominations are greater than 2 and less than 9.
∴ Their probability = \(\frac{24}{52}\) × \(\frac{23}{51}\).

Question 16.
From a bag containing 5 black and 7 white balls, 3 balls are drawn in succession. Find the probability that
(i) all three are of the same colour.
(ii) each colour is represented.
Solution:
From a bag containing 5 black and 7 white balls, 3 balls are drawn in succession.
(i) The 3 balls drawn are of same colour.
∴ Probability of drawing 3 balls of black colour
= \(\frac{5}{12}\) × \(\frac{4}{11}\) × \(\frac{3}{10}\) = \(\frac{1}{22}\)
Probability of drawing 3 white balls
= \(\frac{7}{12}\) × \(\frac{6}{11}\) × \(\frac{5}{10}\) = \(\frac{7}{44}\)
∴ Probability of drawing 3 balls of same colour
= \(\frac{5}{12}\) × \(\frac{4}{11}\) × \(\frac{3}{10}\) + \(\frac{7}{12}\) × \(\frac{6}{11}\) × \(\frac{5}{10}\) = \(\frac{9}{44}\)

(ii) Balls of both colour will be drawn. If B represents black ball and W represents the white ball.
∴ The possible draws are WWB, WBW, BWW.

Question 17.
A die is rolled until a 6 is obtained. What is the probability that
(i) you end up in the second roll
(ii) you end up in the third roll.
Solution:
A die is rolled until a 6 is obtained
(i) We are to end up in the 2nd roll i.e., we get 6 in the 2nd roll. Let A be the event of getting a 6 in one roll of a die.
∴ P(A) = \(\frac{1}{6}\) ⇒ P(A’) = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
∴ Probability of getting a 6 in the 2nd roll
= \(\frac{5}{6}\) × \(\frac{1}{6}\) = \(\frac{5}{36}\)
(ii) Probability of getting a 6 in the 3rd roll
= \(\frac{5}{6}\) × \(\frac{5}{6}\) × \(\frac{1}{6}\) = \(\frac{25}{216}\)

Question 18.
A person takes 3 tests in succession. The probability of his (her) passing the first test is 0.8. The probability of passing each successive test is 0.8 or 0.5 according as he passes or fails the preceding test. Find the probability of his (her) passing at least 2 tests.
Solution:
A person takes 3 tests in succession. The probability of his passing the 1st test is 0.8. The probability of passing each successive test is 0.8 or 0.5 according as he passes or fails the preceding test.
Let S denotes the success (passing) in a test and F denotes the failure in a test.
∴ P(S) = 0.8
∴ P(F) = 1 – P(S) = 1 – 0.8 = 0.2
We have the following mutually exclusive cases:

Event			Probability
S	S	S	0.8 × 0.8 × 0.8 = 0.512
S	S	F	0.8 × 0.8 × 0.2 = 0.128
S	F	S	0.8 × 0.2 × 0.5 = 0.080
F	S	S	0.2 × 0.5 x 0.8 = 0.080

∴ Probability of atleast 2 successes
= 0.512 + 0.128 + 0.080 + 0.080
= 0.8 = \(\frac{4}{5}\)

Question 19.
A person takes 4 tests in succession. The probability of his passing the first test is p, that of his passing each succeeding test is p or y depending on his passing or failing the preceding test. Find the probability of his passing
(i) at least three test
(ii) just three tests.
Solution:
A person takes 4 tests in succession. The probability of his passing the 1st test is P, that of his passing each succeeding test is P or P/2 depending bn his passing or failing the preceding test. Let S and F denotes the success and failure in the test.
∴ P(S) = P, P(F) = 1 – P
We have the following mutually exclusive tests:

Question 20.
Given that all three faces are different in a throw of three dice, find the probability that
(i) at least one is a six
(ii) the sum is 9.
Solution:
Three dice are thrown once showing different faces in a throw.
|S| = 6³ = 216
Let A be the event that atleast one is a six.
Let B be the event that all three faces are different.
|B| = ⁶6₃
(i) Now A^c is the event that there is no six. A^c ∩ B is the event that all 3 faces are different and 6 does not occur.
|A^c ∩ B| = ⁵C₃
P (A^c | B) = \(\frac{P\left(A^C \cap B\right)}{P(B)}\)
= \(\frac{{ }^5 \mathrm{C}_3 / 216}{{ }^6 \mathrm{C}_3 / 216}\) = \(\frac{1}{2}\)

(ii) Let A be the event that the sum is 9.
A ∩ B = {(1,3, 5), (1,5, 3), (3, 5,1), (3, 1, 5), (5, 1, 3), (5, 3, 1), (1, 2, 6), (1, 6, 2), (2, 1, 6), (2, 6, 1), (6, 2, 1), (6, 1, 2), (2, 3, 4), (2, 4, 3), (3, 2, 4), (2, 3, 4), (3, 4, 2), (4, 3, 2)}
|A ∩ B| = 18
P (A | B) = \(\frac{P(A \cap B)}{P(B)}\)
= \(\frac{18 / 216}{20 / 216}\) = \(\frac{9}{10}\)

Question 21.
From the set of all families having three children, a family is picked at random.
(i) If the eldest child happens to be a girl, find the probability that she has two brothers.
(ii) If one child of the family is a son, find the probability that he has two sisters.
Solution:
A family is picked up at random from a set of families having 3 children.
(i) The eldest child happens to be a girl. We have to find the probability that she has two brothers. Let G denotes a girl and B denotes a boy.
∴ P(B) = \(\frac{1}{2}\), P(G) = \(\frac{1}{2}\)
P(BB | G) = P(B) × P(G) = \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\)

(ii) The one child of the family is a son. We have to find the probability that he has two sisters. We have the following mutually exclusive events:
BGG, GBG, GGB.
∴ The required probability
= P(B) × P(G) × P(G) + P(G) × P(B) + P(G) + P(G) × P(G) × P(B)
= \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) + \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) + \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{3}{8}\)

Question 22.
Three persons hit a target with probability \(\frac{1}{2}\), \(\frac{1}{3}\) and \(\frac{1}{4}\) respectively. If each one shoots at the target once,
(i) find the probability that exactly one of them hits the target
(ii) if only one of them hits the target what is the probability that it was the first person?
Solution:
Three persons hit a target with probability
\(\frac{1}{2}\), \(\frac{1}{3}\) and \(\frac{1}{4}\)
Let A, B, C be the events that the 1st person, 2nd person, 3rd person hit the target respectively.

(i) As the events are independent, the probability that exactly one of them hit the target
= P(AB’C’) + P(A’BC’) + P(A’B’C)
= P(A) P(B’) P(C’) + P(A’) P(B) P(C’) + P(A’) P(B’) P(C)

(ii) Let E₁ be the event that exactly one person hits the target.
∴ P(E₁) = \(\frac{11}{24}\)
Let E₂ be the event that 1st person hits the target
∴ P(E₂) = P(A) = \(\frac{1}{2}\)
∴ E₁ ∩ E₂ = AB’C’
⇒ P(E₁ ∩ E₂)
= P(A) × P(B’) P(C’) = \(\frac{6}{24}\)
∴ P(E₂ | E₁) = \(\frac{6 / 24}{11 / 24}\) = \(\frac{6}{11}\)

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Additional Exercise Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Additional Exercise

(A) Multiple Choice Questions (Mcqs) With Answers

Question 1.
Write the value of cos^-1 cos(3π/2).
(a) π
(b) \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{4}\)
(d) 2π
Solution:
(b) \(\frac{\pi}{2}\)

Question 2.
Sets A and B have respectively m and n elements. The total number of relations from A to B is 64. If m < n and m ≠ 1, write the values of m and n respectively.
(a) m = 3, n = 2
(b) m = 2, n = 2
(c) m = 2, n = 3
(d) m = 3, n = 3
Solution:
(c) m = 2, n = 3

Question 3.
Write the principal value of
sin^-1 (\(-\frac{1}{2}\)) + cos^-1 cos(\( -\frac{\pi}{2}\))
(a) \(\frac{\pi}{2}\)
(b) \(\frac{\pi}{3}\)
(c) \(\frac{\pi}{4}\)
(d) π
Solution:
(b) \(\frac{\pi}{3}\)

Question 4.
Write the maximum value of x + y subject to: 2x + 3y < 6, x > 0, y > 0.
(a) 3
(b) 1
(c) 2
(d) 0
Solution:
(a) 3

Question 5.
Let A has 3 elements and B has m elements. Number of relations from A to B = 4096. Find the value of m.
(a) 2
(b) 4
(c) 1
(d) 3
Solution:
(b) 4

Question 6.
Let A is any non-empty set. Number of binary operations on A is 16. Find |A|.
(a) 2
(b) 1
(c) 3
(d) 4
Solution:
(a) 2

Question 7.
Give an example of a relation which is reflexive, transitive but not symmetric.
(a) x < y on Z
(b) x = y on Z
(c) x > y on Z
(d) None of the above
Solution:
(a) x < y on Z

Question 8.
Find the least positive integer r such that – 375 ∈ [r]₁₁
(a) r = 5
(b) r = 6
(c) r = 3
(d) r = 10
Solution:
(d) r = 10

Question 9.
Find three positive integers x_i, i = 1, 2, 3 satisfying 3x ≡ 2 (mod 7)
(a) x = 1, 3, 9…
(b) x = 2, 4, 6…
(c) x = 3, 10, 17…
(d) x = 2, 10, 18…
Solution:
(c) x = 3, 10, 17…

Question 10.
If the inversible function f is defined as f(x) = \(\frac{3 x-4}{5}\) write f^-1(x)
(a) \(\frac{5 x+4}{3}\)
(b) \(\frac{4 x+5}{3}\)
(c) \(\frac{5 x-4}{3}\)
(d) \(\frac{5 x+4}{2}\)
Solution:
(a) \(\frac{5 x+4}{3}\)

Question 11.
Let f : R → R and g : R → R defined as f(x) = |x|, g(x) = |5x – 2| then find fog.
(a) |5x + 2|
(b) |5x – 2|
(c) |2x – 2|
(d) |2x – 5|
Solution:
(b) |5x – 2|

Question 12.
Let ∗ is a binary operation defined by a ∗ b = 3a + 4b – 2, find 4 ∗ 5.
(a) 20
(b) 12
(c) 30
(d) 36
Solution:
(c) 30

Question 13.
Let the binary operation on Q defined as a ∗ b = 2a + b – ab, find 3 ∗ 4.
(a) -2
(b) -1
(c) 2
(d) 1
Solution:
(a) -2

Question 14.
Let ∗ is a binary operation on Z defined as a ∗ b = a + b – 5 find the identity element for ∗ on Z.
(a) e = 1
(b) e = 5
(c) e = -5
(d) e = -1
Solution:
(b) e = 5

Question 15.
Find the number of binary, operations on the set {a, b}.
(a) 12
(b) 14
(c) 15
(d) 16
Solution:
(d) 16

Question 16.
Let ∗ is a binary operation on [0, ¥) defined as a ∗ b = \(\sqrt{a^2+b^2}\) find the identity element.
(a) e = 0
(b) e = 2
(c) e = 1
(d) e = 3
Solution:
(a) e = 0

Question 17.
Find least non-negative integer r such that 7 × 13 × 23 × 413 ≡ r (mod 11).
(a) r = 13
(b) r = 49
(c) r = 7
(d) r = 23
Solution:
(c) r = 7

Question 18.
Find least non-negative integer r such that 1237(mod 4) + 985 (mod 4) ≡ r (mod 4).
(a) r = 1
(b) r = 2
(c) r = -2
(d) r = -1
Solution:
(b) r = 2

Question 19.
Let ∗ is a binary operation on R – {0} defined as a ∗ b = \(\frac{a b}{5}\). If 2 ∗ (x ∗ 5) = 10, then find x:
(a) x = 25
(b) x = -5
(c) x = 5
(d) x = 1
Solution:
(a) x = 25

Question 20.
Find the principal value of cos^-1 (\( -\frac{1}{2}\)) + 2sin^-1 (\( \frac{1}{2}\)).
(a) \(\frac{5 \pi}{6}\)
(b) \(\frac{5 \pi}{2}\)
(c) \(\frac{5 \pi}{4}\)
(d) \(\frac{\pi}{6}\)
Solution:
(a) \(\frac{5 \pi}{6}\)

Question 21.
Evaluate sin^-1 (\(\frac{1}{\sqrt{5}}\)) + cos^-1 (\(\frac{3}{\sqrt{10}}\))
(a) \(\frac{\pi}{2}\)
(b) \(\frac{\pi}{5}\)
(c) \(\frac{\pi}{4}\)
(d) π
Solution:
(c) \(\frac{\pi}{4}\)

Question 22.
Evaluate cos^-1 (\(\frac{1}{2}\)) + 2sin^-1 (\(\frac{1}{2}\)).
(a) \(\frac{2 \pi}{5}\)
(b) \(\frac{2 \pi}{3}\)
(c) π
(d) \(\frac{\pi}{3}\)
Solution:
(b) \(\frac{2 \pi}{3}\)

Question 23.
Find the value of tan^-1 √3 – sec^-1 (-2).
(a) –\(\frac{\pi}{3}\)
(b) \(\frac{2 \pi}{3}\)
(c) \(\frac{\pi}{3}\)
(d) \(\frac{3 \pi}{2}\)
Solution:
(a) –\(\frac{\pi}{3}\)

Question 24.
Evaluate tan (2 tan^-1 \(\frac{1}{3}\))
(a) 2
(b) 1
(c) 0
(d) -1
Solution:
(a) 2

Question 25.
Evaluate : sin^-1 (sin \(\frac{3 \pi}{5}\)).
(a) \(-\frac{2 \pi}{3}\)
(b) \(\frac{\pi}{3}\)
(c) \(\frac{2 \pi}{5}\)
(d) \(\frac{\pi}{5}\)
Solution:
(c) \(\frac{2 \pi}{5}\)

Question 26.
tan^-1 (2cos\(\frac{\pi}{3}\)) is ________.
(a) \(\frac{\pi}{2}\)
(b) \(\frac{\pi}{4}\)
(c) \(\frac{\pi}{5}\)
(d) \(\frac{\pi}{3}\)
Solution:
(b) \(\frac{\pi}{4}\)

Question 27.
Evaluate : sin^-1 (sin \(\frac{2 \pi}{3}\)) is ________.
(a) \(\frac{\pi}{2}\)
(b) \(\frac{\pi}{4}\)
(c) \(\frac{\pi}{5}\)
(d) \(\frac{\pi}{3}\)
Solution:
(d) \(\frac{\pi}{3}\)

Question 28.
The value of sin(tan^-1 x + tan^-1 \( \frac{1}{x}\)), x > 0 = ________.
(a) -1
(b) 1
(c) 0
(d) 2
Solution:
(b) 1

Question 29.
2sin^-1 \( \frac{4}{5}\) + sin^-1 \( \frac{24}{25}\) = ________.
(a) 12
(b) 15
(c) 16
(d) 20
Solution:
(b) 15

Question 30.
Evaluate: tan^-1 1 = (2cos\(\frac{\pi}{3}\))
(a) \(\frac{\pi}{4}\)
(b) \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{3}\)
(d) \(\frac{2 \pi}{3}\)
Solution:
(a) \(\frac{\pi}{4}\)

Question 31.
If sin^-1 (\( \frac{\pi}{5}\)) + cosec^-1 (\(\frac{5}{4}\)) = \(\frac{5}{2}\) then find the value of x.
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(c) 3

Question 32.
Evaluate:
tan^-1 (\( \frac{-1}{\sqrt{3}}\)) + cot^-1 (\( \frac{1}{\sqrt{3}}\)) + tan^-1(sin(\( -\frac{\pi}{2}\))).
(a) \(\frac{- \pi}{12}\)
(b) \(\frac{2 \pi}{5}\)
(c) \(\frac{\pi}{12}\)
(d) \(\frac{\pi}{6}\)
Solution:
(a) \(\frac{- \pi}{12}\)

Question 33.
Evaluate sin^-1 (cos(\( \frac{33 \pi}{5}\)))
(a) \(\frac{\pi}{10}\)
(b) \(\frac{- \pi}{10}\)
(c) \(\frac{\pi}{5}\)
(d) \(\frac{\pi}{2}\)
Solution:
(b) \(\frac{- \pi}{10}\)

Question 34.
Express the value of the following in simplest form. tan (\( \frac{\pi}{4}\) + 2cot^-1 3)
(a) 7
(b) 12
(c) 3
(d) 6
Solution:
(a) 7

Question 35.
Express the value of the following in simplest form sin cos^-1 tan sec^-1 √2
(a) cos 0
(b) cot 0
(c) tan 0
(d) sin 0
Solution:
(d) sin 0

Question 36.
tan \(\left\{\frac{1}{2} \sin ^{-1} \frac{2 x}{1+x^2}+\frac{1}{2} \cos ^{-1} \frac{1-y^2}{1+y^2}\right\}\)
(a) \(\frac{x-y}{1+x y}\)
(b) \(\frac{x+y}{1-x y}\)
(c) \(\frac{x-y}{1+y}\)
(d) \(\frac{x+y}{x y}\)
Solution:
(b) \(\frac{x+y}{1-x y}\)

Question 37.
The relation R on the set A = [1, 2, 3] given by R = {(1, 1), (1, 2), (2, 2), (2, 3), (3, 3)} is:
(a) Reflexive
(b) Symmetric
(c) Transitive
(d) Equivalence
Solution:
(a) Reflexive

Question 38.
Let f : R → R be defined as f(x) = 3x – 2. Choose the correct answer.
(a) f is one-one onto
(b) f is many-one onto
(c) f is one-one but not onto
(d) f is neither one-one nor onto
Solution:
(a) f is one-one onto

Question 39.
Let R be a relation defined on Z as R = {(a, b) ; a² + b² = 25 }, the domain of R is:
(a) {3, 4, 5}
(b) {0, 3, 4, 5}
(c) {0, 3, 4, 5, -3, -4, -5}
(d) None of the above
Solution:
(c) {0, 3, 4, 5, -3, -4, -5}

Question 40.
let R be the relation in the set N given by R={(a, b) : a = b – 2, b > 6}. Choose the correct answer.
(a) (2, 4) • R
(b) (3, 8) • R
(c) (6, 8) • R
(d) (8, 10) • R
Solution:
(d) (8, 10) • R

Question 41.
Set A has 3 elements and set B has 4 elements. Then the number of injective functions that can be defined from set A to set B is
(a) 144
(b) 12
(c) 24
(d) 64
Solution:
(c) 24

Question 42.
Let R be a relation on set of lines as L₁ R L₂ if L₁ is perpendicular to L₂. Then
(a) R is Reflexive
(b) R is transitive
(c) R is symmetric
(d) R is an equivalence relation
Solution:
(c) R is symmetric

Question 43.
A Relation from A to B is an arbitrary subset of:
(a) A × B
(b) B × B
(c) A × A
(d) B × B
Solution:
(a) A × B

Question 44.
Let T be the set of all triangles in the Euclidean plane, and let a relation R on T be defined as aRb if a is congruent to b ∀ a, b ∈ T. Then R is
(a) reflexive but not transitive
(b) transitive but not symmetric
(c) equivalence
(d) None of these
Solution:
(c) equivalence

Question 45.
The maximum number of equivalence relations on the set A = {1, 2, 3} are
(a) 1
(b) 2
(c) 3
(d) 5
Solution:
(d) 5

Question 46.
Let A = {1, 2, 3} and consider the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}. Then R is
(a) reflexive but not symmetric
(b) reflexive but not transitive
(c) symmetric and transitive
(d) neither symmetric nor transitive
Solution:
(a) reflexive but not symmetric

Question 47.
Which of the following functions from Z into Z are bijective?
f(x) = x³
f(x) = x + 2
f(x) = 2x + 1
f(x) = x² + 1
Solution:
f(x) = x + 2

Question 48.
Let R be a relation on the set N of natural numbers denoted by nRm <=> n is a factor of m (i.e. n | m). Then, R is
(a) Reflexive and symmetric
(b) Transitive and symmetric
(c) Equivalence
(d) Reflexive, transitive but not symmetric
Solution:
(c) Equivalence

Question 49.
Let S = {1, 2, 3, 4, 5} and let A = S × S. Define the relation R on A as follows: (a, b) R (c, d) iff ad = cb. Then, R is
(a) reflexive only
(b) Symmetric only
(c) Transitive only
(d) Equivalence relation
Solution:
(d) Equivalence relation

Question 50.
Let X = {-1, 0, 1}, Y = {0, 2} and a function f : X → Y defined by y = 2x⁴, is
(a) one-one onto
(b) one-one into
(c) many-one onto
(d)many-one into
Solution:
(c) many-one onto

Question 51.
Let A = R – {3}, B = R – {1}. Let f : A → B be defined by f(x) = (x-2)/(x-3). Then,
(a) f is bijective
(b) f is one-one but not onto
(c) f is onto but not one-one
(d) None of these
Solution:
(a) f is bijective

Question 52.
The function f : R → R given by f(x) = x³ – 1 is
(a) a one-one function
(b) an onto function
(c) a bijection
(d) neither one-one nor onto
Solution:
(c) a bijection

Question 53.
Let f : [0, ∞) → [0, 2] be defined by f(x) = 2x/1+x, then f is
(a) one-one but not onto
(b) onto but not one-one
(c) both one-one and onto
(d) neither one-one nor onto
Solution:
(a) one-one but not onto

Question 54.
If N be the set of all natural numbers, consider f : N → N such that f(x) = 2x, ∀ × ∈ N, then f is
(a) one-one onto
(b) one-one into
(c) many-one onto
(d) None of these
Solution:
(b) one-one into

Question 55.
Let f : R → R be a function defined by f(x) = x³ + 4, then f is
(a) injective
(b) surjective
(c) bijective
(d) none of these
Solution:
(c) bijective

Question 56.
Given set A = {a, b, c}. An identity relation in set A is
(a) R = {(a, b), (a, c)}
(b) R = {(a, a), (b, b), (c, c)}
(c) R = {(a, a), (b, b), (c, c), (a, c)}
(d) R= {(c, a), (b, a), (a, a)}
Solution:
(b) R = {(a, a), (b, b), (c, c)}

Question 57.
Set A has 3 elements and the set B has 4 elements. Then the number of injective functions that can be defined from set A to set B is
(a) 144
(b) 12
(c) 24
(d) 64
Solution:
(c) 24

Question 58.
sin(sec^-1 x + cosec^-1 x) =
(a) 1
(b) -1
(c) π/2
(d) π/3
Solution:
(a) 1

Question 59.
The principle value of sin^-1 (√3/2) is:
(a) 2π/3
(b) π/6
(C) π/4
(d) π/3
Solution:
(d) π/3

Question 60.
Simplified form of cos^-1 (4x³ – 3x)
(a) 3 sin^-1 x
(b) 3 cos^-1 x
(c) π – 3 sin^-1 x
(d) None of these
Solution:
(b) 3 cos^-1 x

Question 61.
tan^-1 √3 – sec^-1 (-2) is equal to
(a) π
(b) -π/3
(c) π/3,
(d) 2π/3
Solution:
(b) -π/3

Question 62.
If y = sec^-1 x then
(a) 0 ≤ y ≤ π
(b) 0 ≤ y ≤ π/2
(c) -π/2 ≤ y ≤ π/2
(d) None of these
Solution:
(d) None of these

Question 63.
If x + (1/x) = 2 then the principal value of sin^-1 x is
(a) π/4
(b) π/2
(c) π
(d) 3π/2
Solution:
(b) π/2

Question 64.
The principle value of sin^-1 (sin 2π/3) is :
(a) 2π/3
(b) π/3
(c) -π/6
(d) π/6
Solution:
(b) π/3

Question 65.
The value of cos^-1 (1/2) + 2sin^-1 (1/2) is equal to
(a) π/4
(b) π/6
(c) 2π/3
(d) 5π/6
Solution:
(b) π/6

Question 66.
Principal value of tan^-1 (-1) is
(a) π/4
(b) -π/2
(c) 5π/4
(d) -π/4
Solution:
(d) -π/4

Question 67.
A Linear function, which is minimized or maximized is called
(a) an objective function
(b) an optimal function
(c) A feasible function
(d) None of these
Solution:
(a) an objective function

Question 68.
The maximum value of Z = 3x + 4y subject to the constraints : x+ y ≤ 4, x ≥ 0, y ≥ 0 is:
(a) 0
(b) 12
(c) 16
(d) 18
Solution:
(c) 16

Question 69.
The maximum value of Z = 2x +3y subject to the constraints : x + y ≤ 1, 3x + y ≤ 4, x, y ≥ 0 is
(a) 2
(b) 4
(c) 5
(d) 3
Solution:
(c) 5

Question 70.
The point in the half plane 2x + 3y – 12 ≥ 0 is:
(a) (-7,8)
(c) (-7,-8)
(b) (7, -8)
(d) (7, 8)
Solution:
(d) (7, 8)

Question 71.
Any feasible solution which maximizes or minimizes the objective function is Called:
(a) A regional feasible solution
(b) An optimal feasible solution
(c) An objective feasible solution
(d) None of these
Solution:
(b) An optimal feasible solution

Question 72.
Objective function of a LPP is
(a) a constraint
(b) a function to be optimized
(c) a relation between the variables
(d) none of these
Solution:
(b) a function to be optimized

(B) Very Short Type Questions With Answers

Question 1.
If R is a relation on A such that R = R^-1, then write the type of the relation R.
Solution:
We know that (a, b) ∈ R ⇒ (b, a) ∈ R^-1
As R = R^-1, so R is symmetric. [2019(A)

Question 2.
Write the value of cos^-1 cos (\(\frac{3 \pi}{2}\)). [2019(A)
Solution:
cos^-1 cos (\(\frac{3 \pi}{2}\)) = cos^-1 (0) = \(\frac{\pi}{2}\)

Question 3.
Sets A and B have respectively m and n elements. The total number of relations from A to B is 64. If m < n and m ≠ 1, write the values of m and n respectively. [2018(A)
Solution:
|A| = m and |B| = n
Number of relations from A to B = 2^mn.
A.T.Q. 2^mn = 64 = 2⁶.
⇒ mn = 6, m < n with m ≠ 1.
∴m = 2, n = 3

Question 4.
Write the principal value of sin^-1 (\(\frac{- 1}{2}\)) + cos^-1 cos(\(\frac{- \pi}{2}\)) [2018(A)
Solution:
sin^-1 (\(\frac{- 1}{2}\)) + cos^-1 (cos\(\frac{- \pi}{2}\)) = \(\frac{- \pi}{6}\) + \(\frac{\pi}{2}\) = \(-\frac{\pi}{3}\)

Question 5.
Write the maximum value of x + y subject to: 2x + 3y ≤ 6, x ≥ 0, y ≥ 0. [2011(A)
Solution:
2x + 3y = 6 intersects the axes at (3, 0) and (0, 2)
∴ The maximum value of x + y = 3.

Question 6.
Define ‘feasible’ solution of an LPP. [2009(A)
Solution:
The solutions of LPP which satisfy all the constraints and non-negative restrictions are called feasible solutions.

Question 7.
Mention the quadrant in which the solution of an LPP with two decision variables lies when the graphical method is adopted. [2008(A)
Solution:
The solution lies in XOY or 1st quadrant.

Question 8.
Write the smallest equivalence relation on A = {1, 2, 3}.
Solution:
The relation R = {(1, 1), (2, 2), (3, 3)} is the smallest equivalence relation on set A.

Question 9.
Congruence modulo 3 relation partitions the set Z into how many equivalence classes?
Solution:
The relation congruence modulo 3 on the set Z partitions Z into three equivalence classes.

Question 10.
Give an example of a relation which is reflexive, symmetric but not transitive.
Solution:
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.

Question 11.
Give an example of a relation which is reflexive, transitive but not symmetric.
Solution:
‘‘The relation x ≤ y on Z” is reflexive, transitive but not symmetric.

Question 12.
Give an example of a relation which is reflexive but neither symmetric nor transitive.
Solution:
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.

Question 13.
Find three positive integers x_i, i =1, 2, 3 satisfying 3x ≡ 2 (mod 7)
Solution:
3x ≡ 2 mod 7
Least positive value of x = 3
Each member of [3] is a solution
∴ x = 3, 10, 17…

Question 14.
State the reason for relation R on {1, 2, 3} defined as {(1, 2), (2, 1)} is not transitive.
Solution:
(1, 2), (2, 1) ∈ R but (1, 1) ∉ R
∴ R is not transitive.

Question 15.
Give an example of a function which is injective but not surjective.
Solution:
f(x) = \(\frac{x}{2}\) from Z → R is injective but not surjective.

Question 16.
Let X = {1, 2, 3, 4}. Determine whether
f : X → X defined as given below have inverses. Find f^-1 if it exists:
f = {(1, 2), (2, 2), (3, 2), (4, 2)}
Solution:
f is not injective hence not invertible.

Question 17.
Let ∗ is a binary operation defined by a ∗ b = 3a + 4b – 2, find 4 ∗ 5.
Solution:
4 ∗ 5 = 3 × 4 + 4 × 5 – 2
= 12 + 20 – 2
= 30

Question 18.
Let the binary operation on Q defined as a ∗ b = 2a + b – ab, find 3 ∗ 4.
Solution:
3 ∗ 4 = 6 + 4 – 12 = -2

Question 19.
Let ∗ is a binary operation on Z defined as a ∗ b = a + b – 5 find the identity element for ∗ on Z.
Solution:
Let e is the identity element.
⇒ a ∗ e = e ∗ a = a
⇒ a + e – 5 = a
⇒ e = 5

Question 20.
Find the number of binary operations on the set {a, b}.
Solution:
Number of binary operations on
{a, b} = 2²² = e⁴ =16.

Question 21.
Let * is a binary operation on [0, ¥) defined as a * b = \(\sqrt{\mathbf{a}^2+\mathbf{b}^2}\) find the identity element.
Solution:
Let e is the identity element
⇒ a * e = \(\sqrt{\mathbf{a}^2+\mathbf{e}^2}\) = a
⇒ a² + e² = a²
⇒ e = 0

Question 22.
Evaluate cos^-1 (\(\frac{1}{2}\)) + 2 sin^-1 (\(\frac{1}{2}\)).
Solution:
cos^-1 (\(\frac{1}{2}\)) + 2 sin^-1 (\(\frac{1}{2}\))
= \(\frac{\pi}{3}\) + 2 × \(\frac{\pi}{6}\) = \(\frac{2 \pi}{3}\)

Question 23.
Find the value of tan^-1 √3 – sec^-1 (-2)
Solution:
tan^-1 √3 – sec^-1 (-2)
= \(\frac{\pi}{3}\) – \(\frac{2 \pi}{3}\) = – \(\frac{\pi}{3}\).

Question 24.
Evaluate tan (2  tan^-1 \(\frac{1}{3}\))
Solution:
tan (2  tan^-1 \(\frac{1}{3}\)) = tan tan^-1 \(\left(\frac{\frac{2}{3}}{1-\frac{2}{3}}\right)\)
= tan tan^-1 (2) = 2

Question 25.
Evaluate: sin^-1 (sin \(\frac{3 \pi}{5}\)).
Solution:
sin^-1 (sin \(\frac{3 \pi}{5}\)) = sin^-1 sin (\(\pi \frac{-2 \pi}{5}\))
= sin^-1 sin \(\frac{2 \pi}{5}\) = \(\frac{2 \pi}{5}\)

Question 26.
Evaluate tan^-1 1 = (2 cos \(\frac{\pi}{3}\))
Solution:
tan^-1 (2 cos \(\frac{\pi}{3}\))
= tan^-1 (2 × 1/2) = tan^-1 1 = \(\frac{\pi}{4}\)

Question 27.
Define the objective function.
Solution:
If C₁, C₂, C₃ …. Cn are constants and x₁, x₂, …… x_n are variables then the linear function z = C₁x₁ + C₂x₂ +…… C_nx_n which is to be optimized is called an objective function.

Question 28.
Define feasible solution.
Solution:
A set of values of the variables x₁, x₂, …… x_n is called a feasible solution of LPP if it satisfies the constraints and non-negative restrictions of the problem.

Question 29.
Define a convex set.
Solution:
A set is convex if every point on the line segment joining any two points lies on it.

Question 30.
State extreme point theorem.
Solution:
Let S is a convex polygon bounded by the straight lines. The linear function z = Ax + By attains its optimum value at the vertices of S.

(C) Short Type Questions With Answers

Question 1.
Construct the multiplication table X₇ on the set {1, 2, 3, 4, 5, 6}. Also find the inverse element of 4 if it exists. [2019(A)
Solution:
Given set A = { 1, 2, 3, 4, 5, 6} Binary operation ∗ defined on A is X₇.
i.e. a ∗ b = a × b mod 7
= The remainder on dividing a × b by 7
The composition table for this operation is:

∗	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	1	3	5
3	3	6	2	5	1	4
4	4	1	5	2	6	3
5	5	3	1	6	4	2
6	6	5	4	3	2	1

As the second row is identical to first row, we have ‘1’ as the identity element.
As 4 ∗ 2 = 2 ∗ 4 = 1 we have 4^-1 = 2

Question 2.
Let R be a 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 \(\frac{1}{2}\). [2019(A)
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
\(\frac{1}{2}\) = { x ∈ R : x – \(\frac{1}{2}\) ∈ Z}
= {x ∈ R : x = \(\frac{2 k+1}{2}\), k ∈ Z}

Question 3.
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 1 kg of food as per the table given below: [2019(A)

Vitamin
Food	A	B	C
X	1	2	3
Y	2	2	1

1 kg of food X costs ₹16 and 1 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 kg of food X and Y kg of food y are to be mixed to prepare the mixture.
Total cost = 16x + 20y to be minimum.
According to the question
Total vitamin A = x + 2y ≥ 10 units
Total vitamin B = 2x + 2y ≥ 12 units
Total vitamin C = 3x + y ≥ 8 units.
∴ The required LPP is minimize
Z= 16x + 20y
Subject to : x + 2y ≥ 10
x + y ≥ 6
3x + y ≥ 8
x, y ≥ 0

Question 4.
Let ~ be defined by (m, n) ~ (p, q) if mq = np, where m, n, p, q e Z – {0}. Show that it is an equivalence relation. [2018(A)
Solution:
Let A = z – {0}
~ is a relation on A x A defined as (m, n) ~ (p, q) ⇔ mq = np
Reflexive : For all (m, n) ∈ A × A
We have mn = nm
⇒ (m, n) ~ (m, n)
∴ ~ is reflexive.
Symmetric: Let {m, n), (p, q) ∈ A × A and (m, n) ~ (p, q)
⇒ mq = np
⇒ pn = qm
(p, q) ~ (m, n)
∴ ~ is symmetric.
Transitive: Let (m, n), (p, q), (x, y) ∈ A × A
and (m, n) ~ (p, q), (p, q) ~ (x, y)
⇒ mq = np and py = qx
⇒ mqpy = npqx
⇒ my = nx
⇒ (m, n) ~ (x, y)
∴ ~ is transitive.
Thus ~ is an equivalence relation.

Question 5.
Solve the following LPP graphically:
Minimize Z = 4x + 3y
subject to 2x + 5y ≥ 10
x, y ≥ 0. [2018(A)
Solution:
Given LPP is:
Minimize: Z = 4x + 3y
Subject to: 2x + 5y ≥ 10
x, y ≥ 0
Step – 1 Considering the constraints as equations we have 2x + 5y = 10

x	5	0
y	0	2

Let us draw the graph.

Step – 2 As 0(0, 0) does not satisfy 2x + 5y > 0 and x, y > 0 is the first quadrant, we have the shaded region is the feasible region whose vertices are A(5, 0) B(0, 2).
Step – 3 Z (5, 0) = 20
Z (0, 2) = 6 … Minimum
As the feasible region is unbounded. Let us draw the half plane.
4x + 3y < 6

x	0	\(\frac{3}{2}\)
y	2	0

Step – 4 As there is no point common to the feasible region and the half plane 4x + 3y < 6, we have Z is minimum for x = 0, y = 2 and Z(min) = 6

Question 6.
Find the feasible region of the system 2y – x > 0, 6y – 3x < 21, x > 0, y > 0. [2017 (A)
Solution:
Step – 1: Treating the constraints as equations we have
2y – x = 0
6y – 3x = 21
⇒ 2y – x = 0
2y – x = 7
Step – 2: Let us draw the lines.
Table – 1

x	0	2
y	0	1

x	1	37
y	4	5

Step – 3: Clearly A(1, 3) Satisfies both the constraints, x > 0, y > 0 is the first quadrant. Thus the shaded region is the feasible region.

Question 7.
Solve the following LPP graphically:
Maximize: Z = 20x + 30y
Subject to: 3x + 5y ≤ 15
x, y ≥ 0. [2014 (A), 2016 (A), 2017 (A)
Solution:
Given LPP is
Maximize: Z = 20x + 30y
Subject to: 3x + 5y ≤ 15
x, y ≥ 0
Step – 1 Treating the constraints as equations we get 3x + 5y = 15.
Step- 2 Let us draw the graph

x	5	0
y	0	3

Step – 3
As 0(0, 0) satisfies 3x + 5y ≤ 15 the shaded region is the feasible region.
Step – 4
The vertices ofthe feasible region are 0(0, 0), A(5, 0) and B(0, 3).
Z(0) = 0, Z(A) = 100, Z(B) = 90
Z attains maximum at A for x = 5 and y = 0.
The given LPP has a solution, x = 5, y = 0 and Z(max) = 100.

Question 8.
Find the feasible region of the following system:
2x + y ≥ 6, x – y ≤ 3, x ≥ 0, y ≥ 0. [2016 (A)
Solution:
Given system of inequations are
2x + y ≥ 6, x – y ≤ 3, x ≥ 0, y ≥ 0
Step- 1: Consider 2x + y = 6
x – y = 3
Step – 2: Let us draw the graph
Table- 1

x	3	0
y	0	6

Table- 2

x	3	0
y	0	-3

Step – 3: 0(0, 0) satisfies x – y < 3, does not satisfy 2x + y > 6 and x > 0, y > 0 is the first quadrant. Thus the shaded region is the feasible region.

Question 9.
Solve the following LPP graphically:
Minimize: Z = 6x₁ + 7x₂
Subject to: x₁ + 2x₂ ≥ 1
x₁, x₂ ≥ 0. [2015 (A)
Solution:
Given LPP is
Minimize: Z = 6x₁+ 7x₂
Subject to: x₁ + 2x₂ ≥ 2
x₁, x₂ ≥ 0
Let us draw the line x₁ + 2x₂ = 2

x1	0	2
x2	1	0

Clearly (0, 0) does not satisfy x₁ + 2x₂ ≥ 2 and x₁, x₂ ≥ 0 is the first quadrant.
The shaded region is the feasible region.
The coordinates of vertices are A(2, 0) and B(0, 1).

Point	Z = 6x1 + 7x2
A (2, 0)	12
B (0,1)	7 → Minimum

As there is no point common to the half plane 6x₁ + 7x₂ < 7 and the feasible region.
Z is minimum when x₁ = 0, y₁ =1 and the minimum value of z = 7

Question 10.
Find the feasible region of the following system:
2y – x ≥ 0, 6y – 3x ≤ 21, x ≥ 0, y ≥ 0. [2015 (A)
Solution:
Given system is
2y – x ≥ 0
6y – 3x ≤ 21
x, y ≥ 0.
Considering the constraints as equations we have
2y – x = 0
and 6y – 3x = 21

x	0	2
x	0	1

⇒ 3y – x = 7

x	-7	2
x	0	3

Let us draw the lines

Clearly (2, 0) does not satisfy 2y – x ≥ 0 and satisfies 6y – 3x ≤ 21
∴ The shaded region is the feasible region.

Question 11.
Find the maximum value of z = 50x₁ + 60x₂
subject to 2x₁ + 3x₂ ≤ 6
x₁, x₂ ≥ 0. [2013 (A)
Solution:
Let us consider the constraints as equations.
2x₁ + 3x₂ = 6 … (1)
The table of some points on (1) is

x1	0	3
x2	2	0

Let us draw the line 2x₁ + 3x₂ = 6

As (0, 0) satisfies the inequality 2x₁ + 3x₂ ≤ 6 and x₁, x₂ ≥ 0 is the first quadrant, the shaded region is the feasible region with corner points O(0, 0), A(3, 0) and B(0, 2).

Corner point	z = 50x1 + 60x2
O(0, 0)	0
A(3, 0)	150
B(0, 2)	120

Thus Z_(max) = 150 for x₁ = 3, x₂ = 0

Question 12.
Shade the feasible region satisfying the inequations 2x + 3y ≤ 6, x ≥ 0, y ≥ 0 in a rough sketch. [2011(A)
Solution:
Let us consider the line 2x + 3y = 6

x1	0	3
x2	2	0

Let us draw the line on the graph

The feasible region is shaded in the figure.

Question 13.
Show the feasible region for the following constraints in a graph:
2x + y ≤ 4, x ≥ 0, y ≥ 0. [2010(A)
Solution:
Let us draw the graph of 2x + y = 4.

The shaded region shows the feasible region.

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

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

Question 1.
Evaluate the following determinants.
(i) \(\left|\begin{array}{ll}
1 & 1 \\
2 & 3
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ll}
1 & 1 \\
2 & 3
\end{array}\right|\) = 3 – 2 = 1

(ii) \(\left|\begin{array}{ll}
2 & -3 \\
1 & -4
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ll}
2 & -3 \\
1 & -4
\end{array}\right|\) = -8 + 3 = -5

(iii) \(\left|\begin{array}{ll}
\sec \theta & \tan \theta \\
\tan \theta & \sec \theta
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ll}
\sec \theta & \tan \theta \\
\tan \theta & \sec \theta
\end{array}\right|\) = sec² θ – tan² θ = 1

(iv) \(\left|\begin{array}{ll}
0 & x \\
2 & 0
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ll}
0 & x \\
2 & 0
\end{array}\right|\) = 0 – 2x = -2x

(v) \(\left|\begin{array}{cc}
1 & \omega \\
-\omega & \omega
\end{array}\right|\)
Solution:
\(\left|\begin{array}{cc}
1 & \omega \\
-\omega & \omega
\end{array}\right|\) = ω + ω² = -1

(vi) \(\left|\begin{array}{cc}
4 & -1 \\
3 & 2
\end{array}\right|\)
Solution:
\(\left|\begin{array}{cc}
4 & -1 \\
3 & 2
\end{array}\right|\) = 8 + 3 = 11

(vii) \(\left|\begin{array}{ll}
\cos \theta & \sin \theta \\
\sin \theta & \cos \theta
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ll}
\cos \theta & \sin \theta \\
\sin \theta & \cos \theta
\end{array}\right|\) = cos² θ – sin² θ = cos 2θ

(viii) \(\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right|\) = 0
as the rows are identical.

(ix) \(\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|\) = 1\(\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\) = 1 – 0 = 1

(x) \(\left|\begin{array}{ccc}
2 & 3 & 1 \\
0 & 0 & 0 \\
-1 & 2 & 0
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
2 & 3 & 1 \\
0 & 0 & 0 \\
-1 & 2 & 0
\end{array}\right|\) = 0
as all the entries in the 2nd row are zero.

(xi) \(\left|\begin{array}{ccc}
1 & x & y \\
0 & \sin x & \sin y \\
0 & \cos x & \cos y
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
1 & x & y \\
0 & \sin x & \sin y \\
0 & \cos x & \cos y
\end{array}\right|\) = 1\(\left|\begin{array}{cc}
\sin x & \sin y \\
\cos x & \cos y
\end{array}\right|\)
= sin x cos y – cos x sin y = sin (x – y)

(xii) \(\left|\begin{array}{lll}
1 & 2 & 3 \\
1 & 2 & 3 \\
3 & 4 & 5
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
1 & 2 & 3 \\
1 & 2 & 3 \\
3 & 4 & 5
\end{array}\right|\) = 0 (∵ R₁ = R₂)

(xiii) \(\left|\begin{array}{lll}
0.2 & 0.1 & 3 \\
0.4 & 0.2 & 7 \\
0.6 & 0.3 & 2
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
0.2 & 0.1 & 3 \\
0.4 & 0.2 & 7 \\
0.6 & 0.3 & 2
\end{array}\right|\)
= 2\(\left|\begin{array}{lll}
0.2 & 0.1 & 3 \\
0.4 & 0.2 & 7 \\
0.6 & 0.3 & 2
\end{array}\right|\) = 0 (∵ C₁ = C₂)

(xiv) \(\left|\begin{array}{ccc}
1 & \omega & \omega^2 \\
\omega & \omega^2 & 1 \\
\omega^2 & 1 & \omega
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
1 & \omega & \omega^2 \\
\omega & \omega^2 & 1 \\
\omega^2 & 1 & \omega
\end{array}\right|\)

(xv) \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 2 & 2 \\
3 & 3 & 3
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 2 & 2 \\
3 & 3 & 3
\end{array}\right|\) = 0 (∵ C₁ = C₂)

(xvi) \(\left|\begin{array}{ccc}
-6 & 0 & 0 \\
3 & -5 & 7 \\
2 & 8 & 11
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
-6 & 0 & 0 \\
3 & -5 & 7 \\
2 & 8 & 11
\end{array}\right|\)
= (-6) \(\left|\begin{array}{cc}
-5 & 7 \\
8 & 11
\end{array}\right|\) = = (-6) (- 55 – 56)
= (-6) (-111) = 666

(xvii) \(\left|\begin{array}{lll}
1 & 0 & 0 \\
2 & 3 & 5 \\
4 & 1 & 3
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
1 & 0 & 0 \\
2 & 3 & 5 \\
4 & 1 & 3
\end{array}\right|\)
= 1 \(\left|\begin{array}{ll}
3 & 5 \\
1 & 3
\end{array}\right|\) = 9 – 5 = 4

(xviii) \(\left|\begin{array}{ccc}
-18 & 17 & 19 \\
3 & 0 & 0 \\
-14 & 5 & 2
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
-18 & 17 & 19 \\
3 & 0 & 0 \\
-14 & 5 & 2
\end{array}\right|\)
= -3 \(\left|\begin{array}{cc}
17 & 19 \\
5 & 2
\end{array}\right|\)
(Expanding along 2nd row)
= – 3 (34 – 95)
= (-3) (-61) = 183

Question 2.
State true or false.
(i) If the first and second rows of a determinant be interchanged then the sign of the determinant is changed.
Solution:
True

(ii) If first and third rows of a determinant be interchanged then the sign of the determinent does not change.
Solution:
False

(iii) If in a third order determinant first row be changed to second column. Second row to 1st column and third row to third column, then the value of the determinant does not change.
Solution:
False

(iv) A row and a column of a determinant can have two or more common elements.
Solution:
False

(v) The minor and the co-factor of the element a₃₂ of a determinant of third order are equal.
Solution:
False

(vi) \(\left|\begin{array}{lll}
3 & 1 & 3 \\
0 & 4 & 0 \\
1 & 3 & 1
\end{array}\right|\) = 0
Solution:
True

(vii) \(\left|\begin{array}{lll}
6 & 4 & 2 \\
4 & 0 & 7 \\
5 & 3 & 4
\end{array}\right|\) = \(\left|\begin{array}{lll}
6 & 4 & 5 \\
4 & 0 & 3 \\
2 & 7 & 3
\end{array}\right|\)
Solution:
True

(viii) \(\left|\begin{array}{lll}
2 & 3 & 4 \\
5 & 6 & 7 \\
1 & 2 & 3
\end{array}\right|\) = \(\left|\begin{array}{lll}
4 & 2 & 3 \\
7 & 5 & 6 \\
3 & 1 & 2
\end{array}\right|\)
Solution:
True

Question 3.
Fill in the blanks with appropriate answer from the brackes.
(i) The value of \(\left|\begin{array}{ccc}
0 & 8 & 0 \\
25 & 520 & 25 \\
1 & 410 & 0
\end{array}\right|\) = _______. (0, 25, 200, -250)
Solution:
200

(ii) If ω is the cube root of unity, then \(\left|\begin{array}{ccc}
1 & \omega & \omega^2 \\
\omega & \omega^2 & 1 \\
\omega^2 & 1 & \omega
\end{array}\right|\) = _______. (1, 0, ω, ω²)
Solution:
0

(iii) The value of the determinant \(\left|\begin{array}{lll}
1 & a & b+c \\
1 & b & c+a \\
1 & c & a+b
\end{array}\right|\) = _______. (a + b – c, (a + b + c)², 0, 1 + a + b + c)
Solution:
0

(iv) If \(\left|\begin{array}{lll}
a & b & c \\
b & a & b \\
x & b & c
\end{array}\right|\) = 0, then x = _______. (a, b, c, a + b + c)
Solution:
a

(v) \(\left|\begin{array}{lll}
a_1+a_2 & a_3+a_4 & a_5 \\
b_1+b_2 & b_3+b_4 & b_5 \\
c_1+c_2 & c_3+c_4 & c_5
\end{array}\right|\) can be expressed at the most as _______, different 3rd order determinants. (1, 2, 3, 4)
Solution:
4

(vi) Minimum value of \(\left|\begin{array}{cc}
\sin x & \cos x \\
-\cos x & 1+\sin x
\end{array}\right|\) is _______. (-1, 0, 1, 2)
Solution:
0

(vii) The determinant \(\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 3 \\
1 & 3 & 6
\end{array}\right|\) is not equal to _______. \(\left(\left|\begin{array}{lll}
2 & 1 & 1 \\
2 & 2 & 3 \\
2 & 3 & 6
\end{array}\right|,\left|\begin{array}{lll}
2 & 1 & 1 \\
3 & 2 & 3 \\
4 & 3 & 6
\end{array}\right|,\left|\begin{array}{lll}
1 & 2 & 1 \\
1 & 5 & 3 \\
1 & 9 & 6
\end{array}\right|,\left|\begin{array}{ccc}
3 & 1 & 1 \\
6 & 2 & 3 \\
10 & 3 & 6
\end{array}\right|\right)\)
Solution:
\(\left|\begin{array}{lll}
2 & 1 & 1 \\
2 & 2 & 3 \\
2 & 3 & 6
\end{array}\right|\)

(viii) With 4 different elements we can construct _______ number of different determinants of order 2. (1, 6, 8, 24)
Solution:
6

Question 4.
Solve the following:
(i) \(\left|\begin{array}{cc}
4 & x+1 \\
3 & x
\end{array}\right|\)
Solution:
\(\left|\begin{array}{cc}
4 & x+1 \\
3 & x
\end{array}\right|\) = 5
or, 4x – 3x – 3 = 5 or, x = 8

(ii) \(\left|\begin{array}{ccc}
\boldsymbol{x} & a & a \\
m & m & m \\
b & x & b
\end{array}\right|\) = 0
Solution:

(Replacing C₁ and C₂ by C₁ – C₃ and C₂ – C₃ respectively)
⇒ m |(x – a) (-x + b)| = 0
⇒ m (x – a) (b – x) – 0 ⇒ x = a, b

(iii) \(\left|\begin{array}{lll}
7 & 6 & x \\
2 & x & 2 \\
x & 3 & 7
\end{array}\right|\) = 0
Solution:

or, (x – 7) (7x + x² – 1 8) = 0
or, (x – 7) (x² + 7x – 18) = 0
or, (x – 7) (x + 9) (x – 2) = 0
∴ x = -9, 2, 7

(iv) \(\left|\begin{array}{ccc}
0 & x-a & x-b \\
x+a & 0 & x-c \\
x+b & x+c & 0
\end{array}\right|\) = 0
Solution:

or, – (x – a) {0 – (x + b) (x – c)} + (x – b) (x + a) (x + c) = 0
or, (x – a) (x + b) (x – c) + (x – b) (x + a) (x + c) = 0
or, (x² + bx – ax – ab) (x – c) + (x² + ax – bx – ab) (x + c) = 0
or, x³ – cx² + bx² – bcx – ax² + acx – abx + abc + x³ + cx² + ax² + acx- bx² – bcx – abx – abc = 0
or, 2x³ – 2abx – 2bcx + 2acx = 0
or, 2x (x² – ab – bc + ac) = 0
x = 0, x² = ab + bc – ca
∴ x = 0, x = \(\sqrt{a b+b c-c a}\)

(v) \(\left|\begin{array}{ccc}
\boldsymbol{x}+\boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c} \\
\boldsymbol{b} & \boldsymbol{x}+\boldsymbol{c} & \boldsymbol{a} \\
\boldsymbol{c} & \boldsymbol{a} & \boldsymbol{x}+\boldsymbol{b}
\end{array}\right|\) = 0
Solution:

⇒ (x + a + b + c) {x² + bx + cx + bc – a² – bx – b² + ca + ab – cx – c² = 0}
⇒ (x + a + b + c) {x² – a² – b² – c² + ab + bc + ca} = 0
⇒ x + a + b + c = 0
or, x² – a² – b² – c² + ab + bc + ca = 0
⇒ x = – (a + b + c)
∴ or x = \(\sqrt{a^2+b^2+c^2-a b-b c-c a}\)

(vi) \(\left|\begin{array}{ccc}
1+x & 1 & 1 \\
1 & 1+x & 1 \\
1 & 1 & 1+x
\end{array}\right|\) = 0
Solution:

(vii) \(\left|\begin{array}{ccc}
1 & 4 & 20 \\
1 & -2 & 5 \\
1 & 2 x & 5 x^2
\end{array}\right|\) = 0
Solution:

⇒ -30x² + 30 + 30x + 30 = 0
⇒ -30x² + 30x + 60 = 0
⇒ x² – x – 2 = 0
⇒ x² – 2x + x + 2 = 0
⇒ (x – 2) (x + 1) = 0
⇒ x = 2, -1

(viii) \(\left|\begin{array}{ccc}
x+1 & \omega & \omega^2 \\
\omega & x+\omega^2 & 1 \\
\omega^2 & 1 & x+\omega
\end{array}\right|\) = 0
Solution:

⇒ x(x² + xω – xω² + xω² + ω³ – ω⁴ – xω – ω² + ω³ – 1 + ω² + ω – ω³) = 0
⇒ x(x² + ω³ – ω⁴ – ω² + ω³ – 1 + ω² + ω – ω³) = 0
⇒ x(x² + ω³ – ω + ω – 1) = 0
⇒ x(x² + 1 – ω + ω – 1) = 0 (∵ ω³ = 1)
⇒ x³ = 0
⇒ x = 0

(ix) \(\left|\begin{array}{ccc}
2 & 2 & x \\
-1 & x & 4 \\
1 & 1 & 1
\end{array}\right|\) = 0
Solution:

or, 10 + 10x – x² – 8x – x – 8 = 0
or, x² – 2x + x – 2 = 0
or, (x – 2) (x + 1) = 0
x = 2, x = -1

(x) \(\left|\begin{array}{lll}
x & 1 & 3 \\
1 & x & 1 \\
3 & 6 & 3
\end{array}\right|\) = 0
Solution:

or, x(3x – 6) – 0 + 3(6 – 3x) = 0
or, 3x² – 6x + 18 – 9x = 0
or, 3x² – 15x + 18 = 0
or, x² – 5x + 6 = 0
or, (x – 3) (x – 2) = 0
x = 3 or, x = 2

Question 5.
Evaluate the following
(i) \(\left|\begin{array}{ccc}
2 & 3 & 4 \\
1 & -1 & 3 \\
4 & 1 & 10
\end{array}\right|\)
Solution:

(ii) \(\left|\begin{array}{lll}
\boldsymbol{x} & \mathbf{1} & 2 \\
\boldsymbol{y} & \mathbf{3} & 1 \\
z & 2 & 2
\end{array}\right|\)
Solution:

(iii) \(\left|\begin{array}{ccc}
x & 1 & -1 \\
2 & y & 1 \\
3 & -1 & z
\end{array}\right|\)
Solution:

= x (yz + z – z + 1) – (2z – 2 – 3y – 3)
= xyz + x – 2z + 3y + 5
= xyz + x + 3y – 2z + 5

(iv) \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|\)
Solution:

= a(bc – f²) – h (ch – fg) + g (hf – bg)
= abc – af² – ch² + fgh + fgh – bg²
= abc + 2fgh – af² – bg² – ch²

(v) \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|\)
Solution:

(vi) \(\left|\begin{array}{ccc}
\sin ^2 \theta & \cos ^2 \theta & 1 \\
\cos ^2 \theta & \sin ^2 \theta & 1 \\
-10 & 12 & 2
\end{array}\right|\)
Solution:

(vii) \(\left|\begin{array}{ccc}
-1 & 3 & 2 \\
1 & 3 & 2 \\
1 & -3 & -1
\end{array}\right|\)
Solution:

(viii) \(\left|\begin{array}{ccc}
11 & 23 & 31 \\
12 & 19 & 14 \\
6 & 9 & 7
\end{array}\right|\)
Solution:

(ix) \(\left|\begin{array}{ccc}
37 & -3 & 11 \\
16 & 2 & 3 \\
5 & 3 & -2
\end{array}\right|\)
Solution:

(x) \(\left|\begin{array}{ccc}
2 & -3 & 4 \\
-4 & 2 & -3 \\
11 & -15 & 20
\end{array}\right|\)
Solution:

= 2(40 – 45) + 3(-80 + 33) + 4(60 – 22)
= -10 – 141 + 152 = -151 + 152 = 1

Question 6.
Show that x = 1 is a solution of \(\left|\begin{array}{ccc}
x+1 & 3 & 5 \\
2 & x+2 & 5 \\
2 & 3 & x+4
\end{array}\right|\) = 0
Solution:

or, (x + 9) {(x – 1)²} – 0
or, x = -9, 1
∴ x = 1 is a solution of the given equation.

Question 7.
Show that (a + 1) is a factor of \(\left|\begin{array}{ccc}
a+1 & 2 & 3 \\
1 & a+1 & 3 \\
3 & -6 & a+1
\end{array}\right|\) = 0
Solution:

= (a+ 1) {(a + 1)² + 18} – 2(a + 1 – 9) + 3(- 6 – 3a – 3)
= (a + 1) (a² + 2a + 1 + 18) – 2(a – 8) + 3(- 9 – 3a)
= (a + 1) (a² + 2a + 19) – 2a + 16 – 27 – 9a
= (a + 1) (a² + 2a + 19) – 11a – 11
= (a + 1) (a² + 2a + 19) – 11(a + 1)
= (a + 1) (a² + 2a + 19 – 11)
= (a + 1) (a² + 2a + 8)
∴ (a + 1) is a factor of the above determinant.

Question 8.
Show that \(\left|\begin{array}{ccc}
a_1 & b_1 & -c_1 \\
-a_2 & b_2 & c_2 \\
a_3 & b_3 & -c_3
\end{array}\right|=\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|\)
Solution:

Question 9.
Prove the following
(i) \(\left|\begin{array}{lll}
a & b & c \\
\boldsymbol{x} & y & z \\
\boldsymbol{p} & q & r
\end{array}\right|=\left|\begin{array}{lll}
\boldsymbol{y} & \boldsymbol{b} & \boldsymbol{q} \\
\boldsymbol{x} & \boldsymbol{a} & p \\
z & c & r
\end{array}\right|=\left|\begin{array}{lll}
\boldsymbol{x} & \boldsymbol{y} & z \\
\boldsymbol{p} & \boldsymbol{q} & r \\
a & b & c
\end{array}\right|\)
Solution:

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

(iii) \(\left|\begin{array}{lll}
b+c & c+a & a+b \\
q+r & r+p & p+q \\
y+z & z+x & x+y
\end{array}\right|=2\left|\begin{array}{lll}
a & b & c \\
p & q & r \\
x & y & z
\end{array}\right|\)
Solution:

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

(v) \(\left|\begin{array}{ccc}
a+d & a+d+k & a+d+c \\
c & c+b & c \\
d & d+k & d+c
\end{array}\right|\) = abc
Solution:

(vi) \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
b+c & c+a & c+a \\
b^2+c^2 & c^2+a^2 & a^2+b^2
\end{array}\right|\) = (b – c) (c – a) (a – b)
Solution:

(vii) \(\left|\begin{array}{lll}
a & a^2 & a^3 \\
b & b^2 & b^3 \\
c & c^2 & c^3
\end{array}\right|\) = abc (a – b) (b – c) (c – a)
Solution:

(viii) \(\left|\begin{array}{ccc}
\boldsymbol{b}+\boldsymbol{c} & \boldsymbol{a} & \boldsymbol{a} \\
\boldsymbol{b} & \boldsymbol{c}+\boldsymbol{a} & \boldsymbol{b} \\
\boldsymbol{c} & \boldsymbol{c} & \boldsymbol{a}+\boldsymbol{b}
\end{array}\right|\) = 4abc
Solution:

= (b + c – a) {(a + b) (c + a – b) – b (c – a – b)} + a (b – c – a) (c – a – b)
= (b + c – a)(ca + a² – ab + bc + ab – b² – bc + ab + b²) + a(bc – ab – b² – c² + ca + bc – ac + a² + ab)
= (b + c – a) (a² + ab + ca) + a (a² – b² – c² + 2bc)
= a²b + ab² + abc + ca² + abc + c²a – a³ – a²b – ca² + a³ – b²a – c²a + 2abc = 4abc

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

(x) \(\left|\begin{array}{ccc}
a & b & c \\
a^2 & b^2 & c^2 \\
b c & c a & a b
\end{array}\right|\) = (b – c) (c – a) (a – b) (bc + ca + ab)
Solution:

= (a – b) (b – c) (c – a) – (- ab + c²) + c (a + b + c)
= (a – b) (b – c) (c – a) (ab – c² + ca + bc + c²)
= (a – b) (b – c) (c – a) (ab + bc + ca)

(xi) \(\left|\begin{array}{ccc}
a-b-c & 2 a & 2 a \\
2 b & b-c-a & 2 b \\
2 c & 2 c & c-a-b
\end{array}\right|\) = (a + b+ c)³
Solution:

(xii) \(\left|\begin{array}{ccc}
(v+w)^2 & u^2 & u^2 \\
v^2 & (w+u)^2 & v^2 \\
w^2 & w^2 & (u+v)^2
\end{array}\right|\) = 2uvw (u + v + w)³
Solution:

Question 10.
Factorize the following
(i) \(\left|\begin{array}{ccc}
x+a & b & c \\
b & x+c & a \\
c & a & x+b
\end{array}\right|\)
Solution:

= (x + a + b + c) [(x + c – b) (x + b – a) – (a – c) (a – x – c)]
= (x + a + b + c) (x² + xb – ax + cx +bc – ca – bx – b² + ab – a² + ax + ac + ac – cx – c²)
= (x + a + b + c) (x² – a² – b² – c² + ab + bc + ca)

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

(iii) \(\left|\begin{array}{ccc}
x & 2 & 3 \\
1 & x+1 & 3 \\
1 & 4 & x
\end{array}\right|\)
Solution:

Question 11.
Show that by eliminating α and from the equations.
a_i α + b_i β + c_i = 0, i = 1, 2, 3 we get \(\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_2 & b_3 & c_3
\end{array}\right|\) = 0
Solution:
We have
a₁ α + b₁ β + c₁ = 0    …..(1)
a₂ α + b₂ β + c₂ = 0    …..(2)
a₃ α + b₃ β + c₃ = 0    …..(3)
Solving (2) and (3) by cross-multiplication method we have

Question 12.
Prove the following:
(i) \(\left|\begin{array}{lll}
1 & b c & a(b+c) \\
1 & c a & b(c+a) \\
1 & a b & c(a+b)
\end{array}\right|\) = 0
Solution:

(ii) \(\left|\begin{array}{ccc}
x+4 & 2 x & 2 x \\
2 x & x+4 & 2 x \\
2 x & 2 x & x+4
\end{array}\right|\) = (5x + 4) (4- x)²
Solution:

(iii) \(\left|\begin{array}{l}
\sin \alpha \cos \alpha \cos (\alpha+\delta) \\
\sin \beta \cos \beta \cos (\beta+\delta) \\
\sin \alpha \cos \gamma \cos (\gamma+\delta)
\end{array}\right|\) = 0
Solution:

(iv) \(\left|\begin{array}{ccc}
1 & x & x^2 \\
x^2 & 1 & x \\
x & x^2 & 1
\end{array}\right|\) = (1 -x³)²
Solution:

Question 13.
Prove that the points (x₁, y₁), (x₂, y₂), (x₃, y₃) are collinear if \(\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|\) = 0
Solution:
From geometry, we know that, if the points A, B, C, are collinear, then the area of the triangle ABC with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is zero.
\(\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|\) = 0

Question 14.
If A + B + C = π, prove that \(\left|\begin{array}{lll}
\sin ^2 A & \cot A & 1 \\
\sin ^2 B & \cot B & 1 \\
\sin ^2 C & \cot C & 1
\end{array}\right|\) = 0
Solution:

Question 15.
Eliminate x, y, z from a = \(\frac{x}{y-z}\), b = \(\frac{y}{z-x}\), c = \(\frac{z}{x-y}\)
Solution:
We have
a = \(\frac{x}{y-z}\), b = \(\frac{y}{z-x}\), c = \(\frac{z}{x-y}\)
ay – az – x = 0, bz – bx – y = 0, cx – cy – z = 0
x – ay + az = 0
bx + y – bz = 0
cx – cy – z = 0
Now eliminating x, y, z from the above equations we have,

or, – 1 – bc + a(-b + bc) + a(-bc – c) = 0
or, – 1 – bc – ab + abc – abc – ac = 0
or, ab + bc + ca + 1 = 0

Question 16.
Given the equations
x = cy + bz, y = az + ex and z = bx + ay where x, y and z are not all zero, prove that a² + b² + c² + 2abc = 1 by determinant method.
Solution:
x = cy + bz, y = az + cx and z = bx + ay

or, 1 – a² + c(-c – ab) – b(ca + b) = 0
or, 1 – a² – c² – abc – abc – b² = 0
or, a² + b² + c² + 2abc = 1

Question 17.
If ax + hy + g = 0, hx + by +f = 0 and gx + fy + c = λ, find the value of λ, in the form of a determinant.
Solution: