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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 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 BSE Odisha 7th Class English Solutions Follow-Up Lesson 8 The Fisherman and The Jinni Textbook Exercise Questions and Answers.

BSE Odisha Class 7 English Solutions Follow-Up Lesson 8 The Fisherman and The Jinni

BSE Odisha 7th Class English Follow-Up Lesson 8 The Fisherman and The Jinni Text Book Questions and Answers

Session – 1

I. Pre-Reading

• Socialisation (ସାମାଜିକୀକରଣ)
• The teacher prepares a good pre-reading activity on his / her own as done for the main lesson

II. While-Reading

• Read the following text silently and answer the questions that follow.
(ନିମ୍ନ ବିଷୟଟିକୁ ନୀରବରେ ପଢ଼ ଏବଂ ନିମ୍ନ ପ୍ରଶ୍ନଗୁଡିକର ଉତ୍ତର ଦିଅ ।)
There lived a poor fisherman with his wife and children in a hut near the sea. Everyday he used to go to the sea and catch fish. ’
One day, as usual, he went to the sea. He threw his net into the sea, and when he pulled it out, he felt it very heavy. But there was no fish in it. He saw a log of wood in the net. He felt sad. He threw the net for the second time. This time he got a few shells and big stones. When he tried it for the third time, he found it heavier. This time too he did not see any fish in it but a big brass jar with a lid.
Taking the jar out of the net, he opened it. A lot of smoke came out of it. Suddenly a jinni appeared in the smoke. On seeing the jinni, he screamed. “I’ll kill you now” shouted the jinni.
“Why ?” asked the fisherman and said, “I haven’t done you any harm. Please don’t kill me.”
“I was in the jar. I was not free. Now I am free. I will kill you and eat you up” said the jinni. The fisherman was afraid of the jinni. He did not know what to do. Suddenly he got a clever idea and said, “Alright, you can kill me but I don’t believe what you say. You are very big. How could you come out of this jar ?”
The jinni got angry. “How dare you say I cannot come out of this jar ? I can take any form. I can take the form of an elephant or become even a small ant,” said the jinni.
“Is it so ? Can you become an ant ?” said the fisherman cleverly.
“Surely,” said the jinni and immediately took the form of an ant and got into the jar again. The fisherman immediately shut the jar tight and said with a grin, “Now I believe your story, my friend, you cannot come out any more and kill me ?” Then he threw the jar into the sea and returned home happily.

ଓଡ଼ିଆ ଅନୁବାଦ :
ସମୁଦ୍ର ପାଖରେ ଗୋଟିଏ କୁଡ଼ିଆ ଘରେ ଜଣେ ବିଚାରା କେଉଟ ତା’ର ସ୍ତ୍ରୀ ଏବଂ ପିଲାମାନଙ୍କୁ ନେଇ ବାସ କରୁଥିଲା । ପ୍ରତ୍ୟେକ ଦିନ ସେ ସମୁଦ୍ରକୁ ଯାଇ ମାଛ ଧରିବାରେ ଅଭ୍ୟସ୍ତ ଥିଲା ।
ଦିନେ ସ୍ୱାଭାବିକ ଭାବରେ ସେ ସମୁଦ୍ରକୁ ଗଲା । ସେ ତା’ ଜାଲକୁ ସମୁଦ୍ର ଭିତରକୁ ଫିଙ୍ଗିଲା ଏବଂ ଯେତେବେଳେ ସେ ତାହାକୁ ବାହାରକୁ ଟାଣିଲା ତାକୁ ବହୁତ ଓଜନିଆ ଜଣାଗଲା । କିନ୍ତୁ ତା’ ଭିତରେ ମାଛ ନଥିଲା । ସେ ଖଣ୍ଡେ କାଠଗଣ୍ଡି ଜାଲ ଭିତରୁ ପାଇଲା । ସେ ଦୁଃଖ୍ ହେଲା । ସେ ଜାଲକୁ ଦ୍ଵିତୀୟବାର (ସମୁଦ୍ର ଭିତରକୁ) ଫିଙ୍ଗିଲା ।
ତାହା ଅଧ୍ଵକ ଓଜନିଆ ଲାଗିଲା । ଏଥର ମଧ୍ୟ ସେ ସେଥ‌ିରେ କୌଣସି ମାଛ ଦେଖିବାକୁ ପାଇଲା ନାହିଁ; କିନ୍ତୁ ଗୋଟେ ବଡ଼ ଠିପି ଲାଗିଥିବା ପିତ୍ତଳ ଜାର୍ ପାଇଲା ।
ଜାର୍‌ଟିକୁ ଜାଲରୁ ପଦାକୁ ଆଣି ସେ ତାହାକୁ ଖୋଲିଲା । ତା’ ମଧ୍ୟରୁ ବହୁତ ଧୂଆଁ ବାହାରିଲା । ହଠାତ୍ ସେଥ୍ ମଧ୍ୟରୁ ଗୋଟେ ଭୂତ ଆବିର୍ଭାବ ହେଲା । ଭୂତକୁ ଦେଖୁ ସେ ଚିହିଁକି |ଚିତ୍କାର କରି ଉଠିଲା । ଭୂତ ତାକୁ ଚିତ୍କାର କରି କହିଲା, ‘ବର୍ତ୍ତମାନ ମୁଁ ତୋତେ ମାରିଦେବି ।’’
“‘କାହିଁକି ?’’ କେଉଟ ପଚାରିଲା ଏବଂ କହିଲା, ‘ମୁଁ ତୁମର କିଛି କ୍ଷତି କରିନାହିଁ । ଦୟାକରି ମୋତେ ମାରନାହିଁ ।’’
ଭୂତ କହିଲା, ‘‘ମୁଁ ଜାର୍ ଭିତରେ ଥୁଲି । ମୁଁ ମୁକ୍ତ ନଥୁଲି । ବର୍ତ୍ତମାନ ମୁଁ ମୁକ୍ତ । ମୁଁ ତୁମକୁ ମାରିଦେବି ଏବଂ ଖାଇବି ।’’ କେଉଟ ଭୂତକୁ ଭୟ କରିଗଲା । ସେ କ’ଣ କରିବାକୁ ହେବ ଜାଣିପାରିଲା ନାହିଁ । ହଠାତ୍ ତା’ ମୁଣ୍ଡରେ ଗୋଟେ ଚାଲାକି ବୁଦ୍ଧି ଜୁଟିଲା ଏବଂ ସେ କହିଲା, ‘ଠିକ୍ ଅଛି, ତୁମେ ମୋତେ ମାରିଦେଇ ପାରିବ; କିନ୍ତୁ ତୁମେ ଯାହା କହୁଛ ମୁଁ ବିଶ୍ବାସ କରିପାରୁନି । ତୁମେ ତ ବହୁତ ବଡ଼ । ତୁମେ କିପରି ଏହି (ଛୋଟ) ଜାରରୁ ଆସିପାରିଲ ?’’
ଭୂତ ରାଗିଗଲା । ସେ କହିଲା, ‘ତୁମର କିପରି କହିବାକୁ ସାହସ ହେଲା ଯେ ମୁଁ ଏହି (ଛୋଟ) ଜାର୍ ଭିତରୁ ବାହାରକୁ ଆସିପାରି ନଥା’ନ୍ତି। ମୁଁ ଯେକୌଣସି ଆକାର ପ୍ରକାରର ହୋଇପାରେ । ମୁଁ ଗୋଟେ ହାତୀ ଆକୃତିର ହୋଇପାରେ ଏବଂ ଏପରିକି ଗୋଟେ ଛୋଟ ପିମ୍ପୁଡ଼ି ଆକାରର ମଧ୍ୟ ହୋଇପାରେ ।’’
କେଉଟ ବଡ଼ ଚାଲାକିରେ କହିଲା, ‘ଏହା କ’ଣ ତାହାହେଲେ ସେଇଆ ? ତୁମେ ଗୋଟେ ପିମ୍ପୁଡ଼ି ବି ହୋଇପାରିବ ?’’
ଭୂତ କହିଲା, ‘ନିଶ୍ଚୟ’ ଏବଂ ତତ୍‌କ୍ଷଣାତ୍ ଗୋଟିଏ ପିମ୍ପୁଡ଼ିର ଆକାର ଧାରଣ କରି ଜାର୍ ମଧ୍ୟକୁ ପୁନର୍ବାର ପ୍ରବେଶ କଲା । କେଉଟ ତତ୍‌କ୍ଷଣାତ୍ ଜାର ଠିପିଟିକୁ ଜୋର୍‌ରେ ବନ୍ଦ କରିଦେଲା ଏବଂ ଅଳ୍ପ ହସିକରି କହିଲା, ‘ବର୍ତ୍ତମାନ ମୁଁ ତୁମ କାହାଣୀକୁ ବିଶ୍ୱାସ କଲି, ହେ ବନ୍ଧୁ । ତୁମେ ଆଉ କେବେ ପଦାକୁ ଆସିପାରିବ ନାହିଁ କି ମୋତେ ମାରି ପାରିବ ନାହିଁ ।’’ ତା’ପରେ ସେ ଜାର୍‌ଟିକୁ ସମୁଦ୍ର ମଧକୁ ଫୋପାଡ଼ି ଦେଲା ଏବଂ ମନଖୁସିରେ ଘରକୁ ଫେରିଲା ।

Notes And Glossary:
fisherman (ଫିସର ମ୍ୟାନ୍ ) – କେଉଟ
hut (ହଟ୍) – କୁଡ଼ିଆଘର
sea – ସମୁଦ୍ର
used to (ଇଉଡ଼ ଟୁ) – ବରାବର । ନିୟମିତ ଭାବରେ
catch (କ୍ୟାଚ୍) – ଧରିବା
as usual (ଆଜ୍ ୟୁଜୁଆଲ୍) – ସବୁଦିନ ପରି
threw – ପକାଇଲା | ଫିଙ୍ଗିଲା
net (ନେଟ୍) – ଜାଲ
pulled (ପୁଲ୍‌) – ଟାଣିଲା | ଓଟାରିଲା
felt (ଫେଲ୍‌ଟ) – ଅନୁଭବ କଲା
heavy – ମୋଟା
log (ଲଗ) – କାଠଗଣ୍ଠି
wood (ଉଡ୍) – କାଠ
shells (ସେଲ୍‌ସ୍) – ଶାମୁକାସବୁ
stones (ଷ୍ଟୋନ୍‌ସ) – ଗୋଡ଼ିପଥର ସବୁ
tried (ଟ୍ରାଏଡ୍) – ଚେଷ୍ଟା କଲା
found (ଫାଉଣ୍ଡ୍) – ଦେଖୁବାକୁ ପାଇଲା
brass (ବ୍ରାସ୍ ) – ପିତ୍ତଳ
jar – ପାତ୍ର
lid – ଠିପି | ଢାଙ୍କୁ ଣି
smoke (ସ୍ପୋକ୍) – ଧୁଆଁ
suddenly (ସଡ଼ଲି) – ହଠାତ୍
jini – ଜିନି
appeared (ଆପିଅର୍‌ଡ଼) – ଉପସ୍ଥିତ ହେଲା
screamed (ସ୍କ୍ରିମ୍‌ଡ୍) – ଚିତ୍କାର କଲା
harm (ହାର୍ମ) – କ୍ଷତି
free – ମୁକ୍ତ
afraid (ଆଫ୍ରେଡ୍‌) – ଭୟଭୀତ
clever (କ୍ଲେଭର) – ଚତୁର
idea (ଆଇଡ଼ିଆ ) – ବୁଦ୍ଧି / ଯୋଜନା
believe (ବିଲିଭ୍ ) – ବିଶ୍ଵାସ କରିବା
angry (ଆଙ୍ଗ୍ରି) – ରାଗିଯିବା
dare (ଡେୟାର) – ସାହସ କରିବା-
form – ରୂପ / ଆକାର
elephant (ଏଲିଫାଣ୍ଟ୍‌) – ହାତୀ
ant (ଆଣୁ) – ପିମ୍ପୁଡ଼ି
small (ସ୍ମଲ୍) – ଛୋଟ
surely (ସିଓର୍‌ଲି) – ନିଶ୍ଚିତ ଭାବରେ
immediately (ଇମିଡ଼ିଏଟ୍‌ଲି) – ଖୁବ୍ ଶୀଘ୍ର
shut (ସଟ୍) – ବନ୍ଦ କରିଦେବା
tight (ଟାଇଟ୍) – କଠିନ ଭାବରେ
grin (ଗ୍ରିନ୍) – ଚିକ୍କଣ କରିବା
any more (ଏନି ମୋର) – ଆଉ କେବେ
kill – ହତ୍ୟା କରିବା
returned (ରିଟର୍ଣ୍ଣଡ୍) – ଫେରି ଆସିଲା
home (ହୋମ୍) – ଘରକୁ
happily (ହାପିଲି) – ଖୁସି | ଆନନ୍ଦରେ

Comprehension Questions

Question 1.
Who are there in this story ?
(ଏହି ଗପଟିରେ କେଉଁମାନେ ଅଛନ୍ତି ?)
Answer:
A fisherman and a jinny are there in the story.

Question 2.
Where did the fisherman live ?
(କେଉଟ କେଉଁଠାରେ ବାସ କରୁଥିଲା ?)
Answer:
The fisherman lived in a hut near the sea with his wife and children.

Question 3.
What did the fisherman do at the sea everyday ?
(କେଉଟ ପ୍ରତିଦିନ ସମୁଦ୍ରରେ କ’ଣ କରୁଥିଲା ?)
Answer:
The fisherman went to the sea everyday and caught fish.

Question 4.
What did the fisherman see in his net when he pulled it out third time ?
(କେଉଟ ତା’ ଜାଲରେ କ’ଣ ଦେଖିଲା ଯେତେବେଳେ ତୃତୀୟବାର ସେ ତାହା (ଜାଲ)କୁ ପଦାକୁ ଟାଣି ଆଣିଲା ?)
Answer:
When the fisherman pulled his net out third time, he saw a big brass jar with a lid in it.

Question 5.
What did the fisherman do with the brass jar ?
(କେଉଟ ପିତ୍ତଳ ଜାର୍‌ଟିକୁ କ’ଣ କଲା ?)
Answer:
The fisherman took the brass jar out of the net and opened it.

Question 6.
What happened when he opened the lid of the brass jar ?
(ସେ ପିତ୍ତଳ ଜାର୍‌ଟିର ଘୋଡ଼ଣି ଖୋଲିଦେଲା ପରେ କ’ଣ ହେଲା ? )
Answer:
When he opened the lid of the brass jar, a lot of smoke came out of it. Suddenly a jinni appeared in the smoke.

Question 7.
What did the jinni say to the fisherman when he came out of the jar ?
(ଭୂତଟି ଜାର୍ ମଧ୍ୟରୁ ବାହାରକୁ ଆସିଲା ପରେ କେଉଟକୁ କ’ଣ କହିଲା ?)
Answer:
When the jinni came out of the jar, he said to the fisherman to kill him.

Question 8.
What idea did the fisherman get to get rid of the jinni ?
(କେଉଟ ଭୂତ କବଳରୁ ରକ୍ଷା ପାଇବାକୁ କି ଉପାୟ ପାଞ୍ଚିଲା ?)
Answer:
Suddenly the fisherman got a clever idea. He planned to get the jinni into the brass jar again by his cleverness.

Question 9.
What form did the jinni take to get into the jar again ?
(ପୁନର୍ବାର ଜାର୍ ମଧ୍ୟରେ ପ୍ରବେଶ କରିବାପାଇଁ ଭୂତ କି ରୂପ ଧାରଣ କଲା ?)
Answer:
Jinni took the form of a small ant to get into the jar again.

Question 10.
What did the fisherman do when the jinni got into the jar ?
(ଭୂତ ଜାର୍ ମଧ୍ୟକୁ ପ୍ରବେଶ କଲାମାତ୍ରେ କେଉଟ କ’ଣ କଲା ?)
Answer:
When the jinni got into the jar, the fisherman suddenly shut the jar tight and threw it into the sea.

Question 11.
What did he do to the jar and the jinni in the end ?
(ସେ ସର୍ବଶେଷରେ ଜାର୍ ଏବଂ ଭୂତକୁ କ’ଣ କଲା ?)
Answer:
In the end, he threw the jar containing jinni into the sea.

Session – 2

III. Post-Reading

1. Visual Memory Development Technique (VMDT) :
(ଦୃଶ୍ୟ ସ୍ମୃତି ବିକାଶ କୌଶଳ)
The teacher prepares this activity on his / her own.

2. Comprehension Activities :
Given below are some sentences from the story. They are not in order. Arrange them in right order. Write serial numbers in brackets. (Question with Answer)

(i) Then he threw the jar into the sea. (8)
(ii) This time he got some stones in the sea. (4)
(iii) There appeared a jinni in the smoke. (5)
(iv) Once there lived a fisherman in a hut with his family near the sea. (1)
(v) Soon the jinni became an ant and got into the jar. (7)
(vi) One day when he pulled out the net, he saw a log of wood in it. (2)
(vii) He became sad and threw the net into the sea for the second time. (3)
(viii) The jinni said, “I’ll kill you.” (6)
Note to the teachers : [Frame other post-reading activities on your own]

Word Note:
Jinni (ଜିନ୍ନି |) – ghost (ଭୂତ)
beast – animal (ପଶୁ)
clever – wise (ଚତୁର, ଚାଲାକ)
distrust (ଡିସ୍‌ଟ୍ରଷ୍ଟ) – do not believe (ଅବିଶ୍ବାସ କରିବା)
disappointed (ଡିସାପଏଣ୍ଟେଡ୍) – became unhappy (ଦୁଃଖୀ ହେଲେ)
expect (ଏକ୍ସପେକୁ) – something you hope to get ( ଆଶା କରିବା )
form (ଫର୍ମ) – shape, body (ଆକାର)
gratitude – feeling of thankfulness (କୃତଜ୍ଞତା)
grin (ଗ୍ରିନ୍) – smile broadly (ହସ)
idiot (ଇଡ଼ିଅଟ୍) – one who fails to understand simple things (ନିର୍ବୋଧ, ବୋକା)
interrupting (ଇଣ୍ଟେରପ୍‌ଟିଙ୍ଗ୍) – breaking the continuity (ମଝିରେ ବାଧା ଦେଇ)
judgement (ଜଜ୍ମେଣୁ) – decision (ନିଷ୍ପଭି, ମତାମତ, ବିଚାର)
mixed up (ମିକ୍ସଡ୍ ଅପ୍) – became very confusing ( ହୋଇଯିବା, ବୁଝା ନ ପଡ଼ିବା)
patience (ପେସେନ୍ସ) – ability to wait for something for long time or to deal with something without getting angry ( ଧୈର୍ଯ୍ୟ, ସହନଶୀଳଭାବ)
pity (ପିଟି) – sadness that you feel when someone else is hurt or in trouble (ଦୟା, ଅନୁକମ୍ପା)
rage (ରେଜ୍) – anger (କ୍ରୋଧ)
repay (ରିପେ) – to give back something (ଫେରାଇବା, ପ୍ରତିଦାନ)
screamed (ସ୍କ୍ରିମ୍‌ଡ୍) – gave a cry with fear (ଚିତ୍କାର କରିବା)
seize (ସିଜ୍) – to take hold of forcibly (ହଠାତ୍ ଜବରଦସ୍ତ ଧରି)
shells (ସେଲ୍‌ସ୍) – water creatures like snail, oyster etc. having harder outer covering
shelter – place to live ( ଆଶ୍ରୟସ୍ଥଳ)
slave – servant (କ୍ରୀତଦାସ)
starve – to have no food (ଅନାହରରେ ରହିବା)
tale – story
trap (ଟ୍ରାପ୍) – an instrument for catching animals
tremble (ଟ୍ରିମ୍ବଲ୍) – shake in fear
trust (ଟ୍ରଷ୍ଟ୍) – faith (ବିଶ୍ୱାସ )
ungrateful (ଅଗ୍ରେଟ୍‌ଫୁଲ୍) – (a negative quality) not being thankful to a person who does some favour to you
witness (ଉଇଟ୍‌ନେସ୍) – a person who sees something happen

BSE Odisha 7th Class English Solutions Follow-Up Lesson 8 The Fisherman and The Jinni Important Questions and Answers

(A) Choose the right answer from the options.

Question 1.
When the fisherman pulled out the net first time he found .
(a) fish
(b) bottles
(c) a log of wood
(d) a copper bottle.
Answer:
(c) a log of wood

Question 2.
When the fisherman pulled out the net third time he found in it.
(a) log of wood
(b) some shells
(c) big stones
(d) a big brass jar
Answer:
(d) a big brass jar

Question 3.
When the fisherman opened the brass jar he found .
(a) a piece of gold
(b) a bag of sand
(c) a lot of smoke coming out
(d) some stones
Answer:
(c) a lot of smoke coming out

Question 4.
When the Jinni appeared, he wanted.
(a) to his gratitude
(b) to kill the fisherman
(c) to help the fisherman in future
(d) to go away
Answer:
(b) to kill the fisherman

Question 5.
The Jinni entered into the brass jar taking the form of
(a) an ant
(b) smoke
(c) water
(d) a fly
Answer:
(a) an ant

(B) Answer the following questions.
Question 1.
What did the fisherman get when he pulled out his net second time ?
Answer:
The fisherman got a few shells and big stones when he pulled out his net second time.

Question 2.
How did the fisherman let the jinni got into the jar ?
Answer:
The fisherman knew that the Jinni would kill him. Then he pretended to be foolish and asked the jinni that he was so big and he did not believe that he was inside the jar. So he wanted to see it. The Jinni could not understand his trick. He took the form of an ant and entered the jar. The fisherman very soon shut the jar tightly.

Question 3.
Who was clever- the fisherman or the jinni ?
Answer:
The fisherman was clever enough to put the Jinni inside the jar.

(B) Re-arrange the jumbled words to make meaningful sentences.

1. Brahman / O, / let / out / me / cage / of / his / pious
2. and / let / out / I / me / serve /will / you / for/as / slave/a / whole/life
3. by / who / don’t / shade /17 and / give / shelter/everyone / to / who / passes /?
4. services / for / past / the / reward/master / does / me / my /?
5. you / the / is / too / trap / small / hold / to
Answer:
1. Let me out of this cage, O, pious Brahman!
2. Let me out, and I will serve you as a slave for my whole life.
3. Don’t I give shade and shelter to everyone who passes by?
4. Does my master reward me for past services?
5. the trap is too small to hold you.

(C) Find whether True or False.
1. Let me out of the cage, O, pious jackal!
2. You must be a fool to expect gratitude from a hungry beast.
3. Does not my master reward me for past services?
4. The jackal was caught in a trap.
5. The tiger lost patience and at once jumped into the trap.
6. There lived a rich man with his wife and children.
7. every day he used to go to the river and catch fish.
8. Suddenly a jinni appeared in the smoke.
Answer:
(1) False
(2) True
(3) False
(4) False
(5) True
(6) False
(7) False
(8) True

Odisha State Board BSE Odisha 7th Class English Solutions Tail-Piece Lesson Branding Babies by Hot Iron Rods-A Bad Practice

BSE Odisha Class 7 English Solutions Tail-Piece Lesson Branding Babies by Hot Iron Rods-A Bad Practice

BSE Odisha 7th Class English Tail-Piece Lesson Branding Babies by Hot Iron Rods-A Bad Practice

I. Pre-Reading (ପୂର୍ବ ପ୍ରସ୍ତୁତି)
You have already read the lesson “Birsa Munda”. Birsa Munda fought against superstition in the society. Read the following text to know about some other bad practices in our society. We will also know what we have to do.
(‘ବିର୍ସା ମୁଣ୍ଡା’ ବିଷୟ ତୁମେ ପଢ଼ିସାରିଛ । ବିର୍ସା ମୁଣ୍ଡା ସମାଜର କୁସଂସ୍କାର ବିରୋଧରେ ଲଢ଼ୁଥିଲେ । ଆମ ସମାଜରେ ଅନ୍ୟାନ୍ୟ ମନ୍ଦ ପ୍ରଥା ବିଷୟରେ ଜାଣିବାକୁ ନିମ୍ନ ପାଠ୍ୟବିଷୟକୁ ପଢ଼ । ଆମକୁ କ’ଣ କରିବାକୁ ହେବ ତାହା ମଧ୍ୟ ଆମେ ଜାଣିବା ।)

→ Look at the following newspaper headings and pictures.
(ଖବରକାଗଜର ନିମ୍ନ ଶିରୋନାମାକୁ ଏବଂ ଛବିକୁ ଦେଖ ।)

→”Branding by Hot Iron Rods Kills Five Babies in Nawarangpur.”
(‘‘ଶିଶୁମାନଙ୍କୁ ତତଲା ଲୁହାଛଡ଼ରେ ଚେଙ୍କ ନବରଙ୍ଗପୁରର ପାଞ୍ଚଜଣ ଶିଶୁଙ୍କ ଜୀବନ ନେଲା ।’’)

→”Seven Babies Die in Two Months Due to Branding.”
(‘‘ଚେଙ୍କ ଯୋଗୁଁ ଦୁଇ ମାସରେ ସାତଜଣ ଶିଶୁଙ୍କ ପ୍ରାଣ ଗଲା ।’’)

→Do you think these are good practices?
(ତୁମେ କ’ଣ ଭାବୁଛ ଏଗୁଡ଼ିକ ଭଲ ପ୍ରଥା ?)

→Do you have a role to play against such practices ?(ଏହି ପ୍ରଥାଗୁଡ଼ିକ ବିରୋଧରେ ତୁମର କିଛି ଭୂମିକା ଗ୍ରହଣ କରିବାର ଅଛି କି ?)Read the following text to know what you can do to stop such bad things from the society. (ସମାଜରୁ ଏଭଳି ଖରାପ ପ୍ରଥାଗୁଡ଼ିକୁ ବନ୍ଦ କରିବା ନିମିତ୍ତ ତୁମେ କ’ଣ କରିପାରିବ ତାହା ଜାଣିବା ପାଇଁ ନିମ୍ନ ପାଠ୍ୟ ବିଷୟକୁ ପଢ଼ ।)

II. While Reading (ପଠନକାଳୀନ)
Text(ପାଠ୍ୟବସ୍ତୁ)

• SGP-1 (Sense Group Paragraph-1)
• Read the following text silently and answer the questions that follow.
  (ନିମ୍ନ ପାଠ୍ୟ ବିଷୟକୁ ନୀରବରେ ପାଠ କର ଏବଂ ନିମ୍ନସ୍ଥ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

Every month two or three babies die due to branding in our state. When a baby has a fever or diarrhea or any other diseases, the illiterate parents and grandparents call the village quack, the disari. The quack, the village doctor puts hot iron rods on the stomach of the babies as a cure. Instead of getting cured, the babies die. But these blind practices continue. What can we do to check such blind practices? You are studying in schools. You know if we suffer from diseases, we should go to a qualified doctor, not to an illiterate disari or a village quack. As you are educated, you have a role to play in this regard. Tell the people not to go to a disari or a village quack. Ask them to go to a hospital for treatment. Talk to your classmates, form a group to fight against the bad and blind practices of branding in and around your village and locality.

ଓଡ଼ିଆ ଅନୁବାଦ :
ଆମ ରାଜ୍ୟରେ ପ୍ରତ୍ୟେକ ମାସରେ ଦୁଇରୁ ତିନିଜଣ ଶିଶୁ ଚେଙ୍କଦ୍ଵାରା ପ୍ରାଣ ହରାଉଛନ୍ତି । ଗୋଟିଏ ଶିଶୁକୁ କୌଣସି ଜ୍ଵର, ଝାଡ଼ାରୋଗ କିମ୍ବା ଅନ୍ୟ କିଛି ରୋଗ ହେଲେ, ଶିଶୁର ନିରକ୍ଷର ପିତାମାତା କିମ୍ବା ଜେଜେ ଓ ଜେଜେମା’ ଗ୍ରାମ ବୈଦ୍ୟ (ଦିସାରୀ) ଙ୍କୁ ଡକାନ୍ତି । ବୈଦ୍ୟ ବା ଗ୍ରାମ ଡାକ୍ତର ଆରୋଗ୍ୟ ନିମିତ୍ତ ଶିଶୁର ପାକସ୍ଥଳୀରେ । ଏପ୍ରକାର ଅନ୍ଧ ପ୍ରଥା ଚାଲୁ ରହିଛି । ଏ ପ୍ରକାର ଅନ୍ଧ ପ୍ରଥାକୁ ବନ୍ଦ କରିବାକୁ ଆମେ ସବୁ କ’ଣ କରିପାରିବା ? ତୁମେମାନେ ବିଦ୍ୟାଳୟରେ ପାଠ ପଢୁଛ । ତୁମେ ତ ଜାଣ, ଆମକୁ ଯଦି ରୋଗ ହୁଏ, ତେବେ ଆମେ ଜଣେ ଯୋଗ୍ୟ ଡାକ୍ତରଙ୍କ ପାଖକୁ ଯିବା ଉଚିତ ହେବ । ଜଣେ ଅଶିକ୍ଷିତ ଦିସାରୀ ବା ଗ୍ରାମ୍ୟ ବୈଦ୍ୟ ପାଖକୁ ନୁହେଁ । ଯେଣୁ ତୁମେ ଶିକ୍ଷିତ ତୁମର ମଧ୍ୟ ଏଥିପାଇଁ କିଛି ଭୂମିକା ଅଛି । ଲୋକମାନଙ୍କୁ କୁହ – ବୁଝାଅ ଦିସାରୀ ବା ଗ୍ରାମ୍ୟ ବୈଦ୍ୟ ପାଖକୁ ଯିବା ଦରକାର ନାହିଁ । ବୁଝାଇ ଦିଅ (ରୋଗ ହେଲେ) ଡାକ୍ତରଖାନାକୁ ଚିକିତ୍ସା ନିମିତ୍ତ ଯାଆନ୍ତୁ । ଶ୍ରେଣୀର ସାଥୀମାନଙ୍କ ସହିତ କଥା ହୁଅ, ଦଳବାନ୍ଧି ତୁମ ଗ୍ରାମ ବା ଅଞ୍ଚଳ ଅଥବା ଆଖପାଖ ଗ୍ରାମ ଓ ଅଞ୍ଚଳରେ ଏ ପ୍ରକାର ଖରାପ ଅନ୍ଧପ୍ରଥା ବିରୋଧରେ ସଂଗ୍ରାମ କର ।

Notes And Glossary: (ଶବ୍ଦାର୍ଥ) :
die (ଡାଏ) – ମର
due to (ଡିଭ ଟୁ) – ଯୋଗୁ
branding (ବ୍ରାଣ୍ଡିଙ୍ଗ୍‌) – ଚେଙ୍କ
state (ଷ୍ଟେଟ୍) – ରାଜ୍ୟ
fever (ଫିଭର୍) – ଜ୍ୱର
diarrhea (ଡାଇରିଆ) – ହଇଜା
diseases (ଡିଜିଜେସ୍) – ରୋଗ ସବୁ
illiterate (ଇଲିଟ୍‌ରେଟ୍) – ନିରକ୍ଷର
quack – କୋଳାହଳ କରିବା
stomach (ଷ୍ଟୋମାକ୍) – ପେଟ
cure – ଆରୋଗ୍ୟ
instead of (ଇଡ୍‌ ଅଫ୍) – ପରିବର୍ତ୍ତେ
blind practice (ବ୍ଲାଇଣ୍ଡ୍ ପ୍ରାକ୍‌ଟିସ୍) – ଅନ୍ଧ ଅଭ୍ୟାସ
check (ଚେକ୍) – ଯାଞ୍ଚ କରନ୍ତୁ
suffer (ସଫର) – ଯନ୍ତ୍ରଣା ଭୋଗ
qualified (କ୍ଵାଲିଫାଇଡୁ) – ପ୍ରଶିକ୍ଷିତ
educated (ଏଜୁକେଟେଡ୍) – ଶିକ୍ଷିତ
role (ରୋଲ୍) – ଭୂମିକା
play – ଗ୍ରହଣ କରିବା
regard – ସମ୍ମାନ କରିବା
treatment (ଟ୍ରିଟ୍‌ମେଣ୍ଟ୍‌) – ଚିକିତ୍ସା
form – ଫର୍ମ
locality (ଲୋକାଲିଟି) – ଅଞ୍ଚଳ

Comprehension Questions (ବୋଧପରିମାପକ ପ୍ରଶ୍ନବଳୀ) :
Question 1.
What is the topic about?
(ବିଷୟଟି କେଉଁ ବିଷୟରେ ଉଦ୍ଦିଷ୍ଟ ?)
Answer:
The topic is about the blind belief or bad practice of branding babies by hot iron rods for cure in and around our village and locality.

Question 2.
What happens to two to three babies in our state every month?
(ଆମ ରାଜ୍ୟରେ ପ୍ରତ୍ୟେକ ମାସରେ ଦୁଇ ତିନିଜଣ ଶିଶୁଙ୍କର କ’ଣ ହୁଏ | ଘଟେ ?)
Answer:
In our state, every month two or three babies die due to branding.

Question 3.
When do illiterate people or grandparents call a village quack?
(କେତେବେଳେ ଅଶିକ୍ଷିତ ଗ୍ରାମବାସୀମାନେ.କିମ୍ବା ଜେଜେ ବାପା ଓ ଜେଜେମା’ମାନେ ଗ୍ରାମ ବୈଦ୍ୟଙ୍କୁ ଡକାନ୍ତି ?)
Answer:
The illiterate people and grandparents call a village quack when a baby has fever or any diseases.

Question 4.
Do you think it is a good practice?
(ତୁମେ କ’ଣ ଭାବୁଛ ଏହା ଏକ ଠିକ୍ | ଭଲ ପ୍ରଥା ?)
Answer:
No, we think branding babies is not a good practice.

Question 5.
What does the village quack do?
(ଗ୍ରାମ ବୈଦ୍ୟ କ’ଣ କରେ ?)
Answer:
The village quack puts hot iron rods on the stomach of the babies as a cure.

Question 6.
Why does he put the hot iron rods on the stomach of the babies?
(ସେ ଶିଶୁମାନଙ୍କ ପାକସ୍ଥଳୀରେ/ପେଟରେ କାହିଁକି ତତଲା ରଡ଼ର ଚେଙ୍କ ଦେଇଥା’ନ୍ତି ? )
Answer:
He puts hot iron rods on the stomach of the babies to cure them from the disease.

Question 7.
Do the babies get well? What happens to them?
(ଶିଶୁମାନେ କ’ଣ ଆରୋଗ୍ୟ ଲାଭ କରନ୍ତି ? ସେମାନଙ୍କର କ’ଣ ହୁଏ ?)
Answer:
No, the babies do not get well. Instead of getting well they die.

Question 8.
Where should we go to if we suffer from a -disease?
(ଆମକୁ କୌଣସି ରୋଗ ହେଲେ ଆମେ କେଉଁଠାକୁ ଯିବା ଉଚିତ ?)
Answer:
If we suffer from a disease we should go to a qualified doctor.

Question 9.
Who shouldn’t we go to?
(ଆମେ କାହା ପାଖକୁ ନଯିବା ଉଚିତ ?)
Answer:
We should not go to a village quack, the disari.

Question 10.
What have you got to do? Why?
(ତୁମର କ’ଣ କରିବାକୁ ଅଛି ? କାହିଁକି ?)
Answer:
We have to work to check such blind practices because these practices do harm to our society.

Question 11.
What should you tell the people?
(ତୁମେ ଲୋକଙ୍କୁ କ’ଣ କହିବା | ବୁଝାଇବା ଉଚିତ ?)
Answer:
We should tell the people to go to a qualified doctor not to a village quack or disari, if anybody suffers from illness.

Question 12.
What do you think is a better place to go – a village quack or a doctor ?
କେଉଁଠିକୁ ଯିବା ଭଲ ହେବ – ଗୁଣିଆ ନା ଚିକିତ୍ସକ (ଡାକ୍ତର) ପାଖକୁ ?)
Answer:
We think it is better to go to a qualified doctor not to an illiterate village quack or disari.

Question 13.
What should you do?
(ତୁମେମାନେ କ’ଣ କରିବା ଉଚିତ ?)
Answer:
We should talk to our classmates and form a group to fight against the blind or bad practice of branding babies in and around our village and locality.

Notes And Glossary: (ଶବ୍ଦାର୍ଥ) :
battle (ବ୍ୟାଟଲ୍) – fight (ସଂଘର୍ଷ, ଯୁଦ୍ଧ)
beat – no one can do better in arrow shooting, defeat (ହରାଇ ପାରିବା)
beaten (ବିଟେନ୍) – punished with heavy thrashing (ମାଡ଼ ଖାଇଥିଲେ )
branding (ବ୍ରାଣ୍ଡିଙ୍ଗ୍‌) – giving marks with hot iron
captured (କ୍ୟାପ୍‌ଚର୍‌ଡ୍) – caught (him) (ଧରି ନେଇଥିଲେ)
cowboy (କାଓବଏ) – someone who looks after cows (ଗାଈ ଜଗୁଆଳ)
defeated (ଡିଫିଟେଡ୍) – were beaten (ପରାସ୍ତ ହୋଇଥିଲେ)
documentary (ଡକ୍ୟୁମେଣ୍ଟାରୀ – a film giving facts (ତଥ୍ୟଭିଭିକ, ପ୍ରାମାଣିକ ଚଳଚ୍ଚିତ୍ର)
evil (ଏଭିଲ) – bad spirit (ଖରାପ ଆତ୍ମା, ପ୍ରେତାତ୍ମା)
gathered (ଗ୍ୟାଦର୍‌ଡ଼) – came in large number to one place (ଏକାଠି ହୋଇଥିଲେ )
innocent (ଇନୋସେଣ୍ଟ) – good and harmless people (ନିରୀହ, ନିଘୋଷ ବ୍ଯକ୍ତି)
mercilessly (ମର୍ସଲେସ୍‌ଲି) – cruelly (ଭୀଷଣ ନିର୍ଭୟ ଭାବରେ)
money lender (ମନି ଲେଣ୍ଡର) – a person who gives money to people in their need and collects it afterwards with interest (ଟଙ୍କା ସୁଧ କାରବାର କରୁଥିବା ବ୍ୟକ୍ତି)
movement (ମୁଭମେଣୁ) – mass fight to achieve something
pattas (ପଟ୍ଟାଜ୍)– land ownership papers (ଜମି ପଟ୍ଟା)
poisoned (ପଏଜନ୍‌) – (He was) given poison (ବିଷ ଦିଆଯାଇଥିଲା)
poverty (ପଭର୍ଟି) – a condition of having no money, no wealth or basic needs of life (ଦାରିଦ୍ର୍ୟ, ଗରିବ ଅବସ୍ଥା)
property (ପ୍ରପର୍ଟି) – wealth (ଧନସମ୍ପତ୍ତି)
quack – a person who does treatment of people without proper knowledge, especially in villages (ଗାଁ ବଇଦ, ଗୁଣିଆ)
religious (ରିଲିଜିଅସ୍ ) – one who shows strong faith in religion and obey its rules (ଧର୍ମନିଷ୍ଠ, ଧାର୍ମିକ)
reward (ରିୱାର୍ଡ) – wealth or money given to somebody for good work (ପୁରସ୍କାର)
sacrificed (ସାକ୍ରିଫାଇସ୍‌ଡ୍) – gave life for the cause of his country (ପାଇଁ ଜୀବନ ଉତ୍ସର୍ଗ କରିଥିଲେ)
sacrifice (ସାକ୍ରିଫାଇସ୍ ) – to give a gift of animal (goat) to god or goddess to win their favour
Santal (ସାନ୍ତାଲ) – a class of triabl (ଏକ ଆଦିବାସୀ ସମ୍ପ୍ରଦାୟ)
superstition (ସୁପରଷ୍ଟିସନ୍) – belief without based on facts, blind belief (ଅନ୍ଧବିଶ୍ଵାସ )
suspected (ସସ୍‌କୁ ଡ୍) – doubted ( ସନ୍ଦେହ ପ୍ରକଟ କରିବା)
tearful (ଟିୟରଫୁଲ୍) – sorrowful way (ଅଶୁଳ, ଦୁଃଖଦ)
weapons (ଉଇପନ୍ସ୍) – instruments used for fight like sword, gun etc. (ଅସ୍ତ୍ରଶସ୍ତ୍ର)
worship (ଓରସିପ୍ ) – pray (ପୂଜା କରିବା ବା ପ୍ରାର୍ଥନା କରିବା)
wounded (ୱିଣ୍ଡ୍ଡ୍) – injured, cut (his leg) (ହାଣି ହୋଇଯିବା, କ୍ଷତାକ୍ତ ହେବା, ଆହତ ହେବା )

BSE Odisha 7th Class English Solutions Tail-Piece Lesson Branding Babies by Hot Iron Rods-A Bad Practice Important Questions and Answers

(A) Answer the following questions.
Question 1.
Why did the people call him Tilka Baba?
Answer:
Tilka was worshiping Marang Burn. When days went by he became a religious man. The people of all religions, loved him and respected him. So they called him Tilka Baba.

Question 2.
How did Bhagalpur come under the control of the British?
Answer:
After the Plassey battle, British became the rulers of Bengal, Bihar and Odisha. So Bhagalpur of Bihar came under their control.

Question 3.
How did Cleareland tortured the natives?
Answer:
The new collector Cleareland appointed soldiers from other tribes to fight against the Santals. In many ways he tried to harash the people.

(B) Multiple Choice Questions (MCQs) with Answers.
Question 1.
Birsa Munda was born in a _________.
(a) rich family
(b) poor family
(c) joint family
(d) aristocratic family
Answer:
(b) poor family

Question 2.
His father’s name was _________.
(a) Sugana Munda
(b) Laxman Munda
(c) Sovan Munda
(d) None of these
Answer:
(a) Sugana Munda

(C) Re-arrange the jumbled words to make meaningful sentences.
Question 1.
superstition / from / days / very / Birsa / young / was / against
Answer:
From very young days Birsa was against superstition.

Question 2.
the / was / by / evil / the / said / quack / that / wound / caused / an / spirit
Answer:
The quack said that the wound was caused by an evil spirit:

Question 3.
wound / himself / his / of / to / a / cure / goat / Birsa / sacrifice / would / to / have
Answer:
Birsa would have to sacrifice a goat to cure himself of his wound.

(D) Find whether True or False.
Question 1.
Birsa Munda was born in a rich family in Odisha in 1975.
Answer:
False

Question 2.
From the very young age he worked as a cowboy of landlord.
Answer:
True

Question 3.
Birsa would have to sacrifice a hen to cure himself of his wound.
Answer:
False

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:

Odisha State Board BSE Odisha 7th Class English Solutions Follow-Up Lesson 2 A Tiny Warrior When I Grow Up Textbook Exercise Questions and Answers.

BSE Odisha Class 7 English Solutions Follow-Up Lesson 2 A Tiny Warrior

BSE Odisha 7th Class English Follow-Up Lesson 2 A Tiny Warrior Text Book Questions and Answers

Session – 1 (ଶବ୍ଦାବଳୀ)

I. Pre-Reading (ପୂର୍ବ ପ୍ରସ୍ତୁତି)

• Socialization (ସାମାଜିକୀକରଣ):
• In the main lesson the father learned a lesson from his son. The son taught the father that the poor people are. in fact, rich. In this follow-up lesson you will read about a small girl of class-V leading a movement against bazzar notebooks. Let’s read and see how cheap bazaar notebooks do more harm than good.

II. While Reading (ପଠନକାଳୀନ)
Text(ପାଠ୍ୟବସ୍ତୁ)

• SGP-1 (Sense Group Paragraph-1)
• Read paragraph 1 and 2 and answer the questions.
  (ଅନୁଚ୍ଛେଦ ୧ ଓ ୨କୁ ନୀରବରେ ପାଠ କର ଏବଂ ନିମ୍ନ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

1. One evening Mitu and Situ, two sisters, were studying in their room. Mitu studies in class VII and Situ in Class V. The two sisters were studying in two different schools. Their uncle, Mr. Giri, was watching them from a little distance. He was a retired teacher. He had come to visit them on a week-end. Mr. Giri saw Mitu reading one paragraph from her English textbook then reading the meaning of this paragraph in Odia from a bazaar notebook (Meaning book), which disturbed him. ‘If children study English like this, they will never learn English’, he thought.

2. The uncle was eager to help Mitu learn English on her own without the help of this meaning book. He came close to her and asked her to read the first paragraph of the lesson silently. The lesson was “The Story of Cricket”. The first paragraph was: The shape and the size of a cricket ground are not fixed. They are different. The cricket ground of Melbourne in Australia is bigger than that of Feroz Shah Kotla in New Delhi. Similarly the shape of Chepauk Cricket Ground in Chennai is circular. But the Adelaide Cricket Ground in Australia is oval.

ଓଡ଼ିଆ ଅନୁବାଦ :
୧. ଦିନେ ସନ୍ଧ୍ୟାରେ ମିତୁ ଏବଂ ସିତୁ ଦୁଇ ଭଉଣୀ ସେମାନଙ୍କ ପଢ଼ାଘରେ ପଢ଼ା କାର୍ଯ୍ୟ କରୁଥିଲେ । ମିତୁ ୭ମ ଶ୍ରେଣୀରେ ଏବଂ ସିତୁ ୫ମ ଶ୍ରେଣୀରେ ପଢ଼େ । ଦୁଇ ଭଉଣୀ ଦୁଇଟି ଅଲଗା ସ୍କୁଲରେ ପାଠ ପଢୁଥିଲେ । ସେମାନଙ୍କ ଦାଦା/ ମାମୁ ଗିରିବାବୁ ସେମାନଙ୍କୁ କିଛି ଦୂରରୁ ଲକ୍ଷ୍ୟ କରୁଥିଲେ । ସେ ଜଣେ ଅବସରପ୍ରାପ୍ତ ଶିକ୍ଷକ । ସପ୍ତାହର ଶେଷ ଦିନରେ ସେ ତାଙ୍କୁ ଦେଖିବାକୁ ଆସିଥିଲେ । ଗିରିବାବୁ ଦେଖ‌ିଲେ ମିତ୍ରୁ ତା’ ଇଂରାଜୀ ପାଠ୍ୟପୁସ୍ତକର ଏକ ଅନୁଚ୍ଛେଦ ପଢ଼ୁଥିଲା ଏବଂ ପରେ ପରେ ସେହି ଅନୁଚ୍ଛେଦର ଓଡ଼ିଆ ଅନୁବାଦ ବଜାରରୁ ଖରିଦ କରିଥିବା ଏକ ମାନେ ବହିରୁ ପଢ଼ୁଥିଲା । ତାହା ତାଙ୍କୁ ବଡ଼ ଅଡୁଆ ଲାଗିଲା । ସେ ଭାବିଲେ- ଯଦି ପିଲାମାନେ ଇଂରାଜୀ ଏଭଳି ଢଙ୍ଗରେ ପଢ଼ିବେ, ସେମାନେ ଇଂରାଜୀ ଶିକ୍ଷା କରିପାରିବେ ନାହିଁ ।

୨.ଦାଦା/ମାମୁ ଜଣକ ମିତୁ କିଭଳି ଆପେ ଆପେ ମାନେ ବହି ବିନା ଇଂରାଜୀ ଶିକ୍ଷା କରିପାରିବ, ସେ ସମ୍ବନ୍ଧରେ ସାହାଯ୍ୟ କରିବାକୁ ବ୍ୟଗ୍ର ହୋଇଉଠିଲେ । ସେ ତା’ ପାଖକୁ ଆସିଲେ ଏବଂ ପ୍ରଥମ ଅନୁଚ୍ଛେଦଟିକୁ ନୀରବରେ ପାଠ କରିବାକୁ କହିଲେ । ପାଠ୍ୟବସ୍ତୁଟି ଥୁଲା ‘କ୍ରିକେଟ୍‌ର ଏକ କାହାଣୀ ।’ ପ୍ରଥମ ଅନୁଚ୍ଛେଦଟି ଏହିପରି ଥୁଲା ମେଲ୍‌ବର୍ଷ କ୍ରିକେଟ୍ ପଡ଼ିଆ, ନୂଆଦିଲ୍ଲୀର ଫିରୋଜ ଶାହା କୋଟଲା କ୍ରିକେଟ୍‌ ପଡ଼ିଆ ତୁଳନାରେ ବଡ଼ । ସେହିଭଳି ଚେନ୍ନାଇର ଚେପକ୍ କ୍ରିକେଟ୍ ପଡ଼ିଆର ଆକାର ଗୋଲାକାର । କିନ୍ତୁ ଅଷ୍ଟ୍ରେଲିଆର ଆଡିଲେଡ୍ କ୍ରିକେଟ୍‌ ପଡ଼ିଆ ଅଣ୍ଡାକୃତିର ।

Notes And Glossary: (ଶବ୍ଦାର୍ଥ) :
studying (ଷ୍ଟଡ଼ିଙ୍ଗ୍) – ପଢ଼ିବା
uncle (ଅଙ୍କଲ୍) – ଦାଦା | ମାମୁ
watching (ଚିଙ୍ଗ୍) – ଦେଖିବା
little distance (ଲିଟିଲ୍‌ ଡିଷ୍ଟାନ୍‌ସ୍) – ଅଳ୍ପ ଦୂର
retired teacher (ରିଟାୟାର୍ଡ ଟିଚର) – ଅବସରପ୍ରାପ୍ତ ଶିକ୍ଷକ
paragraph (ପାରାଗ୍ରାଫ୍) – ଅନୁଚ୍ଛେଦ
disturbed (ଡିଷ୍ଟବର୍ଡ) – ବିଚଳିତ ହେଲେ
eager to help (ଇଗର୍ ଟୁ ହେଲ୍ପ) – ସାହାଯ୍ୟ ପାଇଁ ଆଗ୍ରହ ପ୍ରକାଶ କରିବା
shape (ସେପ୍) – ଆକାର
size (ସାଇଜ୍) — ପ୍ରକାର, ଲମ୍ବ ପ୍ରସ୍ଥର ଦୂରତ୍ୱ
circular (ସର୍କୁଲାର) – ଗୋଲାକାର
oval (ଓଭାଲ୍) – ଅଣ୍ଡାକୃତି

Comprehension Questions (ବୋଧପରିମାପକ ପ୍ରଶ୍ନବଳୀ)

Question 1.
Who are there in paragraph X?
(ପ୍ରଥମ ଅନୁଚ୍ଛେଦରେ କିଏ କିଏ ଅଛନ୍ତି ?)
Answer:
In paragraph 1, there are two children Mitu and Situ. Their uncle Mr. Giri was with them.

Question 2.
Who are Mitu and Situ?
(ମିତୁ ଏବଂ ସିତୁ କିଏ ?)
Answer:
Mitu and Situ are two sisters. Mitu studies in class VII and Situ in class V. They were reading in different schools.

Question 3.
Who visited them on a week-end?
(ସପ୍ତାହ ଶେଷରେ କିଏ ସେମାନଙ୍କୁ ସାକ୍ଷାତ କରିବାକୁ ଆସିଥିଲେ ?)
Answer:
Mr. Giri, their uncle, visited them on a weak-end.

Question 4.
What was Mr. Giri?
(ମି. ଗିରି କ’ଣ ଥିଲେ ?)
Answer:
Mr. Giri was a retired teacher. He was the uncle of Mitu and Situ.

Question 5.
What did Mr Giri see?
(ମି. ଗିରି କ’ଣ ଦେଖ‌ିଲେ ?)
Answer:
Mr. Giri saw Mitu reading a paragraph from English textbook and then reading the meaning of the paragraph from a bazaar meaning book.

Question 6.
What was Mitu reading?
(ମିତ୍ର କ’ଣ ପଢ଼ୁଥିଲା ?).
Answer:
Mitu was reading a paragraph from English textbook.

Question 7.
Why was her uncle disturbed?
(ମାମୁଙ୍କୁ କାହିଁକି ଅଡୁଆ ଲାଗିଲା ?)
Answer:
Her uncle was disturbed because he did not like learning English taking help of bazaar notebook.

Question 8.
What did her uncle ask her?
(ତା’ ମାମୁ ତାକୁ କ’ଣ କହିଲେ ?)
Answer:
Her uncle asked her to read the first paragraph of the lesson silently.

Question 9.
What was the title of the lesson?
(ପାଠର ଶିରୋନାମା କ’ଣ ଥିଲା ?)
Answer:
The title of the lesson was “The Story of Cricket”.

Question 10.
What was the first paragraph of the lesson about?
(ଅଧ୍ୟାୟର ପ୍ରଥମ ଅନୁଚ୍ଛେଦଟି କେଉଁ ବିଷୟରେ ଥିଲା ?)
Answer:
The first paragraph of the lesson was about the shape and size of cricket ground.

Question 11.
How is this paragraph in your English book different from the paragraph in the meaning book?
(ତୁମ ଇଂରାଜୀ ବହିର ଏହି ଅନୁଚ୍ଛେଦଟି କିପରି ମାନେ ବହିର ଅନୁଚ୍ଛେଦଠାରୁ ଭିନ୍ନ ଥିଲା ?)
Answer:
This paragraph from the English book teaches how to learn and improve the knowledge in English but the paragraph from meaning book only teaches what is the meaning of the paragraph.

Session – 2 (ଦ୍ଵିତୀୟ ପର୍ଯ୍ୟାୟ)

• SGP-2 (Sense Group Paragraph-2)
• Read paragraphs – 3 to 5 and answer the questions that follow.
  (ଅନୁଚ୍ଛେଦ ୩ ରୁ ୫କୁ ପାଠ କର ଏବଂ ନିମ୍ନପ୍ରଦତ୍ତ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

3. Next what happened between the uncle and the niece is stated below :
Mr Giri: Every paragraph has an idea or a topic. Can you tell me the line where the topic is? What is the paragraph about? (Mitu was silent) It is about the shape and size of a cricket ground.
Mitu: The first sentence.
Mr. Giri Very Good: After telling the topic, the writer gives examples/ facts to explain the topic. Can you say what example does the writer give?
Mitu: Melbourne and Feroz Shah Kotla.
Mr Giri: Where is Melbourne cricket ground and where is Feroz Saha Kotla?
Mitu: (Reading the paragraph again) In Australia and Delhi.
Mr Giri: Very Good. Which ground is bigger in size?
Mitu: Melbourne.
Mr Giri: Good. This is about the size. What about the shape? Which ground is circular-like a circle?
Mitu: Chepauk Ground.
Mr Giri: Where is Chepauk Ground?
Mitu: In Chennai.
Mr Giri: Good. Which ground is oval shaped?
Mitu: Adelaide Ground.
Mr Giri: Where is Adelaide Ground?
Mitu: In Australia.

4. Next, the uncle asked his niece Mitu to give her notebook and on her notebook he made a note on the paragraph. The note was as follows:

5. “Now you can understand the paragraph much better. The meaning book only gives the meaning in Odia. It is not useful in learning English properly.” said Mr. Giri.

(ଓଡ଼ିଆ ଅନୁବାଦ)
୩. ତା’ପରେ ମାମୁ ଭାଣିଜୀଙ୍କ ମଧ୍ୟରେ ଯାହା ଘଟିଲା ତାହା ନିମ୍ନରେ ଉଲ୍ଲେଖ କରାଯାଇଛି :
ଶ୍ରୀ ଗିରି: ପ୍ରତ୍ୟେକ ଅନୁଚ୍ଛେଦର ଏକ ଧାରଣା ବା ବିଷୟବସ୍ତୁ ଥାଏ । କେଉଁ ଧାଡ଼ିରେ ବିଷୟବସ୍ତୁ ଅଛି ତୁମେ ମୋତେ କହିପାରିବ ବି ? ଅନୁଚ୍ଛେଦଟି କେଉଁ ବିଷୟରେ ?
ମିଟୁ: ପ୍ରଥମ ବାକ୍ୟ ।
ଶ୍ରୀ ଗିରି: ବହୁତ ଭଲ । ବିଷୟବସ୍ତୁ କହିସାରିବା ପରେ ଲେଖକ ବିଷୟବସ୍ତୁକୁ ବୁଝାଇବା ନିମନ୍ତେ ତୁମେ କହିପାରିବ କି ?
ମିଟୁ: ମେଲବୋର୍ଡ଼ ଏବଂ ଫିରୋହ ଶାହା କୋଟ୍‌ ।
ଶ୍ରୀ ଗିରି: ମେଲବୋର୍ଡ଼ କ୍ରିକେଟ୍ ପଡ଼ିଆ କେଉଁଠାରେ ଏବଂ ଫିରୋଜ ଶାହା କୋଟ୍‌ଲା କେଉଁଠାରେ ?
ମିଟୁ: (ଅନୁଚ୍ଛେଦକୁ ଆଉଥରେ ପଢ଼ି) ଅଷ୍ଟ୍ରେଲିଆ ଏବଂ ଦିଲ୍ଲୀରେ ।
ଶ୍ରୀ ଗିରି: ବହୁତ ଭଲ । କେଉଁ ପଡ଼ିଆଟି ଆକାରରେ ବଡ଼ ?
ମିଟୁ: ମେଲବୋର୍ଡ଼ ।
ଶ୍ରୀ ଗିରି: ଭଲ । ଏହା ଆକାର ବିଷୟରେ । ଆକୃତି ବିଷୟରେ କ’ଣ ? କେଉଁ ପଡ଼ିଆଟି ବୃତ୍ତାକାର – ଏକ ବୃତ୍ତ ଭଳି ।
ମିଟୁ: ଚେପକ୍ ପଡ଼ିଆ ।
ଶ୍ରୀ ଗିରି: ଚେପକ୍ ପଡ଼ିଆ କେଉଁଠାରେ ?
ମିଟୁ: ଚେନ୍ନାଇରେ ।
ଶ୍ରୀ ଗିରି:ଭଲ । କେଉଁ ପଡ଼ିଆଟି ଅଣ୍ଡାକୃତି ?
ମିଟୁ:ଆଡ଼ିଲେଡ଼ ପଡ଼ିଆ ।
ଶ୍ରୀ ଗିରି:ଆଡ଼ିଲେଡ଼ ପଡ଼ିଆ କେଉଁଠାରେ ?
ମିଟୁ:ଅଷ୍ଟ୍ରେଲିଆରେ ।

୪.ତା’ପରେ ମାମୁ ତାଙ୍କ ଭାଣିଜୀ ମିତ୍ରୁକୁ ତା’ ଖାତା ମାଗିଲେ ଏବଂ ତା’ ଖାତାରେ ଅନୁଚ୍ଛେଦର ସାରାଂଶ ଲେଖୁ ଦେଲେ । ସାରାଂଶଟି ନିମ୍ନ ପ୍ରକାର ଥିଲା :

୫. ଗିରିବାବୁ କହିଲେ, ‘ତୁମେ ବର୍ତ୍ତମାନ ଅନୁଚ୍ଛେଦଟିକୁ ଭଲ ଭାବରେ ବୁଝିପାରିଲ । ମାନେ ବହି କେବଳ ଓଡ଼ିଆ ଅର୍ଥ ଦେଇଥାଏ । ଏହା ଠିକ୍ ଭାବେ ଇଂରାଜୀ ଶିକ୍ଷା ପାଇଁ ଦରକାରୀ ନୁହେଁ ।

Notes And Glossary:
next ( ନେଷ୍ଟ ) – ପରେ
happened (ହାପେନ୍ସ) – ଘଟିଲା
topic (ଟପିକ୍) – ବିଷୟ/ବିଷୟବସ୍ତୁ
idea (ଆଇଡ଼ିଆ) – ଧାରଣା
silent (ସାଇଲେଣ୍ଟ୍) – ନୀରବ
facts (ଫ୍ରାକୃସ୍) – ତଥ୍ୟାବଳୀ
examples (ଏକ୍‌ଜାମ୍ପୁଲସ୍ ) – ଉଦାହରଣସମୂହ
bigger (ବିଗର୍ ) – ବୃହତ୍ତର
circle (ସର୍କଲ୍) – ବୃତ୍ତ
oval shapes (ଓଭାଲ୍ ସେପ୍‌) – ଅଣ୍ଡା ଆକୃତିର
different (ଡିଫରେଣ୍ଡ୍) – ଭିନ୍ନ/ଅଲଗା
properly (ପ୍ରପର୍‌ଲି) – ଠିକ୍ ଭାବରେ
understand ( ଅଣ୍ଡରଷ୍ଟାଣ୍ଡ୍) – ବୁଝିବା

Comprehension Questions (ବୋଧପରିମାପକ ପ୍ରଶ୍ନବଳୀ)

Question 1.
Who played the role of teacher in this paragraph?
(ଏହି ଅନୁଚ୍ଛେଦରେ କିଏ ଶିକ୍ଷକ ଭୂମିକା ନିର୍ବାହ କରିଥିଲେ ?)
Answer:
Mr. Giri played the role of teacher in this paragraph.

Question 2.
Who played the role of the student in this paragraph?
(ଏହି ଅନୁଚ୍ଛେଦରେ କିଏ ଶିକ୍ଷାର୍ଥୀ ଭୂମିକା ନିର୍ବାହ କରିଥିଲେ ? )
Answer:
Mitu played the role of the student in this paragraph.

Question 3.
How is one paragraph of a text different from another paragraph?
(ପାଠ୍ୟ ବିଷୟର ଗୋଟିଏ ଅନୁଚ୍ଛେଦ ଅନ୍ୟ ଏକ ଅନୁଚ୍ଛେଦଠାରୁ ଭିନ୍ନ କିପରି ? )
Answer:
Every paragraph has an idea or topic. In this respect two paragraphs of a text are different.

Question 4.
What does a writer do after giving the topic of the paragraph?
(ଜଣେ ଲେଖକ ବିଷୟ ସ୍ଥିର କଲାପରେ କ’ଣ କରନ୍ତି ?)
Answer:
After telling the topic, the writer gives some examples to explain the topic.

Question 5.
When Mitu answered the questions, did she keep her textbook open or closed?
(ମିତୁ ଯେତେବେଳେ ଉତ୍ତର ଦେଉଥିଲା, ସେ ପାଠ୍ୟପୁସ୍ତକ ଖୋଲା ରଖୁଥିଲା ନା ବନ୍ଦ କରିଥିଲା ?)
Answer:
When Mitu answered the questions, she kept her textbook open.

Question 6.
Was she able to answer most of the questions?
(ସେ ଅଧିକାଂଶ ପ୍ରଶ୍ନର ଉତ୍ତର ଦେବାରେ ସମର୍ଥ ହେଲା କି ?)
Answer:
Yes, she was able to answer most of the questions.

Question 7.
Is her uncle a good teacher ? How do you know?
(ତାଙ୍କର ମାମୁ ଜଣେ ଭଲ ଶିକ୍ଷକ କି ? ତୁମେ କିପରି ଜାଣୁଛ ?)
Answer:
Yes, her uncle is a good teacher. He explained the paragraph to Mitu bit by bit and made her understand everything.

Question 8.
Do you like the notes that her uncle made on the paragraph?
(ତା’ର ମାମୁ ଅନଚ୍ଛେଦ ଉପରେ ପ୍ରସ୍ତୁତ କରିଥିବା ସଂକ୍ଷିପ୍ତ ସାରାଂଶକୁ ତୁମେ ପସନ୍ଦ କରୁଛ କି ?)
Answer:
Yes, we like the notes that her uncle made on the paragraph.

Question 9.
Will the note help Mitu remember the paragraph?
(ଅନୁଚ୍ଛେଦଟିକୁ ମନେରଖ୍କୁ ମିତୁକୁ ସଂକ୍ଷିପ୍ତ ସାରାଂଶ ସାହାଯ୍ୟ କରିବ କି ?)
Answer:
Yes, the note will definitely help Mitu to remember the paragraph.

Question 10.
Why is meaning book not useful according to Mitu?
(ମିତୁ କହିବା ଅନୁସାରେ ମାନେ ବହି କାହିଁକି ଦରକାରୀ ନୁହେଁ ?)
Answer:
Meaning book is not at all useful, because it only gives the Odia meaning of the topic, but does not give knowledge.

Question 11.
What did her uncle say about meaning book?
(ତା’ର ମାମୁ ମାନେ ବହି ବିଷୟରେ କ’ଣ କହିଲେ ?)
Ans.
Her uncle said that meaning book is useless because it gives the Odia meaning only. It does not help in learning English properly.

Session – 3 (ତୃତୀୟ ପର୍ଯ୍ୟାୟ)

• SGP-3 (Sense Group Paragraph-3)
• Read paragraph-6 and 7 and answer the questions that follow.
  (ଅନୁଚ୍ଛେଦ ୬ ଓ ୭କୁ ପାଠ କର ଏବଂ ନିମ୍ନପ୍ରଦତ୍ତ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

6. Situ was silently sitting and watching what happened between her uncle and her sister. She broke her silence and asked, ‘If our teachers read aloud a paragraph and explain the meaning in Odia, are they doing the right thing ?”No, not at all. They are as harmful as the bazzar note”said Mr. Giri.

7. What happened after this is the story of Situ. Situ said how meaning books have lots of mistakes. One of her teachers did not give her any mark for an answer. He thought the wrong answer in the bazzar note was the right answer. She took this matter to her headmistress. She called the teacher and asked him to give her mark. She also banned the use of bazzar note books in their school. She made Situ the leader of the movement against meaning books in their school.

୬.ସିତୁ ନୀରବରେ ବସି ଲକ୍ଷ୍ୟ କରୁଥିଲା ଯାହା ତା’ର ମାମୁ ଓ ଭଉଣୀ ମଧ୍ଯରେ ଘଟୁଥିଲା । ନୀରବତା ଭାଙ୍ଗି ସେ ପଚାରିଲା, ‘‘ଯଦି ଆମେ ଶିକ୍ଷକମାନେ ଗୋଟିଏ ଅନୁଚ୍ଛେଦକୁ ବଡ଼ପାଟିରେ ପଢ଼ନ୍ତି ଏବଂ ଏହାର ଅର୍ଥ ଓଡ଼ିଆରେ ବୁଝାନ୍ତି, ସେମାନେ କ’ଣ ଠିକ୍ କରନ୍ତି ?’’ କେବେ ନୁହେଁ ! ଗିରିବାବୁ କହିଲେ, ‘ସେମାନେ ବଜାରର ମାନେ ବହି ପରି କ୍ଷତିକାରକ ଅଟନ୍ତି ।’’

୭.ଏହାପରେ ଯାହା ଘଟିଲା ତାହା ସିତୁର କାହାଣୀ ଥିଲା । ସିତୁ କହିଲା କିପରି ମାନେବହିଗୁଡିକରେ ବହୁତ ଭୁଲ ବା ତ୍ରୁଟି ରହୁଛି । ତା’ର ଜଣେ ଶିକ୍ଷକ ଗୋଟିଏ ପ୍ରଶ୍ନର ଉତ୍ତର ପାଇଁ ତାକୁ କିଛି ନମ୍ବର ଦେଇ ନଥିଲେ । ସେ ବଜାର ମାନେ ବହିରେ ଥିବା ଭୁଲ ଉତ୍ତରକୁ ଠିକ୍ ଉତ୍ତର ବୋଲି ଭାବୁଥିଲେ । ସେ (ସିତୁ) ଏହି ଘଟଣାକୁ ପ୍ରଧାନ ଶିକ୍ଷୟିତ୍ରୀଙ୍କ ଦୃଷ୍ଟିକୁ ଆଣିଲେ । ସେ ଶିକ୍ଷକକୁ ଡାକିଲେ ଏବଂ କହିଲେ ତାକୁ ନମ୍ବର ଦେବାକୁ । ସେ ମଧ୍ୟ ବଜାର ମାନେ ବହିଗୁଡିକୁ ତାଙ୍କ ବିଦ୍ୟାଳୟରେ ବ୍ୟବହାର କରିବାକୁ ନିଷେଧ କଲେ । ସେ ତାଙ୍କ ବିଦ୍ୟାଳୟରେ ବଜାର ମାନେ ବହି ବିରୋଧୀ ଆନ୍ଦୋଳନର ସିତୁକୁ ନେତ୍ରୀ କରିଦେଲେ ।

Notes And Glossary: (ଶବ୍ଦାର୍ଥ) :
watching (ଚିଙ୍ଗ୍) – ଦେଖିବା
broke (ବ୍ରୋକ୍) – ଭାଙ୍ଗିଲା
silence (ସାଇଲେନ୍ସ) – ନୀରବତା
not at all (ନଟ୍ ଆଟ୍ ଅଲ୍) – ଆଦୌ ନୁହେଁ
harmful (ହାର୍ମଫୁଲ୍) – କ୍ଷତିକାରକ
bazzar (ବଜାର)
banned – ନିଷେଧ କଲେ

Comprehension Questions (ବୋଧପରିମାପକ ପ୍ରଶ୍ନବଳୀ) :
Question 1.
What are these two paragraphs about-about Mitu, Mr. Giri or Situ ?
(ଏହି ଦୁଇଟି ଅନୁଚ୍ଛେଦ କାହା ବିଷୟରେ – ମିତ୍ରୁ, ମି.ଗିରି ନା ସିତୁ ବିଷୟରେ ?)
Answer:
These two paragraphs are about Situ.

Question 2.
Who said meaning books have lots of mistakes ?
(ମାନେ ବହିରେ ଅନେକ ଭୁଲ୍ ଅଛି ବୋଲି କିଏ କହିଥିଲା ?)
Answer:
Situ said that the meaning books have lots of mistakes.

Question 3.
Even if the answer of Situ was correct, why didn’t her teacher give her any mark?
(ଯଦିଓ ସିତୁର ଉତ୍ତର ଠିକ୍ ଥିଲା, ତା’ର ଶିକ୍ଷକ ତାକୁ କାହିଁକି କୌଣସି ନମ୍ବର ଦେଲେ ନାହିଁ ।)
Answer:
Though the answer of Situ was correct, her teacher didn’t give her any mark, because he thought that the wrong answer in the bazzar note was the right answer.

Question 4.
What did Situ do next?
(ତା’ପରେ ସିତୁ କ’ଣ କଲା ?)
Answer:
Then Situ took the matter to her headmistress.

Question 5.
What did the headmistress ban?
(ପ୍ରଧାନ ଶିକ୍ଷୟିତ୍ରୀ କ’ଣ ନିଷେଧ କରିଥିଲେ ?)
Answer:
The headmistress banned the use of bazzar note books in their school. She also made Situ the leader of the movement against bazzar note books in their school.

Question 6.
What was the movement about?
(ଆନ୍ଦୋଳନ କେଉଁ ବିଷୟରେ ଥିଲା ? )
Answer:
The movement was against the use of meaning books in their school.

Question 7.
Who did the headmistress make the leader of the movement? Why?
(ପ୍ରଧାନ ଶିକ୍ଷୟିତ୍ରୀ କାହାକୁ ଆନ୍ଦୋଳନର ନେତ୍ରୀ କରାଇଲେ ? କାହିଁକି ?)
Answer:
The headmistress made Situ the leader of the movement. Because Situ knew well about the harmful impact of the bazzar note books.

Question 8.
Will you use meaning book after reading this lesson?
(ଅଧ୍ୟାୟ ପଢ଼ି ସାରିବା ପରେ ତୁମେମାନେ ମାନେବହି ବ୍ୟବହାର କରିବ କି ?)
Answer:
No, we will never use the meaning books after reading this les son.

Session – 4 (ଚତୁର୍ଥ ପର୍ୟ୍ୟାୟ)
Post-Reading (ପଠନ ପରବର୍ତ୍ତୀ)

1. Visual Memory Development Technique (VMDT) :
(ଦୃଶ୍ୟ ସ୍ମୃତି ବିଚାର କୌଶଳ) : (Teacher decides)

2. Comprehension Activities (ବୋଧ ପରିମାପକ କାର୍ଯ୍ୟାବଳୀ) :
(a) Teacher frames MCQs. (ଶିକ୍ଷକ MCQ ତିଆରି କରନ୍ତୁ)
(b) The lesson is divided into three SGPs : three parts. The three topics/themes are given.
Write under each paragraph number. (Question with Answer)

3. Listening (ଶୁଣିବା) :
Teacher frames listening activities.

Session – 5 (ଷଷ୍ଠ ପର୍ଯ୍ୟାୟ)
4. Speaking (କଥନ) :

(a) Chain-drill: Meaning books are harmful.
(b) Dialogue: (Follow the steps of previous lessons.)
Mr Giri: Where is Melbourne?
Mitu: In Australia
Mr Giri: Where is Feroz Shah Kotla Cricket Ground?
Mitu: In New Delhi
Mr Giri: Where is Chepauk Ground?
Mitu: In Chennai

5. Vocabulary ଶବ୍ଦାବଳୀ:
Match the words with the shapes. (Question with Answer)

6. Writing (ଲିଖନାତ୍ମକ)

(a) Teacher gives some questions for writing in one sentence each.
(b) See the notes given by Mr. Giri on the paragraph. Now write a paragraph based on the notes. (Do not see the original paragraph while doing this task). Some help is given.

The paragraph is about __________ cricket ground. __________. Some cricket grounds are_____________. Some are ___________. The Melbourne ____________is ___________than the ___________. Some __________ are circular. Some ____________ are _________. The ___________ is ___________. The ___________is oval. The Melbourne _____________is in Australia.
Answer:
The paragraph is about size and shape of cricket ground. Some cricket grounds are bigger. Some are smaller. The Melbourne cricket ground is bigger than the Feroz Shah. Some cricket ground are circular. Some cricket grounds are oval. The Chepauk cricket ground is circular. The Adelaide cricket ground is oval. The Melbourne cricket ground is in Australia.

7. Mental Talk (ମାନସ କଥନ) :

_____________________________________________________

_____________________________________________________.
8. Let’s Think (ଚାଲ ଚିନ୍ତା କରିବା ବା କଳ୍ପନା କରିବା):

_____________________________________________________

_____________________________________________________.

Word Note: (The words/phrases have been defined mostly on their contextual meanings)

acute (ଆକ୍ୟୁଟ୍) – intense, severe (ଉତ୍କଟ)
ban – to say that something must not be done
bull dog (ବୁଲ୍ ଡର)- a type of strong dog (ଏକ ପ୍ରକାର ଶକ୍ତିଶାଳୀ କୁକୁର)
Chinese Great Wall (ଚାଇନିଜ୍ ଗ୍ରେଟ୍ ୱାଲ୍) – Historic Great Wall of Chine (ପ୍ରାଚୀର)
dazzling (ଡାକ୍‌ଲିଙ୍ଗ୍) – very bright (ଚମକୁଥୁବା )
disappear (ଡିସ୍‌ପିୟର) – to go away (ଅଦୃଶ୍ୟ ହେବା । ଦୂର ହେବା)
disturbed (ଡିଷ୍ଟର୍ବଡ୍) – made him worried/unhappy  (କଲା)
encyclopedias (ଏନସାଇକ୍ଲୋପିଡିଆ) – knowledge books (ଜ୍ଞାନ ପୁସ୍ତକ)
flooded (ଫ୍ଲଡେଡ୍) – filled with (moon light) (ଚନ୍ଦ୍ର ଆଲୋକ) ରେ ପରିପୂର୍ଣ୍ଣ
glamour (ଗ୍ଲାମର୍) – beauty (ସୌନ୍ଦର୍ଯ୍ୟ)
in contrast (ଇନ୍ କନ୍‌ଷ୍ଟ୍ରିଟ୍) – in comparison (ତୁଳନାତ୍ମକ ଭାବରେ)
indifferent (ଇନ୍‌ଫରେଣ୍ଡ୍) – lack of interest (ଅନାଗ୍ରହ)
lit by lamps (ପ୍ରଦୀପ ଦ୍ୱାରା ପ୍ରଜ୍ୱଳିତ) – lighted with lamps
palatial building (ପାଲଟିଆ ବିଲଡିଂ) – royal building (ରାଜକୀୟ ଭବନ)
prison (ପ୍ରିଜନ୍) – a building where usually thieves and criminals are kept for punishment
privileged (ପ୍ରିଭିଲିଜ୍) – blest with special benefits, wealth etc. (ବିଶେଷ ଅଧିକାରପ୍ରାପ୍ତ)
richness (ରିନେସ୍ ) – the quality of being rich (ବିତ୍ତଶାଳୀ ଭାବ)
sadness (ସ୍ୟାଡ଼ନେସ୍ ) – unhappiness (ଦୁଃଖ | ବିଷାଦ)
scattered (ସ୍କାଟର୍‌ଡ଼) – seen over a wide area (ବିଛୁରିତ)
surrounded (ସରାଉଣ୍ଡେଡ୍) – covered from all sides
tiny – small (ଛୋଟ)
toiling (ଟଏଲିଙ୍ଗ୍) – doing hard labour (ଅତ୍ୟଧ୍ଵ ପରିଶ୍ରମ କରୁଥିବା)
wretched condition – very poor condition
similarity (ସିମିଲାରିଟି) – ସମାନତା
warrior (ୱାରିଅର୍) – soldier (ଯୋଦ୍ଧା, କିନ୍ତୁ ଏଠାରେ ଭୁଲ ବ୍ୟବସ୍ଥା ବିରୁଦ୍ଧରେ ଲଢ଼ୁଥିବା ବ୍ୟକ୍ତି ପାଇଁ ବ୍ୟବହାର କରାଯାଇଛି)।

BSE Odisha 7th Class English Solutions Follow-Up Lesson 2 A Tiny Warrior Important Questions and Answers

(A) Choose the right answer from the options.
Question 1.
Mitu studies in class VII but Situ studies in class __________.
(i) IX
(iii) IV
(iv) V
(ii) X
Answer:
(iv) V

Question 2.
Mr. giri was their _________.
(i) uncle
(ii) father
(iii) grandfather
(iv) brother
Answer:
(i) uncle

Question 3.
The Melbourne cricket ground is in ___________.
(i) America
(ii) Japan
(iii) Australia
(iv) Korea
Answer:
(iii) Australia

Question 4.
Situ’s answer was correct but the teacher didn’t give her any mark because he ___.
(i) her handwriting is not good
(ii) she copied it from another student
(iii) he thought wrong answer of bazzar notebok to be correct
(iv) his relation with Mitu is not good.
Answer:
(iii) he thought wrong answer of bazzar notebok to be correct

Question 5.
The headmistress made __________the leader of the movement against meaning books in their school.
(i) Situ
(ii) Mitu
(iii) Mr. Giri
(iv) the English teacher
Answer:
(i) Situ

(B) Answer the following questions.
Question 1.
Are bazzar note books useful for students? Why? or Why not?
Answer:
No, the bazzar note books are not useful for students. They have lot of mistakes and a student can’t learn English properly by using them.

Question 2.
Why did Mr. Giri want to help Mitu in learning English?
Answer:
Mr. Giri saw Mitu reading a paragraph of English lesson from textbook and then followed a bazzar note book for its meaning. As this would not help her in learning English properly, so he wanted to help her.

(C) Re-arrange the jumbled words to make meaningful sentences.
1. attacks / the / his / walls / were / protection / outside / from
Answer:
The walls were his protection from outside attacks.

2. the / in / nights/building / moonlit / looked / beautiful / very
Answer:
In moonlit nights, the building looked very beautiful.

3. man / the / lived / rich /, / there / a / like / king / happily / very
Answer:
The rich man lived there like a king very happily.

4. people / to / poor / the / one / the / day/man / rich / his / took / son
Answer:
One day the rich man took his son to the poor people.

5. food / they /, / theirs / buy / me / grow
Answer:
We buy food, they grow theirs.

(D) Find whether True or False.
1. Once upon a time there was a very rich man.
Answer:
True

2. He along with his family lived in a small hut down the hill.
Answer:
False

3. The building, the walls all around and the gardens inside were lit by moonlight.
Answer:
False

4. His only sorrow was that his Only son did not like all his richness and glamour.
Answer:
True

5. One day the rich man took his son to wander in the forest.
Answer:
False

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 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 BSE Odisha 10th Class Life Science Notes Chapter 1 ପୋଷଣ will enable students to study smartly.

BSE Odisha Class 10 Life Science Notes Chapter 1 ପୋଷଣ

प्रश्न और अभ्यास (ପ୍ରଶ୍ନ ଔର୍ ଅଭ୍ୟାସ)

→ ଉପକ୍ରମ (Introduction) :

• ‘‘ଜୀବନ ପ୍ରକ୍ରିୟା’ର ଗୁରୁତ୍ଵପୂର୍ଣ୍ଣ ପ୍ରକ୍ରିୟା ହେଉଛି ପୋଷଣ । ଶରୀରର ଅଭିବୃଦ୍ଧି ଓ ବିଭିନ୍ନ କ୍ଷୟକ୍ଷତିର ପୂରଣ ପାଇଁ ଖାଦ୍ୟର ଆବଶ୍ୟକତା ରହିଛି ।

→ ପୋଷଣ (Nutrition) :

• ଖାଦ୍ୟରୁ ଶକ୍ତି ନିର୍ଗତ ହୋଇଥାଏ । ଆବଶ୍ୟକ ଶକ୍ତି ଜୀବର ବାହ୍ୟ ପରିବେଶରୁ ଖାଦ୍ୟ ପରିବହନ ପ୍ରକ୍ରିୟା ସମ୍ପାଦିତ ହୁଏ, ତାହାକୁ ପୋଷଣ କୁହାଯାଏ ।
• ଖାଦ୍ୟରୁ ଶକ୍ତି ନିର୍ଗତ ହୋଇଥାଏ । ଆବଶ୍ୟକ ଶକ୍ତି ଜୀବର ବାହ୍ୟ ପରିବେଶରୁ ମିଳିଥାଏ । ଯେଉଁ ପ୍ରକ୍ରିୟାଦ୍ଵାରା ଜୀବର ବାହ୍ୟ ପରିବେଶରୁ ଅନ୍ତର୍ଦ୍ଦେଶକୁ  ସରଳୀକୃତ ଖାଦ୍ୟ ପରିବହନ ପ୍ରକ୍ରିୟା ସମ୍ପାଦିତ ହୁଏ, ତାହାକୁ ପୋଷଣ କୁହାଯାଏ।
• ଶରୀରରେ ହେଉଥ‌ିବା ସମସ୍ତ ଜୈବରାସାୟନିକ ପ୍ରକ୍ରିୟା ଯୋଗୁଁ ଜୀବନର ଧାରା ଅବ୍ୟାହତ ରହିଥାଏ ।
• ଜୀବ ଶରୀରରେ ଶକ୍ତି ଆହରଣ ଓ ଉପାଦାନ ସଂଗ୍ରହ, ପୋଷଣ ମାଧ୍ୟମରେ ହୋଇଥାଏ ।
• ସବୁଜ ଉଭିଦ ଆଲୋକଶ୍ଳେଷଣ ପ୍ରକ୍ରିୟାରେ ନିଜ ଖାଦ୍ୟ ନିଜେ ପ୍ରସ୍ତୁତ କରିଥାଏ ଓ ପରିବେଶରୁ ଆବଶ୍ୟକ ପୋଷକ ମଧ୍ୟ ଗ୍ରହଣ କରିଥାଏ ।
• ପ୍ରାଣୀ ଓ ଅନ୍ୟସବୁ ଜୀବ ଖାଦ୍ୟ ପାଇଁ, ଉଭିଦ ଉପରେ ପ୍ରତ୍ୟକ୍ଷ ବା ପରୋକ୍ଷ ଭାବରେ ନିର୍ଭରଶୀଳ ଅଟନ୍ତି । ଉପାଦାନରେ ପରିଣତ ହୋଇ ସବୁ ଅଙ୍ଗ-ପ୍ରତ୍ୟଙ୍ଗରେ ଉପଲବ୍‌ଧ ହୋଇଥାଏ ।
• ସରଳ ଖାଦ୍ୟର ଦହନ ଓ ଜାରଣ ଘଟି ଶକ୍ତି ମୋଚିତ ହୋଇଥାଏ ।
• ଶକ୍ତିମୋଚନ ଏକ ତଥାକଥ୍ ଧ୍ୱଂସାତ୍ମକ ପ୍ରକ୍ରିୟା । ଏହା ‘ଅପଚୟ’ର ଉଦାହରଣ ।
• ଖାଦ୍ୟରୁ ଶରୀର ଗଠନ ପାଇଁ ଆବଶ୍ୟକ ଉପାଦାନ ସୃଷ୍ଟିହେବା ଏକ ଗଠନମୂଳକ ପ୍ରକ୍ରିୟା ଅଟେ । ଏହାକୁ ‘ଚୟ’ କୁହାଯାଏ । ଚୟ ଓ ଅପଚୟର ସମାହାର ହେଉଛି ‘ଚୟାପଚୟ’ (ଚୟାପଚୟ = ଚୟ + ଅପଚୟ) । ଏହା ଜୀବ ଶରୀରରେ ସବୁବେଳେ ଚାଲିଥାଏ ।

→ ଖାଦ୍ୟର ପ୍ରକାରଭେଦ (Types of food) :

• ରାସାୟନିକ ଗଠନ, କାର୍ଯ୍ୟ ଓ ଶକ୍ତି ପ୍ରଦାନକାରୀ ତଥା ଶରୀର ଗଠନକାରୀ କ୍ଷମତା ଉପରେ ନିର୍ଭରକରି ଖାଦ୍ୟକୁ ମୁଖ୍ୟତଃ 6 ଭାଗର ବିଭକ୍ତ କରାଯାଇଛି । ଯଥା – ଶ୍ଵେତସାର, ପୁଷ୍ଟିସାର, ସ୍ନେହସାର, ଧାତୁସାର, ଭିଟାମିନ୍ ଏବଂ ଜଳ ।

→ ଶ୍ଵେତସାର (Carbohydrates) :

• ଆମେ ଖାଉଥ‌ିବା ଖାଦ୍ୟର ପ୍ରଧାନ ଶ୍ଵେତସାର ହେଉଛି ଶର୍କରା ଓ ମଣ୍ଡଦ ।
• ଶ୍ଵେତସାରରୁ ଆମେ ସହଜରେ ଶକ୍ତି ଆହରଣ କରିଥାଉ ।
• ଗୁ କୋଜି (C₆ H₈ O₆)ରେ ରହିଛି କାର୍ବନ, ଉଦଜାନ ଏବଂ ଅମ୍ଳଜାନ ।
• କୋଷୀୟ ଶ୍ୱସନ ବେଳେ ଗ୍ଲୁକୋଜର ଜାରଣ ଫଳରେ ଅଙ୍ଗାରକାମ୍ଳ ଓ ଜଳ ସୃଷ୍ଟି ହେବା ସହିତ ଶକ୍ତି ନିର୍ଗତ ହୋଇଥାଏ ।
  CH₂O + 6O₂ → 6CO₂ + 6HO + ଶକ୍ତି
• ଏକ ଗ୍ରାମ୍ ଶ୍ଵେତସାରରୁ ପ୍ରାୟ 16 କିଲୋ ଜୁଲ (KJ) ଶକ୍ତି ନିର୍ଗତ ହୋଇଥାଏ ।

I. ଆଳୁ, ଭାତ ଓ ରୁଟିରେ ଥ‌ିବା ଶ୍ଵେତସାର ମଣ୍ଡଦ ଅଟେ ।
II. ଚିନି ଓ ଗୁଡ଼ରେ ଥ‌ିବା ଶ୍ଵେତସାର ସୁକ୍ରୋଜ ଅଟେ ।
III. ଫଳରସ ଓ ପନିପରିବାରେ ଥିବା ଶ୍ଵେତସାର ଗ୍ଲୁକୋଜ ଅଟେ ।

→ ପୁଷ୍ଟିସାର (Protein) :

• ମାଛ, ମାଂସ, ଅଣ୍ଡାର ଧଳା ଅଂଶ, ଛେନା ଓ କ୍ଷୀର ପରି ପ୍ରାଣିଜ ଦ୍ରବ୍ୟ ଓ ଡାଲି ଜାତୀୟ ଶସ୍ୟ, ସୋୟାବିନ୍ ଆଦିରୁ ଆମେ ଉଭିଦଜାତ ପୁଷ୍ଟିସାର ପାଇଥାଉ ।
• ବିଭିନ୍ନ କୋଷ ନିକଟରେ ପହଞ୍ଚାଏ । ଶରୀରର ବୃଦ୍ଧି, ନୂତନ କୋଷ ଓ ତନ୍ତୁ ଗଠନ ପାଇଁ ପୁଷ୍ଟିସାର ଏକାନ୍ତ ଆବଶ୍ୟକ ।

→ (Fats / Lipids) :

• ମାଂସ, କ୍ଷୀର, ଛେନା, ଲହୁଣୀ, ଅଣ୍ଡାର ହଳଦିଆ ଅଂଶ ଓ ତେଲ, ଘିଅରେ ସ୍ନେହସାର ଜାତୀୟ ଖାଦ୍ୟ ରହିଥାଏ । ଶରୀରରେ ସ୍ନେହସାର ଚର୍ବି ଭାବରେ
• ସଂଚିତ ହୋଇ ରହେ ଓ ଆବଶ୍ୟକ ସ୍ଥଳେ କୋଷୀୟ ଶ୍ଵାସନଦ୍ୱାରା ଏହାର ଜାରଣ ହୁଏ ଓ ଏହା ଶରୀରକୁ ଶକ୍ତି ଯୋଗାଇଥାଏ ।
• ଚର୍ମତଳେ ଚର୍ବିର ଏକ ଆସ୍ତରଣ ରହିଥାଏ ଓ ଏହା ତାପ ଅପରିବାହୀ ହୋଇଥିବାରୁ ଶରୀରକୁ ଉଷୁମ ରଖିବାରେ ସାହାଯ୍ୟ କରିଥାଏ । କୋଷଝିଲ୍ଲୀ ଗଠନ ପାଇଁ ଓ ଶକ୍ତି ପାଇଁ ଏହାର ମୁଖ୍ୟ ଭୂମିକା ରହିଛି ।

→ ଧାତୁସାର (Minerals) :

• ଶରୀର ଗଠନପାଇଁ ବିଭିନ୍ନ ଧରଣର ଧାତୁସାର ଯଥା-ଲୌହ (Fe), କ୍ୟାଲସିୟମ୍ (Ca), ଆୟୋଡ଼ିନ୍ (I), ଫସ୍‌ଫରସ୍ (P), ସୋଡ଼ିୟମ୍‌(Na), ପୋଟାସିୟମ୍ (K) ଆଦି ଆବଶ୍ୟକ ।
• ଶରୀରର ଆୟନ ସନ୍ତୁଳନ (Ionic balance) ରକ୍ଷା କରିବାରେ ଧାତୁସାରର ପ୍ରମୁଖ ଭୂମିକା ରହିଥାଏ ।

ଶରୀରରେ ଦାନ୍ତ ଓ ହାଡ଼ର ଗଠନ ପାଇଁ କ୍ୟାଲସିୟମ୍ (Ca) ଓ ଲୋହିତ ରକ୍ତ କଣିକା (RBC)ରେ ଥିବା ହିମୋଗ୍ଲୋବିନ୍‌ର ଗଠନ ପାଇଁ ଲୌହ ଆବଶ୍ୟକ ହୋଇଥାଏ ।

→ ଭିଟାମିନ୍ (Vitamins) :

• କୋଷରେ ବିଭିନ୍ନ ପ୍ରକାର ରାସାୟନିକ ପ୍ରକ୍ରିୟା ଏନଜାଇମ୍ ସାହାଯ୍ୟରେ ହୋଇଥାଏ ।
• ଭିଟାମିନ୍ ଅଭାବରୁ ଶରୀରରେ ବିଭିନ୍ନ ପ୍ରକାରର ରୋଗ ହୋଇଥାଏ ।
• ଭିଟାମିନ୍ ଅଭାବରୁ ଶରୀରରେ ବିଭିନ୍ନ ପ୍ରକାରର ରୋଗ ହୋଇଥାଏ ।

→ ଜଳ (Water):

• କୋଷରେ ଥ‌ିବା କୋଷରସ ବା କୋଷ ଜୀବକର ପ୍ରାୟ 70-୨୦ ଭାଗ ଜଳ ଅଟେ ।
• କୋଷର ସ୍ଥିତି ଓ ଏଥରେ ହେଉଥ‌ିବା ବିଭିନ୍ନ ପ୍ରକ୍ରିୟା ପାଇଁ ଜଳ ଏକାନ୍ତ ଆବଶ୍ୟକ ।
• ଝାଳ, ପରିସ୍ରା ଓ ନିଃଶ୍ବାସରେ ଶରୀରରୁ ଜଳ କ୍ଷୟ ହୋଇଥାଏ, ତାହାର ଭରଣା ପାଇଁ ପ୍ରତିଦିନ ପ୍ରାୟ 3 – 4 ଲିଟର ପାଣି ପିଇବା ଉଚିତ ।
• ଶରୀରରେ ଜଳୀୟ ଅଂଶ କମିଗଲେ ଶରୀର ଅବଶ ହୋଇଯାଏ ଓ ବିଭିନ୍ନ ଅସୁସ୍ଥତା ଦେଖାଦିଏ ।
  

→ ପୋଷଣର ପ୍ରକାରଭେଦ (Types of Nutrition) :
→ (Autotrophic nutrition):

• ଯେଉଁ ଜୀବମାନେ ନିଜ ଖାଦ୍ୟ ନିଜେ ପ୍ରସ୍ତୁତ କରିପାରନ୍ତି, ସେମାନଙ୍କୁ ସ୍ଵଭୋଜୀ କୁହାଯାଏ ।
• ପତ୍ରହରିତ୍‌ ଥ‌ିବା ସମସ୍ତ ଉଭିଦ ଓ ନୀଳ ହରିତ ଶୈବାଳ ସୁଭୋଜୀର ଉଦାହରଣ ଅଟନ୍ତି ।
• ସେମାନେ ପରିବେଶରୁ କଞ୍ଚାମାଲ ଓ ଶକ୍ତି ବ୍ୟବହାର କରି ଆବଶ୍ୟକ କରୁଥିବା ସମସ୍ତ ପ୍ରକାର ଖାଦ୍ୟ ସେମାନଙ୍କ ଟିସୁ ମଧ୍ୟରେ ପ୍ରସ୍ତୁତ କରନ୍ତି ।
• ଏମାନଙ୍କ ପୋଷଣକୁ ସ୍ବଭୋଜୀ ପୋଷଣ କୁହାଯାଏ । ଏହା ଦୁଇଟି ଉପାୟରେ ହୋଇଥାଏ । ଯଥା – ଆଲୋକଶ୍ଳେଷଣ ଓ ରସାୟଶ୍ଳେଷଣ ।

→ (Photosynthesis) :

• ସବୁଜ ଉଭିଦ ଓ ନୀଳ ହରିତ୍ ଶୈବାଳ ସୂର୍ଯ୍ୟର ଆଲୋକ ଶକ୍ତିକୁ ଉପଯୋଗ କରି ସବୁଜକଣିକାର ଉପସ୍ଥିତିରେ ଅଙ୍ଗାରକାମ୍ଳ ଓ ଜଳ ମଧ୍ୟରେ ସଂଯୋଗ ଘଟାଇ ଶ୍ଵେତସାର ଜାତୀୟ ଖାଦ୍ୟ ପ୍ରସ୍ତୁତ କରିଥା’ନ୍ତି । ଏହି ପ୍ରକ୍ରିୟାକୁ ଆଲୋକଶ୍ଳେଷଣ କୁହାଯାଏ ।
  

→ (Chemosynthesis):
ନାଇଟ୍ରେଟ୍ ପ୍ରସ୍ତୁତକାରୀ ବ୍ୟାକ୍ଟେରିଆ ଓ ଗନ୍ଧକ ବ୍ୟାକ୍ଟେରିଆ ପରି କେତେକ ରସାୟଶ୍ଳେଷଣ ବ୍ୟାକ୍ଟେରିଆ ଏକ ବିଶେଷ ଅଜୈବ ରାସାୟନିକ ପ୍ରକ୍ରିୟାରୁ ମିଳୁଥିବା ରାସାୟନିକ ଶକ୍ତି ସଂଗ୍ରହ କରିଥା’ନ୍ତି । ଏହି ପ୍ରକ୍ରିୟାକୁ ରସାୟଶ୍ଳେଷଣ କୁହାଯାଏ ।

→ (Heterotrophic Nutrition) :

• ଯେଉଁ ଜୀବମାନେ ନିଜ ଖାଦ୍ୟ ନିଜେ ପ୍ରସ୍ତୁତ କରିନପାରି ପୋଷଣ ପାଇଁ ଅନ୍ୟ ପ୍ରାଣୀ ବା ଉଭିଦ ଉପରେ ନିର୍ଭର କରନ୍ତି ସେମାନଙ୍କୁ ପରଭୋଜୀ କୁହାଯାଏ।
• ସମସ୍ତ ପ୍ରାଣୀ, ମଲାଙ୍ଗ, ନିର୍ମୂଳୀ, ରାଫ୍ଲେସିଆ ଆଦି ପରଜୀବୀ ଉଭିଦ, କବକ ଏବଂ ଅଧିକାଂଶ ବ୍ୟାକ୍ଟେରିଆ ପରଭୋଜୀ
• ଏହି ଜୀବମାନଙ୍କ ପୋଷଣ ପ୍ରଣାଳୀକୁ ପରଭୋଜୀ ପୋଷଣ କୁହାଯାଏ ।

→ ପରଭୋଜୀ ପୋଷଣର ପ୍ରକାରଭେଦ :
ପରଭୋଜୀ ପୋଷଣ ମୁଖ୍ୟତଃ ଚାରିପ୍ରକାର ।

(i) (Holozoic nutrition) :

• ପରଜୀବୀୟ ପ୍ରାଣୀଙ୍କୁ ଛାଡ଼ି ଅନ୍ୟ ସମସ୍ତ ପ୍ରାଣୀ ଏହି ପ୍ରଣାଳୀରେ ଅନ୍ୟ ଉଦ୍ଭଦ ବା ପ୍ରାଣୀଙ୍କୁ ସଂପୂର୍ଣ୍ଣ ଅଥବା ଆଂଶିକ ଭାବରେ ଖାଦ୍ୟରୂପେ ଗ୍ରହଣ କରିଥା’ନ୍ତି ।
• ପରିପାକ ପରେ ସରଳୀକୃତ ଖାଦ୍ୟର ଆତ୍ମୀକରଣ ବା ଅନ୍ତର୍ଗହଣ ହୋଇଥାଏ ।
• ଏହା ଶରୀର ଗଠନରେ ଓ ଶରୀରକୁ କାର୍ଯ୍ୟକ୍ଷମ ରଖିବାରେ ସହାୟକ ହୋଇଥାଏ ।

(ii) ମୃତୋପଜୀବୀୟ ପୋଷଣ (Saprophytic nutrition) :

• ଯେଉଁ ପରଭୋଜୀ, ମୃତ, ଗଳିତ, ପଚାସଢ଼ା ଉଭିଦ ବା ପ୍ରାଣୀରୁ ଖାଦ୍ୟ ସଂଗ୍ରହ କରି ନିଜ ପୁଷ୍ଟିସାଧନ କରିଥା’ନ୍ତି ସେମାନଙ୍କୁ ମୃତୋପଜୀବୀ କୁହାଯାଏ ।
• ଏହି ଜୀବମାନେ କଠିନ ପଦାର୍ଥକୁ ଖାଦ୍ୟରୂପେ ଗ୍ରହଣ କରିପାରନ୍ତି ନାହିଁ ।
• ସାଧାରଣତଃ ଖାଦ୍ୟ ଗ୍ରହଣବେଳେ ଏମାନେ ନିଜ ଶରୀରରୁ ପାଚକ ରସ କ୍ଷରଣ କରି, ଶରୀର ବାହାରେ ହିଁ ଜଟିଳ ଖାଦ୍ୟକୁ ସରଳ ଖାଦ୍ୟରେ ପରିଣତ କରି ଓ ପରେ ସରଳୀକୃତ ଖାଦ୍ୟକୁ ଶରୀର ମଧ୍ୟକୁ ଶୋଷଣ କରି ଶରୀର ଗଠନରେ
• ଛତୁ ଜାତୀୟ କବକ, ଇଷ୍ଟ୍, ବ୍ୟାକ୍ଟେରିଆ ଆଦି ଜୀବମାନଙ୍କଠାରେ ଏହିପରି ପୋଷଣ ଦେଖିବାକୁ ମିଳିଥାଏ ।

(iii) (Parasitic Nutrition) :

• ଯେଉଁ ଜୀବମାନେ ଅନ୍ୟ ଜୀବନ୍ତ ଉଭିଦ ବା ପ୍ରାଣୀଙ୍କ ଶରୀର ଭିତରେ ବା ବାହାରେ ରହି ସେମାନଙ୍କଠାରୁ ଖାଦ୍ୟ ସଂଗ୍ରହ କରି ନିଜର ପୁଷ୍ଟିସାଧନ କରନ୍ତି ସେମାନଙ୍କୁ ପରଜୀବୀ କୁହାଯାଏ ।
• ପରଜୀବୀମାନେ ଭୋଜଦାତା ଉଭିଦ ବା ପ୍ରାଣୀଙ୍କଠାରୁ ସରଳୀକୃତ ଖାଦ୍ୟ ସିଧାସଳଖ ଗ୍ରହଣ କରି ନିଜର ପୁଷ୍ଟିସାଧନ କରିଥା’ନ୍ତି ।
• ଭୋଜଦାତାର ଆଶ୍ରୟରେ ରହି ପରଜୀବୀମାନେ ସାଧାରଣତଃ ତାହାର ଅନିଷ୍ଟ କରିଥା’ନ୍ତି ।
• ମଲାଙ୍ଗ, ନିର୍ମୂଳୀ, ରାଫ୍ଲେସିଆ ଆଦି ଉଭିଦ, ପ୍ଲାସ୍‌ଡ଼ିୟମ୍, ଉକୁଣୀ, ଜୋକ, କେତେକ କୃମି ପରି ପ୍ରାଣୀ ପରଜୀବୀ

କେତେକ ଭୋଜଦାତାର ଶରୀର ଭିତରେ ଅନ୍ତଃପରଜୀବୀ ଭାବେ ( ଉଦାହରଣ- ପ୍ଲାସ୍‌ମୋଡ଼ିୟମ୍) ଓ କେତେକ ଶରୀର ବାହାରେ ବାହ୍ୟପରଜୀବୀ ଭାବେ ( ଉଦାହରଣ – ଉକୁଣୀ) ରହିଥା’ନ୍ତି ।

(iv) (Symbiotic Nutrition) :

• ବେଳେବେଳେ ଦୁଇଟି ସଂପୂର୍ଣ୍ଣ ଭିନ୍ନ ଜାତିର ପ୍ରାଣୀ, ଅଥବା ଉଭିଦ ଓ ପ୍ରାଣୀ ବା ପ୍ରାଣୀ ଓ ଅଣୁଜୀବ ବା ଉଭିଦ ଓ ଅଣୁଜୀବ ଏକାଠି ବାସ କରୁଥିବା ଦେଖାଯାଏ । ଏହାକୁ ସହଜୀବୀତା କୁହାଯାଏ ।
• ଏଥୁରେ କେହି କାହାରି କ୍ଷତି କରନ୍ତି ନାହିଁ, ବରଂ ସେମାନଙ୍କ ଭିତରେ ପୋଷଣର ଆଦାନ ପ୍ରଦାନ ମଧ୍ୟ ହୋଇଥାଏ । ଏହାକୁ ସହଜୀବୀୟ ପୋଷଣ କୁହାଯାଏ।
• ଆମ ଅନ୍ତନଳୀରେ ସହଜୀବୀଭାବେ ରହୁଥ‌ିବା ଇସ୍‌ରିଚିଆ କୋଲାଇ ନାମକ ବ୍ୟାକ୍ଟେରିଆ ନିଜ ଶରୀରରେ ଭିଟାମିନ୍ (B₁₂) (ସାୟନୋକୋବାଲାମିନ୍) ପ୍ରସ୍ତୁତ କରି ଆମକୁ ଯୋଗାଇଥାଏ । ତା’ ପରିବର୍ତ୍ତେ ଆମ ଅନ୍ତନଳୀର ସରଳୀକୃତ ଖାଦ୍ୟଗ୍ରହଣ କରି ନିଜର ପ୍ରତିପାଳନ କରିଥାଏ ।
• 

→ ଖାଦ୍ୟାଭାସକୁ ଆଧାର କରି ମୁଖ୍ୟତଃ ତିନି ଜାତିର ପ୍ରାଣୀ ଅଛନ୍ତି ।
(a) ଶାକାହାରୀ – ଉଭିଦ ବା ଉଭିଦଜାତ ପଦାର୍ଥ ଭକ୍ଷଣ କରୁଥିବା ପ୍ରାଣୀ ।
(b) ମାଂସାହାରୀ – ଅନ୍ୟ ପ୍ରାଣୀ ବା ପ୍ରାଣିଜ ପଦାର୍ଥକୁ ଭକ୍ଷଣ କରୁଥିବା ପ୍ରାଣୀ ।
(c) ସର୍ବାହାରୀ – ଖାଦ୍ୟରେ ବାଛବିଚାର ନ କରି ଯାହା ଖାଦ୍ୟୋପଯୋଗୀ ତାହା ଭକ୍ଷଣ କରୁଥିବା ପ୍ରାଣୀ । ଆଲୋକଶ୍ଳେଷଣ

→ (Photosynthesis) :

• ସବୁଜ ଉଭିଦ ଆଲୋକଶ୍ଳେଷଣ ପ୍ରକ୍ରିୟାରେ ଶ୍ଵେତସାର ଜାତୀୟ ଖାଦ୍ୟ ପ୍ରସ୍ତୁତ କରିଥା’ନ୍ତି । ଏହି ଶ୍ଵେତସାର ଜାତୀୟ ଖାଦ୍ୟ ଜୀବଜଗତର ସବୁ ଜୀବଙ୍କ ପାଇଁ ପ୍ରତ୍ୟକ୍ଷ ବା ପରୋକ୍ଷ ଖାଦ୍ୟ ଅଟେ ।
• ଯେଉଁ ପ୍ରକ୍ରିୟାଦ୍ଵାରା ସବୁଜ ଉଭିଦ ସବୁଜକଣା ବା କ୍ଲୋରୋଫିଲ୍ ମାଧ୍ୟମରେ ଦୃଶ୍ୟମାନ ଆଲୋକ ଶକ୍ତିକୁ ରାସାୟନିକ ଶକ୍ତିରେ ରୂପାନ୍ତରିତ କରିବା ସହିତ ପରିବେଶରୁ ଗ୍ରହଣ କରିଥିବା ଅଙ୍ଗାରକାମ୍ଳ ଓ ଜଳକୁ ଉପଯୋଗ କରି ସରଳ ଶର୍କରା ତିଆରି କରିଥାଏ, ତାହା ଆଲୋକଶ୍ଳେଷଣ ପ୍ରକ୍ରିୟା ଅଟେ ।

→ ଆଧାର ଓ ସଂସ୍ଥା:

• ପ୍ରତ୍ୟେକ ପ୍ରକ୍ରିୟା ପାଇଁ ଏକ ଆଧାର ଓ ସଂସ୍ଥା ଆବଶ୍ୟକ ହୋଇଥାଏ ।
• ସବୁଜ ଉଦ୍ଭଦରେ ଆଲୋକଶ୍ଳେଷଣ ପାଇଁ ବ୍ୟବହୃତ ଆଧାର ସବୁଜ ପତ୍ର ଅଟେ । ସବୁଜ ପତ୍ରର ପୃଷ୍ଠରେ ଥ‌ିବା ଛୋଟ ଛୋଟ ରନ୍ଧ୍ରଗୁଡ଼ିକୁ ସ୍ତୋମ୍ ବା ଷ୍ଟୋମାଟା କୁହାଯାଏ । ଏହି ସ୍ତୋମ୍ ଦେଇ ପରିବେଶ ଓ ପତ୍ର ମଧ୍ୟରେ ଅଙ୍ଗାରକାମ୍ଳ ଓ ଜଳୀୟବାଷ୍ପର ବିନିମୟ ଘଟେ । ସବୁଜପତ୍ରର ଅନ୍ତଃଗଠନ ସବୁଜ ରଙ୍ଗଯୁକ୍ତ ପାଲିସେଡ଼୍ ଓ ସ୍ପଞ୍ଜି ପାରେନ୍‌କାଇମା ଟିସୁଦ୍ୱାରା ହୋଇଥାଏ । ପତ୍ର ଭିତରେ ଥ‌ିବା ଏହି ଦୁଇ ପ୍ରକାର ପାରେନ୍‌କାଇମା ଟିସୁର କୋଷ ଭିତରେ କ୍ଲୋରୋପ୍ଲାଷ୍ଟ ନାମକ ଅଙ୍ଗିକା ରହିଥାଏ । କ୍ଲୋରୋପ୍ଲାଷ୍ଟରେ ଥ‌ିବା ରସକୁ ବ୍ୟୋମା କୁହାଯାଏ ।
• ଥାଇଲାକଏଡ଼ ଦୀର୍ଘ ସରୁ ଚେପଟା ଥଳି ସଦୃଶ । ଥଳିର ଭିତର ସ୍ଥାନକୁ ମ୍ୟୁମେନ କୁହାଯାଏ । କେତେକ ସ୍ଥାନରେ ଛୋଟ ଥାଇଲାକଏଡ଼ ଥାକ ଥାକ ହୋଇ ସଜ୍ଜିତ ହୋଇ ରହିଥା’ନ୍ତି । ଏଭଳି ଥାକକୁ ଗ୍ରାନା କୁହାଯାଏ ।

→ ଆଧାର :
1. ସବୁଜ ପତ୍ରରେ ଥ‌ିବା ସ୍ତୋମ
2. ସବୁଜ ରଙ୍ଗଯୁକ୍ତ ପାଲିସେଡ଼ ଓ ସ୍ପଞ୍ଜି ପାରେନ୍‌କାଇମା ।
3. କ୍ଲୋରୋପ୍ଲାଷ୍ଟ

→ କ୍ଲୋରୋପ୍ଲାଷ୍ଟ :

• ବ୍ୟୋମା (ରସ)ରେ ଏନ୍‌ଜାଇମ୍ ଥାଏ ।
• ଦ୍ୱସ୍ତରୀୟ ଥାଇଲାକଏଡ୍.
• ଭିତର ଙ୍ଗାନ ଲୁମେନ୍‌
• ଥାକକୁ ଗ୍ରାନା କହନ୍ତି ।

ଥାଇଲାକଏଡ଼ର ପ୍ରତ୍ୟେକ ଝିଲ୍ଲୀ ସ୍ତରରେ କ୍ଲୋରୋଫିଲ୍, ପ୍ରୋଟିନ୍ ଓ ଲିପିଡ଼ର ବିଭିନ୍ନ ବୃହତ୍ ଅଣୁ ସଜେଇ ହୋଇ ରହିଥା’ନ୍ତି । ଏହିଭଳି ଗଠନଯୁକ୍ତ ଥାଇଲାକଏଡ଼ ଓ କ୍ଲୋରୋପ୍ଲାଷ୍ଟର ଷ୍ଟ୍ରୋମା ରସ ‘ଆଲୋକଶ୍ଳେଷଣ ସଂସ୍ଥା’ ସୃଷ୍ଟି କରିଥା’ନ୍ତି ।

→ ପ୍ରକ୍ରିୟା :

• ଆଲୋକଶ୍ଳେଷଣ ପ୍ରକ୍ରିୟାକୁ ଜାଣିବାପାଇଁ ବିଭିନ୍ନ ପରୀକ୍ଷା କରାଯାଇଛି ।
• ପ୍ରଥମ ସିଦ୍ଧାନ୍ତ 1905 ମସିହାରେ ଫ୍ରେଡ଼ରିକ୍ ବ୍ଲାକ୍‌ମ୍ୟାନ୍ (Frederick Blackman) ଜଣାଇଥିଲେ ।
• ଗୋଟିଏ ସହ ପ୍ରକ୍ରିୟା ଦୃଶ୍ୟମାନ ଆଲୋକ ଉପରେ ନିର୍ଭର କରେ । ଏହାକୁ ଆଲୋକ ପ୍ରକ୍ରିୟା କୁହାଯାଏ । ଅନ୍ୟଟି ଆଲୋକ ଉପରେ ନିର୍ଭର କରିନଥାଏ । ଏହାକୁ ଅନ୍ଧକାର ପ୍ରକ୍ରିୟା କୁହାଯାଏ ।

• ଆସୁଥ‌ିବା ହାଇଡ୍ରୋଜେନ୍‌ଦ୍ୱାରା ଅଙ୍ଗାରକାମ୍ଳ ବିଜାରିତ ହୁଏ ଓ ସରଳ ଶର୍କରା ସଂଶ୍ଳେଷିତ ହେବା ସହିତ ଅମ୍ଳଜାନ ନିର୍ଗତ ହେବା ଦର୍ଶାଯାଇଥିଲା । ଏହା 1931 ମସିହରେ ଫନ୍‌ ନିଲ୍‌ଙ୍କଦ୍ବାରା ପରିକଳ୍ପନା କରାଯାଇଥିଲା । ରବର୍ଟ ହିଲ୍ 1937 ମସିହାରେ ଉନ୍ନତମାନର ପରୀକ୍ଷଣ ମାଧ୍ୟମରେ ପରିକଳ୍ପନାଟିକୁ ପ୍ରମାଣିତ କରିଥିଲେ । କ୍ଲୋରୋପ୍ଲାଷ୍ଟ କ୍ଲୋରୋଫିଲ୍ ମାଧ୍ୟମରେ ଆଲୋକ ଶକ୍ତିକୁ ବ୍ୟବହାର କରି ଗ୍ଲୁକୋଜ ସଂଶ୍ଳେଷଣ କରିଥାଏ । ଏହା ପାଇଁ 6 ଟି ଅଙ୍ଗାରକାମ୍ଳ (CO₂) ଓ 12 ଟି ଜଳ (H₂O) ଅଣୁ ବ୍ୟବହୃତ ହୁଏ ।

→ ଆଲୋକ ପ୍ରକ୍ରିୟା

• ଆଲୋକ ପ୍ରକ୍ରିୟା ଦୃଶ୍ୟମାନ ଆଲୋକ ଉପସ୍ଥିତିରେ ଥାଇଲାକଏଡ୍ ଝିଲ୍ଲୀରେ ସଂଗଠିତ ହୁଏ ।
• ପ୍ରଥମ ପର୍ଯ୍ୟାୟରେ ଝିଲ୍ଲୀରେ ସଜ୍ଜିତ କ୍ଲୋରୋଫିଲ୍ ଅଣୁଗୁଡ଼ିକର ସମଷ୍ଟି ଆଲୋକ ଶକ୍ତି ଗ୍ରହଣ କରନ୍ତି ।
• ପର୍ଯ୍ୟାୟକ୍ରମେ ଅନ୍ୟାନ୍ୟ କ୍ଲୋରୋଫିଲକୁ ଶକ୍ତି ସ୍ଥାନାନ୍ତରିତ କରିଥା’ନ୍ତି ।
• କ୍ଲୋରୋଫିଲ୍‌କୁ ଆଲୋକ ପ୍ରତିକ୍ରିୟା କେନ୍ଦ୍ର ବା Photosystem I କୁ P700 କୁହାଯାଏ ।
• କ୍ଲୋରୋଫିଲ୍‌ରୁ ଅଧ‌ିକ ଶକ୍ତି ସମ୍ପନ୍ନ ଅସ୍ଥିର ଇଲେକଟ୍ରନ୍ ବାହାରି ଆସେ ।
• ଇଲେକ୍ସନ୍ ବିଭିନ୍ନ ବାହକ ଅଣୁ ମାଧ୍ୟମରେ ଗତିକରି ଏକ ଗ୍ରାହକ ଅଣୁ ପାଖରେ ପହଞ୍ଚେ ।
• ଶେଷ ଗ୍ରାହକ ଅଣୁକୁ ସହକାରକ କୁହାଯାଏ ।
• ନିକୋଟିନାମାଇଡ୍ ଏଡ଼େନାଇନନ୍ ଡାଇନ୍ୟୁକ୍ଲିଟାଇଡ଼ ଫସ୍‌ଫେଟ୍ ବା NADP⁺ ଇଲେକଟ୍ରନ୍ ଗ୍ରହଣ କରି ବିଜାରିତ NADPH ରେ ପରିଣତ ହୁଏ ।
  ଦ୍ୱିତୀୟ ପର୍ଯ୍ୟାୟରେ ଇଲେକ୍ସନ୍ ଶୂନ୍ୟତାକୁ ପୂରଣ କରିବାପାଇଁ ଆଲୋକ ପ୍ରତିକ୍ରିୟା କେନ୍ଦ୍ର P680 ବା Photosystem II କେନ୍ଦ୍ରରୁ ଇଲେକଟ୍ରନ୍ ଆସିଥାଏ ।
• ଏହାଦ୍ଵାରା ଥାଇଲାକଏଡ ପରିବେଶରେ ଜଳ ଅଣୁର ଆଲୋକ ବିଶ୍ଳେଷଣ ହୁଏ । ଫଳରେ e^–, H⁺, Q, ନିର୍ଗତ ହୁଏ ।
• ଥାଇଲାକଏଡ ମଧ୍ଯସ୍ଥିତ ଲ୍ୟୁମେନରେ ପ୍ରୋଟନ ଜମା ହୁଏ ।
• ଏହାଦ୍ୱାରା ସୃଷ୍ଟ ଅବକ୍ରମ ବଳକୁ ଉପଯୋଗ କରି ADP ଅଧ‌ିକ ଶକ୍ତି ସମ୍ପନ୍ନ ATP ରେ ପରିଣତ ହୁଏ ।
• NADPH ଓ ATP ଉଭୟ ମିଶି ଆଲୋକଶ୍ଳେଷଣ ଶକ୍ତି ଗଠନ କରନ୍ତି ।

→ ଅନ୍ଧକାର ପ୍ରକ୍ରିୟା

• ଆଲୋକ ଉପରେ ଅନ୍ଧକାର ପ୍ରକ୍ରିୟା ନିର୍ଭର କରିନଥାଏ ।
• ଅଙ୍ଗାରକାମ୍ଳ ସ୍ତୋମ୍ ଦେଇ ଷ୍ଟୋମାରସରେ ଦ୍ରବୀଭୂତ ହୁଏ ।
• ବ୍ୟୋମା ରସରେ 5-କାର୍ବନ ଯୁକ୍ତ ଏକ ଗ୍ରାହକ ଅଣୁ ଆଲୋକଶ୍ଳେଷଣ ଶକ୍ତି ବ୍ୟବହାର କରି ଏନ୍‌ଜାଇମ୍ ମାଧ୍ୟମରେ ବିବନ୍ଧିତ କରାଏ । ଜୈବିକ କ୍ରିୟା ଦ୍ବାରା 3-କାର୍ବନ ଯୁକ୍ତ ଶର୍କରା ତିଆରି ହୁଏ ।
• ଗ୍ରାହନ ଅଣୁକୁ ରାଇବୋଲୋଜ୍ ବିସ୍‌ଫସ୍‌ଫେଟ୍ ବା RUBP ଓ ଏନ୍‌ଜାଇମ୍କୁ ରାଇବୋଲୋଜ୍ ବିସ୍‌ଫସଫେଟ୍ ରୁବିସ୍କୋ କୁହାଯାଏ ।
• ରୁବିସ୍କୋର ଭୂମିକା ଗୁରୁତ୍ବପୂର୍ଣ୍ଣ ।

• ଏହି ଅନ୍ଧକାର ପ୍ରକ୍ରିୟାକୁ ମେଲଭିନ୍ କେଲଭିନ୍‌ଙ୍କ ନାମାନୁସାରେ କେଲଭିନ୍‌ ଚକ୍ର କୁହାଯାଏ ।
• ପ୍ରଥମ ପର୍ଯ୍ୟାୟରେ ଗ୍ରାହକ ଅଣୁ ସହ ଅଙ୍ଗାରକାମ୍ଳର ବିବନ୍ଧନ ହୁଏ ।
• ଦ୍ଵିତୀୟ ପର୍ଯ୍ୟାୟରେ ଗ୍ଲୁକୋଜର ସଂଶ୍ଳେଷଣ ହୁଏ ।
• ତୃତୀୟ ପର୍ଯ୍ୟାୟରେ ଗ୍ରାହକ ଅଣୁର ପୁନରୁତ୍ପାଦନ ହୁଏ ।

→ (Digestive System of Man)

• ଆମେ ଖାଉଥ‌ିବା ଖାଦ୍ୟ ସମୂହ ସିଧାସଳଖ ଖାଦ୍ୟରେ ପରିଣତ ହୁଏ ଓ ଶେଷରେ ରକ୍ତରେ ମିଶି ଥାଏ । ଏହାକୁ ହଜମ (ପରିପାକ) ବା ଜୀର୍ଣ୍ଣ ହେବା କହିଥାଉ । ଅଦରକାରୀ ଅଂଶ ମଳ ରୂପେ ଶରୀରରୁ ନିଷ୍କାସିତ ହୋଇଥାଏ । ଆମର ପାକତନ୍ତ୍ର ପାକନଳୀ ଓ ପାକଗ୍ରନ୍ଥିକୁ ନେଇ ଗଠିତ ।

→ (Alimentary Canal)

• ପାକନଳୀ ପାଟିରୁ ଆରମ୍ଭ ହୋଇ ମଳଦ୍ଵାରରେ ଶେଷ ହୋଇଥାଏ ।
• ପାକନଳୀର ଗଠନ ଓ କାର୍ଯ୍ୟ ଉପରେ ନିର୍ଭର କରି ଏହାକୁ ବିଭିନ୍ନ ଅଂଶରେ ଭାଗ ଗ୍ରସନୀ, ନିଗଳ ବା ଗ୍ରାସନଳୀ, ପାକସ୍ଥଳୀ, କ୍ଷୁଦ୍ରାନ୍ତ୍ର, ବୃହଦନ୍ତ୍ର, ମଳାଶୟ ଓ ମଳଦ୍ଵାର ।

ପାକନଳୀ ଦେଖିବାକୁ ଗୋଟିଏ ଲମ୍ବ ଟ୍ୟୁବ୍ ପରି । ଏହାର କାନ୍ଥ ବର୍ତ୍ତୁଳ ବା ଚକ୍ରାକୃତି ପେଶୀ ଓ ଲମ୍ବ ଭାବରେ ବିସ୍ତୃତ ବା ଅନୁଦୈର୍ଘ୍ୟ ପେଶୀଦ୍ଵାରା ଗଠିତ ହୋଇଥାଏ । ଏହି ଦୁଇ ପ୍ରକାର ପେଶୀର ସଂକୋଚନ ଓ ଶିଳନ ଫଳରେ କୁହାଯାଏ ।

(i) ପାଟି (Mouth) :

• ଏହା ଚଳନଶୀଳ ଓଠଦ୍ୱାରା ପରିବେଷ୍ଟିତ ।
• ଏହା ଖାଦ୍ୟର ଅନ୍ତର୍ଗହଣ ପାଇଁ ଉଦ୍ଦିଷ୍ଟ ।
• ଏହା ମୁଖଗହ୍ଵରକୁ ଖୋଲି ରହିଛି ।

(ii) ମୁଖଗହ୍ଵର (Buccal Cavity) :

• ପାଟି ଭିତରକୁ ରହିଥାଏ ମୁଖଗହ୍ଵର ।
• ମୁଖଗହ୍ଵରର ଦୁଇ ପଟେ ଗାଲ; ଉପରେ ତାଳୁ ଓ
• ତଳେ ଜିଭ ରହିଥାଏ ।

(a) ଦାନ୍ତ (Tooth):

• ଉଠେ ଓ ଛଅ ବର୍ଷ ବେଳକୁ ଉକ୍ତ ଦାନ୍ତ ଝଡ଼ିବାକୁ ଆରମ୍ଭ କରେ । ଏହି ସ୍ଥାନରେ ସ୍ଥାୟୀ ଦାନ୍ତ ଉଠେ । ବୟସ୍କ ଲୋକର ତଳ ଓ ଉପର ମାଢ଼ିରେ 32ଟି (16ଟି ଲେଖାଏଁ) ଦାନ୍ତ ଥାଏ । ପ୍ରତି ମାଢ଼ିରେ 4ଟି କର୍ଜନ ଦାନ୍ତ, 2 ଟି ଛେଦକ ବା ଶ୍ଵାନଦାନ୍ତ Canine), 4 ଟି ଚର୍ବଣ ଦାନ୍ତ ଓ ଟି ପେଷଣ ଦାନ୍ତ ରହିଅଛି।
• ଖାଦ୍ୟାଭ୍ୟାସ ଉପରେ ନିର୍ଭର କରି ଆମର ଦାନ୍ତ 4 ପ୍ରକାରର ହୋଇଥାଏ ।

(b) && (Tongue) :

• ମୁଖଗହ୍ଵରର ଉପରେ ତାଳୁ ଓ ତଳେ ମୋଟା, ମାଂସଳ ଜିଭ ରହିଥାଏ । ଏଥ‌ିରେ ତିନି ପ୍ରକାର ସ୍ବାଦମୁକୂଳ ରହିଥାଏ ।

କାର୍ଯ୍ୟ :

• ଚର୍ବଣବେଳେ ଏହା ଖାଦ୍ୟକୁ ଦାନ୍ତ ନିକଟରେ ପହଞ୍ଚାଇଥାଏ ।
• ଏହା ଖାଦ୍ୟକୁ ଲାଳ ସହିତ ମିଶାଇଥାଏ ।
• ଖାଦ୍ୟକୁ ଗିଳିବାରେ ସହାୟତା କରିଥାଏ ।
• ଏହା ମିଠା, ଖଟା, ପିତା ଓ ଲୁଣିଆ ଆଦି ସ୍ଵାଦ ବାରିଥାଏ ।
• ଏହା କଥା କହିବାରେ ସହାୟତା କରିଥାଏ ।

(iii) (Pharynx and Oesophagus) :

• ନାସାପଥ ଓ ମୁଖଗହ୍ଵର ମିଶି ଗ୍ରସନୀ ତିଆରି ହୋଇଛି । ଏହା ଉଭୟ ଖାଦ୍ୟ ଓ ଶ୍ୱାସ ବାୟୁ ଯିବାପାଇଁ ଏକ ସାଧାରଣ ପଥ ।
• ଏହାର ଶେଷମୁଣ୍ଡରେ ଦୁଇଟି ଦ୍ଵାର ରହିଛି । ଗୋଟିଏ ଦ୍ଵାର ଖୋଲିଛି ଶ୍ଵାସନଳୀ ଭିତରକୁ, ଅନ୍ୟଟି ଖୋଲିଛି ଖାଦ୍ୟନଳୀ ମଧ୍ୟକୁ ।
• ଖାଦ୍ୟନଳୀର ଦ୍ଵାରକୁ ଗଲେଟ୍ ଓ ଶ୍ଵାସନଳୀର ଦ୍ଵାରକୁ ଗ୍ଲଟିସ୍ କୁହାଯାଏ । ଶ୍ଵାସନଳୀର ଦ୍ଵାରରେ ଏକ ତରୁଣାସ୍ଥିର ପ୍ଲେଟ ରହିଛି । ଏହି ପ୍ଲେଟ୍‌କୁ ଅଧ୍ୱଜିହ୍ଵା ବା ଏପିକ୍ଲଟିସ୍ କୁହାଯାଏ ।
• ପ୍ରବେଶ କରିଥାଏ ।
• ଗ୍ରାସନଳୀ ବେକ ମଧ୍ୟ ଦେଇ ତଳ ଆଡ଼କୁ ଗତି କରିଛି ଏବଂ ମଧ୍ୟଚ୍ଛଦାକୁ ଭେଦ କରି ପାକସ୍ଥଳୀକୁ ଖୋଲିଛି ।
• କାର୍ଯ୍ୟ : ଏହା ମଧ୍ୟଦେଇ ପେରିଷ୍ଟାଲ୍‌ସିସ୍‌ରା ଖାଦ୍ୟ ଆପେ ଆପେ ପାକସ୍ଥଳୀ ମଧ୍ୟକୁ ପ୍ରବେଶ କରେ । ଏଠାରେ ଖାଦ୍ୟର କୌଣସି ପରିବର୍ତ୍ତନ ହୁଏ ନାହିଁ ।

(iv) (Stomach) :

• ଏହା ମୋଟା, ମାଂସଳ ଓ ‘J’ ଆକୃତିର ଏକ ଥଳୀ ଅଟେ ଏବଂ ଏହା ଉଦର ଗହ୍ଵରର ଉପର ଅଂଶର ବାମପଟେ ରହିଥାଏ ।
• ଏହାର ଉପର ଅଂଶ ଚଉଡ଼ା ଓ ତଳ ଅଂଶ କମ୍ ଓସାରିଆ ।
• ଏହାର ଉପର ଅଂଶ ହୃତ୍‌ପିଣ୍ଡ ନିକଟରେ ଥ‌ିବାରୁ ସେହି ଅଂଶକୁ କାର୍ଡିଆକ୍ ଷ୍ଟୋମାକ୍ ଓ ତଳ ଅଂଶକୁ ପାଇଲୋରିକ୍ ଷ୍ଟୋମାକ୍ କୁହାଯାଏ ।
• ଏହାର ମାଂସଳ କାନ୍ଥ ଖାଦ୍ୟକୁ ପାଚକରସ ସହିତ ମିଶିବାରେ ସାହାଯ୍ୟ କରେ ।
• ଅର୍ଦ୍ଧଜୀର୍ଣ୍ଣ ଖାଦ୍ୟ ଅଳ୍ପ ଅଳ୍ପ ପରିମାଣରେ କ୍ଷୁଦ୍ରାନ୍ତ ମଧ୍ୟକୁ ପ୍ରବେଶ କରେ ।

(v) (Small Intestine) :

• ଏହା ପ୍ରାୟ 6 ମିଟର ଦୀର୍ଘ, ସରୁ ଓ କୁଣ୍ଡଳାକାର ନଳୀ ଅଟେ ।
• ଏହାର ପ୍ରଥମ ଅଂଶକୁ ଗ୍ରହଣୀ, ମଧ୍ୟବର୍ତୀ ଅଂଶକୁ ଯେଯୁନମ୍ ଓ ଶେଷ ଅଂଶକୁ ଇଲିୟମ୍ କୁହାଯାଏ ।
• କାର୍ଯ୍ୟ : ଏଠାରେ ପରିପାକ କ୍ରିୟା ସମ୍ପୂର୍ଣ୍ଣ ହୋଇଥାଏ ଏବଂ ପରିପାକ ହୋଇଥ‌ିବା ଖାଦ୍ୟର ଅବଶୋଷଣ ବା ଆତ୍ମୀକରଣ ଘଟିଥାଏ ।

(vi) (Large Intestine) :

• ଏହା କ୍ଷୁଦ୍ରାନ୍ତଠାରୁ ଛୋଟ ଓ ପ୍ରଶସ୍ତ ଅଟେ ।
• ଏହା ସିକମ୍, କୋଲନ୍ ଓ ମଳାଶୟ ରେ ବିଭକ୍ତ ହୋଇଥାଏ ।
• କାର୍ଯ୍ୟ : ଏଠାରେ ଅଜୀର୍ଣ୍ଣ ଖାଦ୍ୟ ମଳରେ ପରିଣତ ହୋଇଥାଏ ।

(vii)

• ଏହା ବୃହଦନ୍ତ୍ରର ଶେଷ ଆଡ଼କୁ ଅବସ୍ଥିତ ଏବଂ ଏହାର ଶେଷ ଭାଗରେ ଥ‌ିବା ଛିଦ୍ରକୁ ମଳଦ୍ଵାର କୁହାଯାଏ ।
• କାର୍ଯ୍ୟ : ମଳାଶୟରେ ମଳ କିଛି ସମୟ ପାଇଁ ଗଚ୍ଛିତ ରହେ ଏବଂ ମଳଦ୍ଵାର ଦେଇ ନିଷ୍କାସିତ ହୁଏ ।

→ ପାକଗ୍ରନ୍ଥି (Digestive Glands) :

• ଖାଦ୍ୟକୁ ସରଳୀକୃତ କରିବା ପାଇଁ ପାକନଳୀ ମଧ୍ଯରେ ଅନେକଗୁଡ଼ିଏ ପାକଗ୍ରନ୍ଥି ଅଛି । ଯଥା –

(i) (Salivary Glands) :

• ମନୁଷ୍ୟର 3 ଯୋଡ଼ା ଲାଳଗ୍ରନ୍ଥି ରହିଛି ଏବଂ ଏହାର ବାହିକା (ducts)ଗୁଡ଼ିକ ମୁଖଗହ୍ଵର ମଧ୍ୟକୁ ଖୋଲି ରହିଥାଏ ।
• ଏହି ଗ୍ରନ୍ଥିରୁ କ୍ଷରିତ ଲାଳ ଖାଦ୍ୟକୁ ପିଚ୍ଛିଳ ଓ ମଣ୍ଡପରି କରିଦିଏ ।
• ଲାଳରେ ଥିବା ପାଚକ ଏନ୍‌ଜାଇମ୍‌କୁ ଟାୟାଲିନ୍ (Ptyalin) ବା ସାଲାଇଭାରି ଆମାଇଲେଜ୍ କୁହାଯାଏ ।

(ii) ଜଠର ଗ୍ରନ୍ଥି (Gastric Glands) :

• ଏହି ଗ୍ରନ୍ଥିରୁ କ୍ଷରିତ ହେଉଥ‌ିବା ରସକୁ ପାଚକ ରସ କୁହାଯାଏ । ଏଥିରେ ଲବଣାମ୍ଳ (HCl) ସହିତ ପେପସିନ୍ ଓ ଲାଇପେଜ୍ ଏନ୍‌ଜାଇମ୍ ରହିଛି ।
• ଏଗୁଡ଼ିକ ପାଚକ ରସ (Gastric juice) ଓ ଲବଣାମ୍ଳ (HCI) କ୍ଷରଣ କରିଥା’ନ୍ତି ।

(iii) (Liver) :

• ଏହା ଶରୀରର ବୃହତ୍ତମ ଗ୍ରନ୍ଥି ଏବଂ ଏହା ଉଦର ଗହ୍ଵରର ଉପରିଭାଗର ଡାହାଣପାର୍ଶ୍ଵରେ ଅବସ୍ଥିତ । ଏହାର ବର୍ଣ୍ଣ ଲୋହିତ ବାଦାମୀ ।
• ଏହା ପିତ୍ତ ରସ କ୍ଷରଣ କରେ ।

(iv) ଅଗ୍ନ୍ୟାଶୟ (Pancreas) :

• ଏହା ଏକ ମିଶ୍ରିତ ଗ୍ରନ୍ଥି ।
• ଏହା ମଧ୍ଯ ଖାଦ୍ୟନଳୀ ବାହାରେ ରହିଛି । ଏଥୁରୁ ଉଭୟ ଏନ୍‌ଜାଇମ୍ ଓ ହରମୋନ୍ କ୍ଷରିତ ହୁଏ । ଅଗ୍ନ୍ୟାଶୟ ରସରେ ଆମାଇଲେଜ, ଲାଇପେଜ୍ ଏବଂ ପ୍ରୋଟିଏଜ୍ ପରି ଖାଦ୍ୟ ହଜମକାରୀ ଏନ୍‌ଜାଇମ୍ ରହିଛନ୍ତି ।

→ ଆନ୍ତ୍ରିକ ଗ୍ରନ୍ଥି (Intestinal gland) :

• କ୍ଷୁଦ୍ରାନ୍ତ୍ରରେ ଥ‌ିବା ଆନ୍ତ୍ରିକ ଗ୍ରନ୍ଥିଗୁଡ଼ିକରୁ ଆର୍ଥିକ ରସ କ୍ଷରିତ ହୁଏ । ଏହି ରସରେ ଥ‌ିବା ବିଭିନ୍ନ ପ୍ରକାରର ଏନ୍‌ଜାଇମ୍ ହଜମକ୍ରିୟା ଶେଷ କରନ୍ତି ।

→ ପାଚନ କ୍ରିୟା (Digestion) :

• ଆମେ ଖାଉଥ‌ିବା ଖାଦ୍ୟରେ ଶ୍ଵେତସାର, ସ୍ନେହସାର, ପୁଷ୍ଟିସାର, ଭିଟାମିନ୍, ଧାତବ ଲବଣ ଓ ଜଳ ରହିଥାଏ । ଖାଦ୍ୟ ହଜମର ବିଭିନ୍ନ ପ୍ରକ୍ରିୟାଗୁନିକ
• ହେଲା : ଖାଦ୍ୟ ଗ୍ରହଣ, ପାକକ୍ରିୟା, ଅବଶୋଷଣ, ଆତ୍ମୀକରଣ ଓ ମଳତ୍ୟାଗ ।

(A) ମୁଖଗହ୍ଵର (Buccal Cavity) :

• ଏଠାରେ ଖାଦ୍ୟ ଚର୍ବିତ ଓ ପେଷିତ ହେବା ସଙ୍ଗେ ସଙ୍ଗେ ଟାୟାଲିନ୍‌ର କାର୍ଯ୍ୟକାରିତା ହାର ବୃଦ୍ଧି କରେ ।
• ଖାଦ୍ୟରେ ଲାଳ ମିଶିବାଦ୍ୱାରା ଟାୟାଲିନ୍ ପ୍ରାୟ 30-40 ଭାଗ ମଣ୍ଡଦ (Starch)କୁ ମାଲ୍‌ଟୋଜ୍ (Maltose)ରେ ପରିଣତ କରେ ।

(B) (Stomach) :
ଭାଙ୍ଗି ଅତି ସୂକ୍ଷ୍ମଖଣ୍ଡରେ ପରିଣତ ହୁଏ । ପାକସ୍ଥଳୀରୁ ନିସୃତ ପାଚକ ରସ ଖାଦ୍ୟ ସହିତ ମିଶି ଏହାକୁ ଏକ ପ୍ରକାର ତରଳ ମଣ୍ଡ ବା ଚାଇମରେ ପରିଣତ କରେ । ଲବଣାମ୍ଳ (HCl) ପାକମଣ୍ଡକୁ ଅମ୍ଳାତ୍ମକ କରିବା ସହିତ ଜୀବାଣୁ ନାଶ କରେ । ପାଚକରସରେ ଦୁଇ ପ୍ରକାର ଏନ୍‌ଜାଇମ୍ ଥାଏ – ପେପ୍‌ସିନ୍ ଓ ଲାଇପେଜ୍ । ପେପ୍‌ସିନ୍ ଲବଣାମ୍ଳ ମାଧ୍ୟମରେ ସକ୍ରିୟ ହୁଏ ଓ ପ୍ରୋଟିନ୍ ଖାଦ୍ୟକୁ ପ୍ରୋଟିଓଜେସ୍ ଓ ପେପ୍‌ନ୍‌ରେ ପରିଣତ କରେ ।

(C) ଗ୍ରହଣୀ :
କ୍ଷୁଦ୍ରାନ୍ତର ଗ୍ରହଣୀଠାରେ ଯକୃତରୁ କ୍ଷରିତ ପିତ୍ତ ଓ ଅଗ୍ନ୍ୟାଶୟରୁ କ୍ଷରିତ ଅଗ୍ନ୍ୟାଶୟ ରସ ଖାଦ୍ୟରେ ଆସି ମିଶେ । ପିତ୍ତରେ ଥ‌ିବା ପିତ୍ତ ଲବଣ ଖାଦ୍ୟର ଅମ୍ଳତ୍ଵ ଦୂର କରେ ଓ ସ୍ନେହସାର ଜାତୀୟ ଖାଦ୍ୟର ଅବଦ୍ରବୀକରଣ ବା ଇମଲ୍ଟିଫିକେସନ୍ (Emulsification) କରାଇଥାଏ ।

(D) ଜେଜୁନମ୍ ଓ ଇଲିୟମ୍ :
ଏହି ସ୍ଥାନରେ ସମସ୍ତ ଖାଦ୍ୟର ହଜମ ପ୍ରକ୍ରିୟା ସମ୍ପୂର୍ଣ୍ଣ ହୋଇଥାଏ । ଜେଜୁନମ୍ ଓ ଇଲିୟମ୍‌ରୁ କ୍ଷରିତ ଆର୍ଥିକ ରସରେ ରହିଥ‌ିବା ବିଭିନ୍ନ ପ୍ରକାର ଏନଜାଇମ୍ ହଜମ ହୋଇନଥୁବା ଅବଶିଷ୍ଟ ଖାଦ୍ୟକୁ ସମ୍ପୂର୍ଣ ହଜମ କରିଥା’ନ୍ତି । ଏଠାରେ ହେଉଥ‌ିବା ହଜମ ପ୍ରକ୍ରିୟାଟି :

(E) ବୃହଦନ୍ତ୍ର :
ଖାଦ୍ୟ ବୃହଦନ୍ତ୍ରଠାରେ ପହଞ୍ଚିଲା ବେଳକୁ ଏହା ହଜମ ହୋଇ ସରଳୀକୃତ ଖାଦ୍ୟରେ ପରିଣତ ହୋଇଥାଏ ।

(v) (Absorption) :
ସମସ୍ତ ସରଳୀକୃତ ଖାଦ୍ୟ, ଭିଟାମିନ୍, ଧାତବ ଲବଣ ଓ ଜଳ ଇତ୍ୟାଦି ଆହାରନଳୀର କାନ୍ଥ ବାଟ ଦେଇ ରକ୍ତ ମଧ୍ୟକୁ ପ୍ରବେଶ କରିବା ପ୍ରକ୍ରିୟାକୁ ଅବଶୋଷଣ କୁହାଯାଏ । ଏହି ପ୍ରକ୍ରିୟା ନିଷ୍କ୍ରିୟ ଅବଶୋଷଣ ଓ ସକ୍ରିୟ ଅବଶୋଷଣଦ୍ୱାରା ହୋଇଥାଏ ।

(vi) (Assimilation) :

• ଅବଶୋଷଣ ପରେ ସରଳୀକୃତ ଖାଦ୍ୟ ରକ୍ତଦ୍ୱାରା ବାହିତ ହୋଇ ଯକୃତରେ ପହଞ୍ଚେ । ଯକୃତରୁ ଏହା ଶରୀରର ବିଭିନ୍ନ ଅଂଶରେ ପହଞ୍ଚିଥାଏ ଏବଂ ଶକ୍ତିମୋଚନ ଓ ଅନ୍ୟ କାର୍ଯ୍ୟରେ ନିୟୋଜିତ ହୋଇଥାଏ । ଖାଦ୍ୟର ଏହି ବିନିଯୋଗ
• ଅଧିକାଂଶ ଏମିନୋ ଅମ୍ଳ ପୁଷ୍ଟିସାର ସଂଶ୍ଳେଷଣରେ ବ୍ୟବହୃତ ହୁଏ ଯାହାକି ଶରୀର ବୃଦ୍ଧି ଓ ଟିସୁର ପୁନଃନିର୍ମାଣରେ ସହାୟକ ହୁଏ । ବଳକା ଏମିନୋ ଅମ୍ଳ ଯକୃତରେ ୟୁରିଆରେ ପରିଣତ ହୁଏ ।

(vii) (Egestion) :
ଏକକାଳୀନ ନିମ୍ନଲିଖ୍ ଘଟଣାବଳୀଦ୍ଵାରା ମଳ ନିଷ୍କାସନ ହୋଇଥାଏ ।

• ମଳଦ୍ଵାର ଚାରିପଟେ ରହିଥ‌ିବା ସଂକୋଚନ ପେଶୀର ଶିଥୁଳନ,
• ମଳାଶୟ ପେଶୀର ସଂକୋଚନ,
• ଉଦରପେଶୀ ଓ ମଧ୍ଯଚ୍ଛଦାର ସଂକୋଚନ ସହିତ ସାମୟିକ ଶ୍ଵାସ ବିରାମ ।

ମାଂସ ହଜମ କରୁଥିବା ଏନଜାଇମ୍ ଆମ ପାକସ୍ଥଳୀକୁ ହଜମ କରିପାରେ ନାହିଁ କାରଣ :

• ପୁଷ୍ଟିସାର ହଜମ କରୁଥିବା ପ୍ରୋଟିଏଜ୍ ଜାତୀୟ ଏନଜାଇମ୍ ନିଷ୍କ୍ରିୟ ଅବସ୍ଥାରେ କ୍ଷରିତ ହୋଇଥାଏ । ପାକସ୍ଥଳୀର ଅମ୍ଳୀୟ ପରିବେଶରେ ଏହା ସକ୍ରିୟ ହୁଏ ଓ ପାକସ୍ଥଳୀରେ ଖାଦ୍ୟ ପହଞ୍ଚିଲେ ସାଧାରଣତଃ ଏନ୍‌ଜାଇମ୍ କ୍ଷରଣ
• ଆମ ପାକସ୍ଥଳୀରେ ଅନେକ ଶ୍ଳେଷ୍ମକ ବା ମ୍ୟୁକସ୍ ଗ୍ରନ୍ଥି ରହିଛି । ସେଥୁରୁ କ୍ଷରିତ ମ୍ୟୁକସ୍ ଅମ୍ଳୀୟ ପରିବେଶ ତଥା ଏନ୍‌ଜାଇମ୍ ପ୍ରଭାବରୁ ପାକସ୍ଥଳୀକୁ ରକ୍ଷାକରେ ।
• ପାକସ୍ଥଳୀର କୋଷମାନଙ୍କ ମଧ୍ୟରେ ଥ‌ିବା ନିବିଡ଼ ବନ୍ଧନ ଯୋଗୁଁ ସହଜରେ ପେପ୍‌ସିନ୍ ପାକସ୍ଥଳୀ କାନ୍ଥ ଭିତରକୁ ଟିସୁ କ୍ଷୟ କରିପାରେ ନାହିଁ ।
• ଏଥ୍ ସହିତ ପାକସ୍ଥଳୀର କୋଷ ପ୍ରତି ଦୁଇ ବା ତିନିଦିନ ବ୍ୟବଧାନରେ ନୂଆ କୋଷଦ୍ଵାରା ପୁନଃସ୍ଥାପିତ ହୁଅନ୍ତି । ଏଥ୍ ଯୋଗୁଁ ଆମ ପାକସ୍ଥଳୀ ପେପ୍‌ସିନ୍ ଏନ୍‌ଜାଇମ୍ଵାରା ହଜମ ହୁଏନାହିଁ ।