Class 12

Odisha State Board CHSE Odisha Class 12 Invitation to English 4 Solutions Grammar Interrogatives Textbook Activity Questions and Answers.

CHSE Odisha 12th Class English Grammar Interrogatives

We usually make questions by changing the word order :

(a) Look at these sentences.

A	B
This is the road to your house.	Is this the road to your house?
He has a car	Has he a car?
There is a canteen in the office	Is there a canteen in the office?

The sentences in ‘A’ are statements and those in ‘B’ are questions. The verb in each sentence in ‘B’ here comes first and the subject is put after it. In other words, there is an inversion of subject and verb. They are all interrogative sentences.
(b) Look at these sentences.

A Statements	B Yes/No Questions
My friend works on a firm	Has he a car?
You sold your car	Is there a canteen in the office?
They go to the cinema every evening.	Do they go to the cinema every evening?

For questions in ‘B’ the auxiliary do or a form of it is provided in questions and it is put before the subject.
(c) 1. (i) Did you see my sister anywhere here?
(ii) Did you see my sister somewhere here?
2. (i) Is anyone absent?
(ii) Is someone absent?

Notice the difference between (1) sentences and (2) sentences. In the latter the speaker tends to believe that the answer is ‘yes’ but not so in the former.

(d) Study those information questions, made with phrases formed by the question word how joined to an adjective or adverb.

A Statements	B Yes/No Questions
There are five colleges in the district	How many colleges are there in the district?
This pen costs twenty rupees.	How much does this pen cost?
The postman comes here two times a day.	How often does the post man come here?

(e) Negative questions (Isn’t it…. ? / didn’t you…. ?) We use negative questions specially to show surprise :
Didn’t you hear the bell? I rang it five times.

Activity -1
Divide the following sentences into two broad categories and state why you have done so.
1. Where’s my pen?
2. Isn’t that my pen?
3. Why haven’t you done your homework?
4. How are you?
5. Would you like something to eat?
6. What’s the time by your watch?
7. Are you listening?
8. You have been to Delhi before, haven’t you?

1. Where	is my pen?
2. Is	not that my pen?
3. Why	(haven’t you done your homework?
4. How	are you?
5. Would	you like something to eat?
6. What	(is) the time by your watch?
7. Are	you listening?
8. You have been to haven’t you? (Yes-No Qn) Delhi before

Activity- 2
Write yes/no questions or wh-questions, using the following expressions :
1 . you / go / to Puri / last year?
2. where / be / Seema / today?
3. you / write / class notes?
4. what / be / the STD code / for Bhubaneswar?
5. where / Samar / live / at the moment?
Answers :
1. Did you go to Puri last year? (Yes – No Qn.)
2. Where is Seema today? (Wh-Qn.)
3. Do you write class notes? (Yes – No Qn.)
4. What is the STD code for Bhubaneswar? (Wh – Qn.)
5. Where does Samar live at the moment? (Wh – Qn.)

Activity -3
Change these statements into wh-questions, using the in the brackets :
1 . Seema put that letter on my desk. (Who)
2. Jatin left his bag on the bus. (Where)
3. That is my friend’s bag. (Whose)
4. He means this one. (Which)
5. All the trains were late yesterday because of heavy rain. (Why)
6. They walked two kilometres before they found the shop. (How far)
7. 54,763 eggs were used in the world’s biggest omelette. (How many)

Answers:
1 . Who put that letter on your desk?
2. Where did Jatin leave his bag?
3. Whose bag is that?
4. Which one does he mean?
5. Why were all the trains late yesterday?
6. How far did they walk before they found the shop?
7. How many eggs were used in the world’s biggest omelette?

Activity- 4
Ask questions to get the following answers.
1. I’m fine, thank you.
2. About 40 kilos.
3. Yesterday evening.
4. The giraffe.
5. Yes, just before my eyes.
Answers :
1. How are you?
2. How much does it weigh?
3. When did you come?
4. Which animal has the longest neck?
5. Did the accident happen before your eyes?

Activity – 5
Work in small groups. One member goes out, and in his / her absence other members choose an object in the classroom. When (s)he comes back, (s)he has to guess the article by asking yes/no questions to the other members, asking them serially, for example :
O: Is it the blackboard?
S1: No, it isn’t.
O: Is it the teacher’s table?
S2: No, it isn’t…

1 . O: Is that the chair?
S1: No, that isn’t
O: Is it the duster in my hand?
S2: No, it isn’t.

2. O: Is it the map of India?
S1: No, it isn’t.
O: Is it a globe?
S2: No, it isn’t

3. O: Is it the blackboard?
S1: Yes, it is.
O: Is it the teacher’s table?
S2: Yes, it is.

4. O: Is that the chair?
S1: Yes, that is.
O: Is it the map of India?
S2: Yes, it is

Activity – 6
One student stands in front of the class. He thinks of a person or a place. Other students ask him yes/no or wh-questions in order to guess the person or the place. The students ask in turn and twenty questions are allowed. If a student guesses correctly, he replaces the first student.

Example :
S, = have you thought of a person.
X = Person.
S2 = Is the person dead?
X = Yes.
S3 = Was the person a man?
X = Yes.
S4 = When did he die?
X = In Twentieth Century. (Here X can refuse to answer, because (s) he may give away
the secret.)
S5 = Was he a filmstar?
X = No.
S6 = Was he famous?
X = Yes.
S7 = In which field was he famous?
X = Politics.
Sg = Was he a freedom fighter?
X = Yes.
S9 = Was he a minister?
X = No.
S10 = Is it Mahatma Gandhi?
X = Yes. [And now S,0 becomes X]

Activity- 7
Look at the questions in the left column and the list of functions in the right column. Decide on a function for each question :

1. What time is it?	a. asking to distinguish
2. Is that poisonous or nonpoisonous?	b. expressing lack of belief.
3. You’re back rather early, aren’t you?	c. offering assistance
4. What do you mean by early?	d. asking for assistance
5. Must you sing at this time of the night?	e. asking for information
6. Shall I do that for you?	f. expressing irritation
7. Would you mind holding this for a moment	g. asking for an opinion
8. Why are you late?	h. expressing mild surprise.
9. What did he look like?	i. asking for an explanation.
10. What do you think of the new bowler?	j. asking for a description

Answers :
1. = e
2. = j
3. = b
4. = h
5. = f
6. = c
7. = d
8. = g
9. = i
10. = a

Activity- 8
In the light of your findings from the previous activity, what is the distinction between the terms question and interrogative?
Questions are used to elicit information or to clear a doubt or problem requiring solution. Interrogative has the force of a question (used formally)

Activity – 9
How many questions can you frame to get the following sentence or a part of it as the answer?
When Amar Nath and his partners took over the place, it was in ruins and was inhabited by bats and mice.
1. What was the place like when Amar Nath and his partners took over it?
2. What was the place inhabited with when Amar Nath and his partners took over it?
3. Who took over the place which was in ruins and was inhabited by bats and mice?
4. Whose partners took over the place?

Multiple-Choice Questions (MCQs)
Transfer the following sentences into interrogative ones

Question 1.
He goes there.
(A) Did he go there?
(B) Does he go there?
(C) Do he go there?
(D) Does he goes there?
Answer:
(B) Does he go there?

Question 2.
My mother cooked well.
(A) Do my mother cook well?
(B) Did my mother cooked well?
(C) Did my mother cook well?
(D) Did my mother cooks well?
Answer:
(C) Did my mother cook well?

Question 3.
You have not seen the Taj Mahal.
(A) Have you not seen the Taj Mahal?
(B) Have you not saw the Taj Mahal?
(C) Have you not see the Taj Mahal?
(D) none of these
Answer:
(A) Have you not seen the Taj Mahal?

Question 4.
My friends go to the cinema.
(A) Does my friends go to the cinema?
(B) Do my friends goes to the cinema?
(C) Do my friends go to the cinema?
(D) Does my friends goes to the cinema?
Answer:
(C) Do my friends go to the cinema?

Question 5.
I like mangoes.
(A) Am I liking mangoes?
(B) Do I liked mangoes?
(C) Does I like mangoes?
(D) Do I like mangoes?
Answer:
(D) Do I like mangoes?

Question 6.
How long you will stay here.
(A) How long will you stay here?
(B) How long you shall stay here?
(C) How long would you stay here?
(D) none of these
Answer:
(A) How long will you stay here?

Question 7.
When you last visited him.
(A) When had you last visited him?
(B) When did you last visit him?
(C) When you had last visited him?
(D) When had you last visit him?
Answer:
(B) When did you last visit him?

Question 8.
This bunch of grapes tastes sour.
(A) Do this bunch of grapes taste sour?
(B) Do this bunch of grapes tastes sour?
(C) Does this bunch of grapes taste sour?
(D) Does this bunch of grapes tastes sour?
Answer:
(C) Does this bunch of grapes taste sour?

Question 9.
I saw you go there.
(A) Do I saw you go there?
(B) Did I saw you go there?
(C) Had I seen you go there?
(D) Did I see you to go there?
Answer:
(D) Did I see you to go there?

Question 10.
We are good friends.
(A) Were we good friends?
(B) Have you been good friends?
(C) Are we good friend?
(D) none of there
Answer:
(C) Are we good friend?

Question 11.
We shall never forget those happy days.
(A) Shall we never forget those happy days?
(B) Should we never forget those happy days?
(C) Shall we ever forget these days?
(D) Can we ever forget those happy days?
Answer:
(C) Shall we ever forget these days?

Question 12.
I have somebody with me.
(A) Have I somebody with me?
(B) Have I anybody with me?
(C) Have I nobody with me?
(D) none of these
Answer:
(B) Have I anybody with me?

Question 13.
I have never been to London.
(A) Have I never been to London?
(B) Have I hardly been to London?
(C) Have I ever been to London?
(D) none of these
Answer:
(C) Have I ever been to London?

Question 14.
A leopard can never change its spot.
(A) Can a leopard change its spot?
(B) Can a leopard never change its spot?
(C) Can a leopard sometimes change its sport?
(D) Can a leopard ever change its spot?
Answer:
(D) Can a leopard ever change its spot?

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 9 Integration Ex 9(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Exercise 9(b)

Integrate the following:
(In some cases suggestions have been given for substitution.)
(i) ∫sin 3x dx
Solution:

(ii) ∫cos ax dx
Solution:

(iii) ∫cos (2 – 7x) dx
Solution:

(iv) ∫sin \(\frac{x}{2}\) dx
Solution:

(v) ∫sec² 4x dx
Solution:

(vi) ∫cosec² \(\frac{x}{3}\) dx
Solution:

(vii) ∫sec (x + 2) tan (x + 2) dx
Solution:
∫sec (x + 2) tan (x + 2) dx
[Put x + 2 = θ
Then dx = dθ]
= ∫sec θ . tan θ dθ
= sec θ + C = sec (x + 2) + C

(viii) ∫cosec (x + \(\frac{\pi}{4}\)) cot (x + \(\frac{\pi}{4}\)) dx (x + \(\frac{\pi}{4}\) = z)
Solution:

(ix) ∫x² cos x³ dx (x³ = z)
Solution:

(x) ∫e^x . sec e^x tan e^x dx (e^x = z)
Solution:
∫e^x . sec e^x . tan e^x dx
[Put e^x = z
Then e^x dx = dz]
= ∫ sec z . tan z dz
= sec z + C = sec e^x + C

(xi) ∫\(\frac{\sec ^2 \sqrt{x}}{\sqrt{x}}\) dx
Solution:

Question 2.
(i) ∫sin x cos x dx (sin x = v)
Solution:
∫sin x cos x dx
[Put sin x = v
Then cos x dx = dv]
= ∫v dv = \(\frac{1}{2}\)v² + C
= \(\frac{1}{2}\)sin² x + C

(ii) ∫tan³ x sec² x dx (tan x = t)
Solution:
∫tan³ x sec² x dx
[Put tan x = t
Then sec² x dx = dt]
= ∫t³ dt = \(\frac{1}{4}\)t⁴ + C
= \(\frac{1}{4}\)tan⁴ x + C

(iii) ∫\(\frac{{cosec}^2 x}{1+\cot x}\) dx
Solution:
∫\(\frac{{cosec}^2 x}{1+\cot x}\) dx
[Put 1 + cot x = t
Then -cosec² x dx = dt
or, cosec² x dx = – dt]
= ∫\(\) = -∫\(\) = -ln |t| + C
= -ln |1 + cot x| + C

(iv) ∫\(\frac{\sin x}{\cos ^3 x}\) dx
Solution:

(v) ∫\(\frac{\cos x}{\sin ^5 x}\) dx
Solution:

(vi) ∫\(\frac{{cosec}^2 x(\ln x)}{x}\) dx (In x = z)
Solution:
∫\(\frac{{cosec}^2 x(\ln x)}{x}\) dx
[Put ln x = z
Then \(\frac{d x}{x}\) = dz]
= ∫cosec² z dz
= -cot z + C = -cot (ln x) + C

(vii) ∫\(\sqrt{1-\sin x}\) cos x dx
Solution:

Question 3.
(i) ∫x\(\sqrt{x^2+3}\) dx (x² + 3 = v)
Solution:

(ii) ∫\(\frac{7 d x}{2-3 x}\) dx
Solution:

(iii) ∫\(\frac{x}{\sqrt{x^2-a^2}}\) dx
Solution:

(iv) ∫\(\frac{x^2+1}{\left(x^3+3 x+7\right)^3}\) dx
Solution:

(v) ∫(x⁴ – 3x² + 1)⁴ (2x³ – 3x) dx
Solution:

Question 4.
(i) ∫e^3x dx
Solution:
∫e^3x dx
[Put 3x = t
Then 3 dx = dt
or, dx = \(\frac{1}{3}\)dt]
= ∫e^t . \(\frac{1}{3}\)dt = \(\frac{1}{3}\)e^t + C
= \(\frac{1}{3}\)e^3x + C

(ii) ∫e^2x+7 dx
Solution:
∫e^2x+7 dx
[Put 2x + 7 = t
Then 2 dx = dt
or, dx = \(\frac{1}{2}\)dt]
= ∫e^t . \(\frac{1}{2}\)dt = \(\frac{1}{2}\)e^t + C
= \(\frac{1}{2}\)e^2x+7 + C

(iii) ∫e^x/3 dx
Solution:
[Put \(\frac{x}{3}\) = t
Then dx = 3 dt
= ∫e^t . 3dt = 3e^t + C = 3e^x/3 + C

(iv) ∫\(e^{x^3}\) x² dx
Solution:

(v) ∫a^2x dx
Solution:

(vi) ∫2\(a^{x^2}\) x dx
Solution:

(vii) ∫e^2tanx sec² x dx
Solution:

(viii) ∫\(\frac{e^x}{\left(e^x-2\right)^2}\) dx
Solution:

(ix) ∫\(e^{\cos ^2 x}\) sin 2x dx
Solution:
∫\(e^{\cos ^2 x}\) sin 2x dx
[Put cos2 x = t
Then 2cos x (-sin x)dx – dt
or, -sin 2x dx = dt
or, sin 2x dx = -dt]
= ∫e^t (-dt) = e^t + C = –\(e^{\cos ^2 x}\) + C

Question 5.
(i)∫\(\frac{\sin ^{-1} x}{\sqrt{1-x^2}}\) dx
Solution:

(ii)∫\(\frac{d x}{\sqrt{1-(x-1)^2}}\)
Solution:

(iii)∫\(\frac{\left(\sec ^{-1} x\right)^2 d x}{x \sqrt{x^2-1}}\)
Solution:

(iv)∫\(\frac{d x}{x\left[1+(\ln x)^2\right]}\)
Solution:

(v)∫\(\frac{d x}{x^2+2 x+2}\) (Integrate \(\frac{d x}{(x+1)^2+1}\))
Solution:

Question 6.
(i) ∫tan 3x dx
Solution:

(ii) ∫cos \(\frac{x}{3}\) dx
Solution:

(iii) ∫sec (2x + 1) dx
Solution:
∫sec (2x + 1) dx
[Put 2x + 1 = t
Then 2 dx = dt
or, dx = \(\frac{1}{2}\)dt]
= ∫sec t . \(\frac{1}{2}\)dt = \(\frac{1}{2}\) ∫sec t dt
= \(\frac{1}{2}\) ln |sec t + tan t| + C
= \(\frac{1}{2}\) ln |sec (2x + 1) + tan (2x + 1)| + C

(iv) ∫cosec 7x dx
Solution:
∫cosec 7x dx
[Put 7x = t
Then 7 dx = dt
or, dx = \(\frac{1}{7}\)dt]
= ∫cosec t . \(\frac{1}{7}\)dt
= \(\frac{1}{7}\) ln |cosec t + cot t| + C
= \(\frac{1}{7}\) ln |cosec 7x + cot x| + C

(v) ∫2x cot (x² + 3) dx    (x² + 3 = z)
Solution:
∫2x cot (x² + 3) dx
[Put x² + 3 = z
Then 2x dx = dz]
= ∫cot dz
= In |sin z| + C
= In |sin (x² + 3)| + C

(vi) ∫e^x tan e^x dx
Solution:
∫e^x tan e^x dx
[Put e^x = t
Then e^x dx = dt]
= ∫tan t dt
= In |sec t| + C
= In |sec e^x| + C

(vii) ∫(sec 2x – 3)² dx
Solution:
∫(sec 2x – 3)² dx
∫sec² 2x dx – 6∫sec 2x dx + 9∫dx
= \(\frac{1}{2}\) tan 2x – 3 ln |sec 2x + tan 2x| + 9x + C

Question 7.
(i) ∫\(\frac{e^x+e^{-x}}{e^x-e^{-x}}\) dx
Solution:
∫\(\frac{e^x+e^{-x}}{e^x-e^{-x}}\) dx
[Put (e^x – e^-x) = t
Then (e^x + e^-x)dx = dt]
= ∫\(\frac{d t}{t}\) = ln |t| + C = ln |(e^x – e^-x| + C

(ii) ∫3^x e^2x dx
Solution:

(iii) ∫\(\frac{(x+1) \ln \left(x^2+2 x+2\right)}{x^2+2 x+2}\) dx
Solution:

Question 8.
(i) ∫\(\frac{\sin x}{\sin (x+\alpha)}\) dx
Solution:

= ∫(cos α – sin α . cot t) dt
= cos α ∫dt – sin α ∫cot dt
= cos α . t – sin α . ln |sin t| + C
= (x + α) cos α – sin α  ln |sin (x + α)| + C
= x cos α – sin α ln |sin (x + α)| + C

(ii) ∫\(\frac{\sin x}{\cos (x-\alpha)}\) dx
Solution:

(iii) ∫\(\frac{\tan x+\tan \alpha}{\tan x-\tan \alpha}\) dx
Solution:

 

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

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

Question 1.
Find the extreme points of the following functions. Specify if the extremum is a maximum or minimum.
Find the extreme values.
(i) y = x² + 2x + 3
Solution:

(ii) y = 5x² – 2x⁵
Solution:

∴ The function attains minimum at x = 0.
and maximum at x = 1 .
The minimum value = 0
The maximum value = 5 – 2 = 3

(iii) y = \(\frac{3 x}{x^2+1}\)
Solution:

(iv) y = x²\(\sqrt{1-x^2}\)
Solution:

(v) y = 2x³ – 15x² – 36x + 18
Solution:

The function attains maximum at x = -1 and minimum at x = 6.
Maximum value
= -2 – 15 + 36 + 18 = 37
Minimum value
= 2 × 216 – 15 × 36 – 36 × 6 + 13
= 432 – 540 – 216 + 18
= 450 – 756
= -306

(vi) y = 60/(x⁴ – x² + 25)
Solution:

(vii) y = (x – 1)³
Solution:

So the point x = 1 is neither a point of maximum nor a point of minimum. It is an inflexion point.

(viii) y = (x – 2)³ (x + 3)⁴
Solution:

(ix) y = x + \(\frac{1}{x}\)
Solution:

∴ Maximum at x = -1
and minimum at x = 1
Maximum value = -1 – 1 = -2
Minimum value = 2.

(x) y = 4 cos 2x – 3 sin 2x, x ∈ (-\(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
Solution:

(xi) y = sin x cos x, x ∈ (\(\frac{\pi}{8}\), \(\frac{\pi}{2}\))
Solution:

But there is no minimum in the domain.

(xii) y = cos x (1 + sin x), x ∈ [0, 2π]
Solution:

(xiii) y = sin^p x cos^q x; p, q > 0, x ∈ [0, \(\frac{\pi}{2}\))
Solution:

(xiv) y = x e^-x ; x ∈ (-2, 2)
Solution:

Question 2.
Show that the following functions do not possess maximum or minimum.
(i) x³
Solution:
y = x³

Thus the function does not possess a maximum or a minimum. (Proved)

(ii) x⁵
Solution:
Processed as in (i)

(iii) 3x³ – 12x² + 16x – 15
Solution:
Let y = 3x³ – 12x2 + 16x – 5

So the function does not possess a maximum or a minimum. (Proved)

(iv) 4 – 3x + 3x² – x³
Solution:
Processed as in (iii).

(v) ln |x|, x ≠ 0
Solution:
y = ln |x|, x ≠ 0

Which have no solution. Hence the function has neither a maximum nor a minimum.
(Proved)

Question 3.
Use the function f(x) = \(x^{\frac{1}{x}}\), x > 0 to show that e^x > π^c.
Solution:
f(x) = \(x^{\frac{1}{x}}\), x > 0

Question 4.
Prove the inequality x² \(e^{-x^2}\) ≤ e^-1, x ∈ R.
Solution:

Question 5.
If f(x) = a ln x + bx² + x has extreme values at x = -1 and x = 2 then find a and b.
Solution:
f(x) = a ln x + bx² + x
f'(x) = \(\frac{a}{x}\) + 2b x + 1
Given that f(x) attains extreme values at x = -1 and x = 2
Then f(-1) = 0 and f(2) = 0
⇒ -a – 2b + 1 = 0 … (1)
\(\frac{a}{2}\) + 4b + 1 = 0 … (2)
From (1) we get a = -2b + 1
Putting it in (2)
we get \(\frac{1-2 b}{2}\) + 4b + 1 = 0
⇒ 1 – 2b + 8b + 2 = 0
⇒ 3 + 6b = 0
⇒ b = \(-\frac{3}{6}\) = \(-\frac{1}{2}\)
Again a = 1 – 2b = 1 + 1 = 2
∴ a = 2, b = \(-\frac{1}{2}\)

Question 6.
Show that \(\frac{x}{1+x \tan x}\), x ∈ (0, \(\frac{\pi}{2}\)) is maximum when x = cos x.
Solution:

Thus the function attains maximum when x = cos x. (Proved)

Question 7.
Determine the absolute maximum and absolute minimum of the following function on [-1, 1]
f(x) = \(\left\{\begin{array}{l}
(x+1)^2, x \leq 0 \\
(x-1)^2, x>0
\end{array}\right.\)
Solution:
f(x) = \(\left\{\begin{array}{l}
(x+1)^2, x \leq 0 \\
(x-1)^2, x>0
\end{array}\right.\)
f(x) = \(\left\{\begin{array}{l}
2(x+1), x \leq 0 \\
2(x-1), x>0
\end{array}\right.\)
For extremum f'(x) = 0
⇒ x + 1 = 0 if x < 0
or, x – 1 = 0 if x > 0
⇒ x = -1 or x = 1
Now f(-1) = 0, f(1) = 0
Again f is not differentiable at x = 0 and f(0) = 1
The function attains absolute minimum at x = ±1. Absolute minimum value = 0.
The function attains absolute maximum at x = 0. Absolute Maximum value = 1.

Question 8.
Find extreme values of
f(x) = \(\left\{\begin{array}{l}
\frac{x}{1-x^2},-1<x<0 \\
x^3-x, 0 \leq x<2
\end{array}\right.\) on (-1, 2)
Solution:
Given that

Question 9.
Find two numbers x and y whose sum is 15 such that xy² is maximum.
Solution:
Let z = xy²
Given that x + y = 15
Then y = 15 – x
Thus z = x (15 – x)²
\(\frac{d z}{d x}\) = (15 – x)² + x² (15 – x) . (-1)
= (15 – x) (15 – x – 2x)
= (15 – x) (15 – 3x)
= 3(15 – x) (5 – x)
\(\frac{d^2 z}{d x^2}\) = -3(5 – x) – 3(15 – x)
= -3 (20 – 2x)
= -6 (10 – x)
For extreme points
\(\frac{d z}{d x}\) = 0
⇒ 3(15 – x) (5 – x) = 0
⇒ x = 15 or 5.
\(\left.\frac{d^2 z}{d x^2}\right]_{x=15}\) = 30 > 0
\(\left.\frac{d^2 z}{d x^2}\right]_{x=5}\) = 30 < 0
∴ z = xy² is maximum when x = 5.
In this case y = 10
∴The two numbers are 5, 10.

Question 10.
If the sum of two positive numbers in constant then show that their product is maximum when they are equal.
Solution:
Let x and y be two positive numbers such that x + y = c (constant)
Then y = c – x
Let p = xy = x(c – x) = cx – x²

∴ The two numbers are equal. (Proved)

Question 11.
Determine a rectangle of area 25 sq. units which has minimum perimeter.
Solution:
Let x and y be the length and breadth of a rectangle of area 25 sq. units.

∴ The perimeter is minimum when the rectangle is a square of side 5 units.

Question 12.
Find the altitude of a right circular cylinder of maximum volume inscribed in a sphere of radius r.
Solution:
Let ABCD be a right circular cylinder inscribed in a sphere with centre at O and radius ‘r’. Let x be the altitude of the cylinder.

Question 13.
Show that the radius of the right circular cylinder of greatest curved surface that can be inscribed in a given cone is half the radius of the base of the cone.

Solution:
Let ABC be a cone, Let r be the radius of the base and h be the height of the cone. Let PQRS be a cylinder inscribed in the cone whose height is x.
In the diagram OT = x.
Then CT = h – x
Now Δ CTR and Δ RQB are similar.

∴ Radius of the right circular cylinder of greatest curved surface is half the base of the cone. (Proved)

Question 14.
Show that the semi vertical angle of a cone of given slant height is tan^-1√2 when its volume is maximum.
Solution:

Question 15.
A cylindrical open water tank with a circular base is to be made out of 30 sq. metres of metal sheet. Find the dimensions so that it can hold maximum water. (Neglect thickness of sheet)
Solution:

Let ABCD be an open cylindrical tank with circular base.
Let r be the radius of the base and x be the height of the tank.
Given that the tank is made out of 30 square metres of metal sheet.
The surface area of the tank
= 27πrx + πr² [∵ The tank is open.
So 2πrx + πr² = 30
⇒ 2πrx = 30 – πr²

Question 16.
A cylindrical vessel of capacity 500 cubic metres open at the top is to be constructed. Find the dimensions of the vessel if the material used is minimum given that the thickness of the material used is 2 cm.
Solution:

Question 17.
Find the coordinates of the point on the curve x²y – x + y = 0 where the slope of the tangent is maximum.
Solution:
Let (x, y) be the point on the curve
x²y – x + y = 0.

The slope is maximum when x = 0. when x = 0, y = 0
The point is (0, 0) at which the slope of the tangent is maximum.

Question 18.
Find the points on the curve y = x² + 1 which are nearest to the point (0, 2).
Solution:
Let A = (0, 2)
Let P (x, y) be any point on the curve
y = x² + 1. …(1)

Question 19.
Show that the minimum distance of a point on the curve \(\frac{a^2}{x^2}\) + \(\frac{b^2}{y^2}\) = 1
from the origin is a + b.
Solution:
Let P(x, y) be any point on the curve

Question 20.
Show that the vertical angle of a right circular cone of minimum curved surface that circumscribes a given sphere is 2 sin^-1(√2 – 1).
Solution:

Let r is the radius of the given sphere, R is the radius of the base, H is the height and l is the C lant height of the cone ABC. Let θ is the semi vertical angle.

Question 21.
Show that the semi-vertical angle of a right circular cone of minimum volume that circumscribes a given sphere is sin^-1(\(\frac{1}{3}\)).
Solution:

Let ABC be a right circular cone which circumscribes a sphere of radius ‘r’.
Let θ be the semi-vertical angle of the cone.
Let D be the centre of the sphere. By symmetry D must lie on AO, where O is the centre of the base.
In the diagram DO = r.
Again in the Δ ADE

Question 22.
Show that the shortest distance of the point (0, 8a) from the curve ax² = y³ is 2a√11.
Solution:
Given curve is 2x² = y³ … (1)
Let any point on it is P(x, y)
Distance of P from A (0, 8a)

Question 23.
Show that the triangle of greatest area that can be inscribed in a circle is equilateral.
Hints: It BC is any chord then for the Δ ABC to have maximum area, the point A must be on the perpendicular bisector of BC so as to have the largest height AD. Let mDBAD = a. Let BO = OA = r. Then area of Δ ABC = \(\frac{1}{2}\)|BC| |AD| = \(\frac{1}{2}\) . 2r sin 2α (r + r + cos 2α) = r² sin 2a . (1 + cos 2a). Then maximise Δ to obtain 2a = \(\frac{\pi}{3}\)
Solution:

Let ABC be a triangle inscribed in a circle with centre at O and radius ‘r’.
The triangle ABC has maximum area if the point A is on the perpendicular bisector of BC.
Let ∠BAD = α, Let BO = AO = r,
Then ∠BOD = 2α.
In Δ OBD, sin 2α = \(\frac{\mathrm{BD}}{\mathrm{OB}}\)
⇒ BD = OB sin 2α = r sin 2α
Again OD = r cos 2α.
Thus AD = r + r cos 2α
Let Z be the area of the triangle ABC.

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

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

Find \(\frac{d y}{d x}\)
Question 1.
xy² + x²y + 1 = 0
Solution:
xy² + x²y + 1 = 0

Question 2.
\(x^{\frac{1}{2}} y^{-\frac{1}{2}}\) + \(x^{\frac{3}{2}} y^{-\frac{3}{2}}\) = 0
Solution:

Question 3.
x² + 3y² = 5
Solution:
x² + 3y² = 5
⇒ \(\frac{d}{d x}\)(x²) + 3 \(\frac{d}{d x}\)(y²) = 0
⇒ 2x + 6y\(\frac{d y}{d x}\) = 0
⇒ \(\frac{d}{d x}\) = –\(\frac{x}{3 y}\)

Question 4.
y² cot x = x² cot y.
Solution:
y² cot x = x² cot y

Question 5.
y = tan xy
Solution:
y = tan xy

Question 6.
x = y In (xy).
Solution:
x = y In (xy).

Question 7.
e^xy + y sin x = 1
Solution:
e^xy + y sin x = 1

Question 8.
In \(\sqrt{x^2+y^2}\) = tan^-1 \(\frac{y}{x}\)
Solution:

Question 9.
y^x = x^{sin y}
Solution:
y^x = x^{sin y}

Question 10.
If sin (x + y) = y cos (x + y) then prove that \(\frac{d y}{d x}\) = –\(\frac{1+y^2}{y^2}\).
Solution:

Question 11.
If \(\sqrt{1-x^4}\) + \(\sqrt{1-y^4}\) = k(x² – y²) then show that \(\frac{d y}{d x}\) = \(\frac{x \sqrt{1-y^4}}{y \sqrt{1-x^4}}\).
Solution:

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

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

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

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

(ii) s = √t +1
Solution:
s = √t +1

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Thus Lagrange’s mean value theorem is verified.

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

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

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

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

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

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

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

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

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

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

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

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

Odisha State Board 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 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: