NCERT Solutions for Maths Class 12 Exercise 6.2

NCERT Solutions for Maths Class 12 Exercise 6.2

Hello Students! In this post, you will find the complete NCERT Solutions for Maths Class 12 Exercise 6.2.

You can download the PDF of NCERT Books Maths Chapter 6 for your easy reference while studying NCERT Solutions for Maths Class 12 Exercise 6.2.

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NCERT Solutions for Maths Class 12 Exercise 6.1

NCERT Solutions for Maths Class 12 Exercise 6.3


NCERT Solutions for Maths Class 12 Exercise 6.2

 

Maths Class 12 Ex 6.2 Question 1.

Show that the function given by f(x) = 3x + 17 is strictly increasing on R.

Solution:
We have, f(x) = 3x + 17
f’(x) = 3 > 0 x R
Therefore, f is strictly increasing on R.


Maths Class 12 Ex 6.2 Question 2.

Show that the function given by f(x) = e2x is strictly increasing on R.

Solution:
We have, f(x) = e2x
f’(x) = 2e2x
Case I   When x > 0, then f’(x) = 2e2x


Maths Class 12 Ex 6.2 Question 3.

Show that the function given by f(x) = sin x is
(a) strictly increasing in (0, π/2) 
(b) strictly decreasing in (π/2, π)
(c) neither increasing nor decreasing in (0, π)

Solution:
We have, f(x) = sin x
f’(x) = cos x
(a) Since f’(x) = cos x is +ve in the interval (0, π/2)
f(x) is strictly increasing in (0, π/2)
(b) Since f’(x) = cos x is -ve in the interval (π/2, π)
f(x) is strictly decreasing in (π/2, π)
(c) Since f’(x) = cos x is +ve in the interval (0, π/2)
while f’(x) is -ve in the interval (π/2, π)
f(x) is neither increasing nor decreasing in (0, π).


Maths Class 12 Ex 6.2 Question 4.

Find the intervals in which the function f given by f(x) = 2x² – 3x is
(a) strictly increasing
(b) strictly decreasing

Solution:
We have, f(x) = 2x² – 3x
f’(x) = 4x – 3
f’(x) = 0 at x = ¾

The point x = ¾ divides the real number line in two disjoint intervals.


Maths Class 12 Ex 6.2 Question 5.

Find the intervals in which the function f given by f(x) = 2x3 – 3x² – 36x + 7 is
(a) strictly increasing
(b) strictly decreasing

Solution:
We have, f(x) = 2x3 – 3x² – 36x + 7

f’(x) = 6x2 – 6x – 36 = 6(x2 – x – 6)
f’(x) = 6(x – 3) (x + 2)
f’(x) = 0 at x = 3 and x = –2
The points x = 3 and x = –2 divide the real number line into three disjoint intervals, i.e., (–∞, –2), (–2, 3), (3, ∞).
Now, f’(x) is +ve in the intervals (–∞, –2) and (3, ∞). Since in the interval (–∞, –2), each factor (x – 3) and (x + 2) is -ve.
f’(x) = +ve.
(a) f is strictly increasing in (–∞, –2) and (3, ∞)
(b) In the interval (–2, 3), (x + 2) is +ve and (x – 3) is -ve.
f’(x) = 6(x – 3)(x + 2) = (+) × (–) = -ve
f is strictly decreasing in the interval (–2, 3).


Maths Class 12 Ex 6.2 Question 6.

Find the intervals in which the following functions are strictly increasing or decreasing:
(a) x² + 2x – 5
(b) 10 – 6x – 2x²
(c) –2x3 – 9x² – 12x + 1
(d) 6 – 9x – x²
(e) (x + 1)3 (x – 3)3

Solution:
(a) Given: f(x) = x² + 2x – 5 

f’(x) = 2x + 2 = 2(x + 1)    …. (i)

Therefore,

f′(x) = 0 x = −1

Now, x = −1 divides the real number line into intervals (−, −1) and (−1, ).

In (−, −1), f′(x) = 2x + 2 < 0 

Thus, f is strictly decreasing in (−, −1).

In (−1, ), f′(x) = 2x + 2 > 0

Thus, f is strictly increasing in (−1, )

 

(b) Given: f(x) = 10 − 6x − 2x2

Therefore, f′(x) = −6 − 4x

f′(x) = 0 x = −3/2

Now, x = −3/2, divides the real number line into two intervals (−, −3/2) and (−3/2, ).

In (−, −3/2), f′(x) = −6 −4x < 0

Hence, f is strictly increasing for x < −3/2.

In (−3/2, ), f′(x) = −6 − 4x > 0

Hence, f is strictly increasing for x > −3/2.

 

(c) Given: f(x) = –2x3 – 9x2 – 12x + 1
f’(x) = –6x2 – 18x – 12 = –6(x2 + 3x + 2)
f'(x) = –6(x + 1)(x + 2)

Now, f’(x) = 0 gives x = –1 or x = –2
The points x = – 2 and x = – 1 divide the real number line into three disjoint intervals namely ( –
, –2) ( –2, –1) and( –1, ).
In the interval (–
, –2), i.e., – < x < -2, (x + 1) and (x + 2) are –ve.
f’(x) = (–) (–) (–) = –ve.
f(x) is decreasing in (–, –2)
In the interval (–2, –1), i.e., – 2 < x < –1, (x + 1) is –ve and (x + 2) is +ve.
f'(x) = (–)(–) (+) = +ve.
f(x) is increasing in (–2, –1)
In the interval (–1,
), i.e., –1 < x < , (x + 1) and (x + 2) are both positive.

f’(x) = (–) (+) (+) = –ve.
f(x) is decreasing in (–1, )
Hence, f(x) is increasing for –2 < x < –1 and decreasing for x < –2 and x > –1.

(d) Given: f(x) = 6 − 9x − x2

Hence, f′(x) = −9 − 2x

f′(x) = 0 x = −9/2

In (−9/2, ∞), f′(x) < 0

Hence, f is strictly decreasing for x > −9/2.

In (−∞, −9/2), f′(x) > 0

Hence, f is strictly decreasing in x > −9/2.

(e) Given: f(x) = (x + 1)3 (x − 3)3

Hence, f′(x) = 3(x + 1)2 (x − 3)3 + 3(x − 3)2 (x + 1)3

                        = 3(x + 1)2 (x − 3)2 [x – 3 + x + 1]

                        = 3(x + 1)2 (x − 3)2 (2x − 2)

                        = 6(x + 1)2 (x − 3)2 (x − 1)

Therefore, f′(x) = 0 x = −1, 3, 1

Now, x = −1, 3, 1 divides the real number line into four intervals (−∞, −1), (−1, 1), (1, 3) and (3, ∞).

In (−∞, −1) and (−1, 1)f′(x) = 6(x + 1)2 (x − 3)2 (x − 1) < 0

Hence, f is strictly decreasing in (−∞, −1) and (−1, 1).

In (1, 3) and (3, ∞)f′(x) = 6(x + 1)2 (x − 3)2 (x − 1) > 0

Hence, f is strictly increasing in (1, 3) and (3, ∞).

 

Maths Class 12 Ex 6.2 Question 7.

Show that , is an increasing function of x throughout its domain.


Solution:


Since, x > −1, x = 0 divides domain (−1, ∞) in two intervals −1 < x < 0 and x > 0

When, −1 < x < 0

Then, x < 0 x2 > 0

x > −1 (2 + x) > 0

(2 + x)2 > 0  

Thus, f is increasing throughout the domain.

 

Maths Class 12 Ex 6.2 Question 8.

Find the values of x for which y = [x (x – 2)]² is an increasing function.

Solution:
We have, y = x4 – 4x3 + 4x2
∴ dy/dx = 4x3 – 12x2 + 8x
For the function to be increasing dy/dx = 0
4x3 – 12x2 + 8x = 0
4x(x – 1) (x – 2) = 0

x = 0, x = 1, x = 2
Now, x = 0x = 1 and x = 2 divide the real number line in the intervals (−∞, 0), (0, 1), (1, 2) and (2,∞).

In (−∞, 0) and (1, 2)dy/dx < 0

Hence, y is strictly decreasing in intervals (−∞, 0) and (1, 2)

In (0, 1) and (2, ∞)dy/dx > 0 

Hence, y is strictly increasing in intervals (0, 1) and (2, ∞).

 

Maths Class 12 Ex 6.2 Question 9.

Prove that  is an increasing function of θ in [0, π/2].

Solution:


Ex 6.2 Class 12 Maths Question 10.

Prove that the logarithmic function is strictly increasing on (0, ∞).

Solution:
Let f(x) = log x
Then, f’(x) = 1/x

For x > 0, 1/x > 0,
Therefore, f’(x) > 0
Hence, f(x) is an increasing function in the interval (0, ∞).

 

Maths Class 12 Ex 6.2 Question 11.

Prove that the function f given by f(x) = x² – x + 1 is neither strictly increasing nor strictly decreasing on (–1, 1).

Solution:
Given: f(x) = x² – x + 1
Then, f’(x) = 2x – 1

f(x) is strictly increasing when f’(x) > 0.

2x – 1 > 0

x > ½

Thus, f(x) is increasing on (1/2, 1).

f(x) is strictly decreasing when f’(x) < 0.

2x – 1 < 0

x < ½

Thus, f(x) is decreasing on (–1, 1/2).

Hence, f(x) is neither strictly increasing nor decreasing on (–1, 1).


Maths Class 12 Ex 6.2 Question 12.

Which of the following functions are strictly decreasing on [0, π/2]
(a) cos x
(b) cos 2x
(c) cos 3x
(d) tan x

Solution:
(a) We have, f(x) = cos x
f’(x) = – sin x < 0 in [0, π/2]
f’(x) is a decreasing function.


Maths Class 12 Ex 6.2 Question 13.

On which of the following intervals is the function f given by f(x )= x100 + sin x – 1 strictly decreasing ?
(A) (0, 1)
(B) [π/2, π]
(C) [0, π/2]
(D) none of these

Solution:
(D) We have, f(x) = x100 + sin x – 1
f’(x)= 100x99+ cos x

(a) For (0, 1) i.e. 0 < x < 1, x99 and cos x are both +ve    
f’ (x) > 0
f(x) is increasing on (0, 1).

Therefore, the option (D) is correct.


Maths Class 12 Ex 6.2 Question 14.

Find the least value of a such that the function f given by f(x) = x² + ax + 1 is strictly increasing on [1, 2].

Solution:
We have, f(x) = x² + ax + 1
f’(x) = 2x + a.
Since f(x) is an increasing function on (1, 2)
f’(x) > 0 for all 1 < x < 2

Now, f”(x) = 2 for all x (1, 2) f”(x) > 0 for all x (1, 2)
f’(x) is an increasing function on (1, 2)
f’(x) is the least value of f’(x) on (1, 2)
But f’(x) > 0
x (1, 2)
f’(1) > 0 2 + a > 0
a > –2

Thus, the least value of a is –2.


Maths Class 12 Ex 6.2 Question 15.

Let I be any interval disjoint from [–1, 1]. Prove that the function f given by f(x) = x + 1/x is strictly increasing on I.

Solution:
Given f(x) = x + 1/x 

Hence, f’ (x) is strictly increasing on I.


Maths Class 12 Ex 6.2 Question 16.

Prove that the function f given by f(x) = log sin x is strictly increasing on (0, π/2) and strictly decreasing on (π/2, π).

Solution:

We have, f(x) = log sin x
f’(x) = 
When 0 < x < π/2, f’(x) is +ve; i.e., increasing
When π/2 < x < π, f’(x) is -ve; i.e., decreasing,
f(x) is decreasing.

Hence, f is increasing on (0, π/2) and strictly decreasing on (π/2, π).


Maths Class 12 Ex 6.2 Question 17.

Prove that the function f given by f(x) = log (cos x) is strictly decreasing on (0, π/2) and strictly increasing on (π/2, π).

Solution:
We have, f(x) = log (cos x)
f’(x) = 
In the interval (0, π/2), f’(x) = -ve
f is strictly decreasing.
In the interval (π/2, π), f’(x) is + ve.
f is strictly increasing in the interval.


Maths Class 12 Ex 6.2 Question 18.

Prove that the function given by f(x) = x3 – 3x2 + 3x – 100 is increasing in R.

Solution:
We have, f(x) = x3 – 3x2 + 3x – 100

f’(x) = 3x2 – 6x + 3
= 3(x2 – 2x + 1)
= 3(x – 1)2
Now x
R, f'(x) = (x – 1)2 ≥ 0
i.e. f'(x) ≥ 0
x R

Hence, f(x) is increasing on R.


Maths Class 12 Ex 6.2 Question 19.

The interval in which y = x2 e-x is increasing in
(A) (–∞, ∞)                    
(B) (2, 0)                (C) (2, ∞)                       (D) (0, 2)

Solution:

(D) We have, y = x2 e-x   or f(x) = x2 e-x

f’(x) = 2xe-x + x2(–e-x) = xe-x(2-x) = e-xx(2 – x)
Now e-x is positive for all x
R; f’(x) = 0 at x = 0, 2
x = 0, x = 2 divide the number line into three disjoint intervals, viz. (-∞, 0), (0, 2) and (2, ∞).
(a) In interval (-∞, 0), x is +ve and (2 – x) is +ve
f’(x) = e-xx(2 – x) = (+) (–) (+) = -ve
f is decreasing in (-∞, 0).
(b) In interval (0, 2), f’(x) = e-x x(2 – x)
= (+) (+) (+) = +ve
f is increasing in (0, 2).
(c) In interval (2, ∞), f’(x) = e-x x(2 – x) = (+) (+) (–) = -ve
f is decreasing in the interval (2, ∞)

Hence, the option (D) is correct.


Related Links:

NCERT Solutions for Maths Class 12 Exercise 6.1

NCERT Solutions for Maths Class 12 Exercise 6.3

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