## NCERT Solutions for Class 12 Maths Chapter 6 – Application of Derivatives Ex 6.2

#### Question 1:

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

Letbe any two numbers in R.
Then, we have:
Hence, is strictly increasing on R.

#### Question 2:

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

Letbe any two numbers in R.
Then, we have:
Hence, f is strictly increasing on R.

#### Question 3:

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

The given function is f(x) = sin x.
(a) Since for eachwe have.
Hence, f is strictly increasing in.
(b) Since for each, we have.
Hence, is strictly decreasing in.
(c) From the results obtained in (a) and (b), it is clear that f is neither increasing nor decreasing in (0, π).

#### Question 4:

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

The given function is f(x) = 2x2 − 3x.
Now, the pointdivides the real line into two disjoint intervals i.e., and
In interval
Hence, the given function (f) is strictly decreasing in interval.
In interval
Hence, the given function (f) is strictly increasing in interval.

#### 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

The given function is f(x) = 2x3 − 3x2 − 36x + 7.
x = − 2, 3
The points x = −2 and = 3 divide the real line into three disjoint intervals i.e.,
In intervalsis positive while in interval
(−2, 3), is negative.
Hence, the given function (f) is strictly increasing in intervals
, while function (f) is strictly decreasing in interval
(−2, 3).

#### 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

(a) We have,
Now,
x = −1
Point x = −1 divides the real line into two disjoint intervals i.e.,
In interval
f is strictly decreasing in interval
Thus, f is strictly decreasing for x < −1.
In interval
∴ f is strictly increasing in interval
Thus, f is strictly increasing for x > −1.
(b) We have,
f(x) = 10 − 6x − 2x2
The pointdivides the real line into two disjoint intervals i.e.,
In interval i.e., when,
f'(x)=-6-4x>0.
∴ f is strictly increasing for .
In interval i.e., when,
∴ f is strictly decreasing for .
(c) We have,
f(x) = −2x3 − 9x2 − 12x + 1
Points x = −1 and x = −2 divide the real line into three disjoint intervals i.e.,
In intervals i.e., when x < −2 and x > −1,
.
∴ f is strictly decreasing for x < −2 and x > −1.
Now, in interval (−2, −1) i.e., when −2 < x < −1, .
∴ f is strictly increasing for .
(d) We have,
The pointdivides the real line into two disjoint intervals i.e., .
In interval i.e., for.
∴ f is strictly increasing for.
In interval i.e., for,
∴ f is strictly decreasing for.
(e) We have,
f(x) = (x + 1)3 (x − 3)3
The points x = −1, x = 1, and x = 3 divide the real line into four disjoint intervals i.e.,, (−1, 1), (1, 3), and.
In intervalsand (−1, 1), .
∴ f is strictly decreasing in intervalsand (−1, 1).
In intervals (1, 3) and.
∴ f is strictly increasing in intervals (1, 3) and.

#### Question 7:

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

We have,
dydx=11+x-(2+x)(2)-2x(1)(2+x)2=11+x-4(2+x)2=x2(1+x)(2+x)2Now,
dydx=0
⇒x2(1+x)(2+x)2=0⇒x2=0      [(2+x)≠0 as x>-1]⇒x=0Since > −1, point = 0 divides the domain (−1, ∞) in two disjoint intervals i.e., −1 < x < 0 and x > 0.
When −1 < < 0, we have:
x<0 x2="">0x>-1⇒(2+x)>0⇒(2+x2)>0∴
y’=x2(1+x)(2+x)2>0Also, when x > 0:
x>0⇒x2>0, (2+x)2>0∴
y’=x2(1+x)(2+x)2>0Hence, function f is increasing throughout this domain.

#### Question 8:

Find the values of x for whichis an increasing function.

We have,
The points x = 0, x = 1, and x = 2 divide the real line into four disjoint intervals i.e.,
In intervals.
∴ y is strictly decreasing in intervals .
However, in intervals (0, 1) and (2, ∞),
∴ y is strictly increasing in intervals (0, 1) and (2, ∞).
y is strictly increasing for 0 < x < 1 and x > 2.

#### Question 9:

Prove that  is an increasing function of θ in.

We have,
Since cos θ ≠ 4, cos θ = 0.
Now,
In interval, we have cos θ > 0. Also, 4 > cos θ ⇒ 4 − cos θ > 0.
Therefore, y is strictly increasing in interval.
Also, the given function is continuous at
Hence, y is increasing in interval.

#### Question 10:

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

It is clear that for x > 0,
Hence, f(x) = log x is strictly increasing in interval (0, ∞).

#### Question 11:

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

The given function is f(x) = x2 − x + 1.
The pointdivides the interval (−1, 1) into two disjoint intervals i.e.,
Now, in interval
Therefore, f is strictly decreasing in interval.
However, in interval
Therefore, f is strictly increasing in interval.
Hence, f is neither strictly increasing nor decreasing in interval (−1, 1).

#### Question 12:

Which of the following functions are strictly decreasing on?
(A) cos (B) cos 2(C) cos 3(D) tan x

(A) Let
In interval
is strictly decreasing in interval.
(B) Let
is strictly decreasing in interval.
(C) Let
The point divides the intervalinto two disjoint intervals
i.e., 0
∴ f3 is strictly increasing in interval.
Hence, f3 is neither increasing nor decreasing in interval.
(D) Let
In interval
∴ f4 is strictly increasing in interval
Therefore, functions cos x and cos 2x are strictly decreasing in
Hence, the correct answers are A and B.

#### Question 13:

On which of the following intervals is the function f given by  strictly decreasing?
(A)  (B)
(C)  (D) None of these

We have,
In interval
Thus, function f is strictly increasing in interval (0, 1).
In interval
Thus, function f is strictly increasing in interval.
∴ f is strictly increasing in interval.
Hence, function f is strictly decreasing in none of the intervals.

#### Question 14:

Find the least value of a such that the function f given is strictly increasing on [1, 2].

We have,
Now, function is increasing on [1,2].
∴ f’x≥0 on 1,2
Now, we have 1⩽x⩽2
⇒2⩽2x⩽4
⇒2+a⩽2x+a⩽4+a
⇒2+a⩽f’x⩽4+a
Since f’x≥0
⇒2+a≥0
⇒a≥-2
So, least value of a is -2.

#### Question 15:

Let I be any interval disjoint from (−1, 1). Prove that the function f given by
is strictly increasing on I.

We have,
The points x = 1 and x = −1 divide the real line in three disjoint intervals i.e., .
In interval (−1, 1), it is observed that:
∴ f is strictly decreasing on .
In intervals, it is observed that:
∴ f is strictly increasing on.
Hence, function f is strictly increasing in interval I disjoint from (−1, 1).
Hence, the given result is proved.

#### Question 16:

Prove that the function f given by f(x) = log sin x is strictly increasing on and strictly decreasing on

We have,
In interval
∴ f is strictly increasing in.
In interval
f is strictly decreasing in

#### Question 17:

Prove that the function f given by f(x) = log cos x is strictly decreasing on  and strictly increasing on

We have,
In interval
f is strictly decreasing on.
In interval
f is strictly increasing on.

#### Question 18:

Prove that the function given by is increasing in R.

We have,
For any xR, (x − 1)2 > 0.
Thus, is always positive in R.
Hence, the given function (f) is increasing in R.

#### Question 19:

The interval in which  is increasing is
(A)  (B) (−2, 0) (C)  (D) (0, 2)