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NCERT solution class 12 chapter 6 Application of Derivatives exercise 6.2 mathematics part 1

EXERCISE 6.2


Page No 205:

Question 1:

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

Answer:

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.

Answer:

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)

Answer:

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

Answer:

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

Answer:

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

Answer:

(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.

Answer:

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.

Answer:

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.

Answer:

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.

Page No 206:

Question 10:

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

Answer:

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).

Answer:

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

Answer:

(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

Answer:

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.

The correct answer is D.

Question 14:

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

Answer:

We have,

Now, function is increasing on [1,2].

∴ f’x≥0 on 1,2Now, we have 1⩽x⩽2⇒2⩽2x⩽4⇒2+a⩽2x+a⩽4+a⇒2+a⩽f’x⩽4+aSince f’x≥0⇒2+a≥0⇒a≥-2So, 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.

Answer:

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

Answer:

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

Answer:

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.

Answer:

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)

Answer:

We have,

The points x = 0 and x = 2 divide the real line into three disjoint intervals i.e.,

In intervalsis always positive.

f is decreasing on

In interval (0, 2),

∴ f is strictly increasing on (0, 2).

Hence, f is strictly increasing in interval (0, 2).

The correct answer is D.


 

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