Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 1:

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

Answer:

Let be any two numbers in R.

Then, we have:

Hence, f is strictly increasing on R.

Question 2:

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

Answer:

Let be 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 each we have .

Hence, f is strictly increasing in .

(b) Since for each , we have .

Hence, f 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 point divides 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 x = 3 divide the real line into three disjoint intervals i.e.,

In intervals is 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 point divides 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 point divides 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 intervals and (−1, 1),  .

∴ f is strictly decreasing in intervals and (−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 x > −1, point x = 0 divides the domain (−1, ∞) in two disjoint intervals i.e., −1 < x < 0 and x > 0.

When −1 < x < 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 which is 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 .

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 point divides 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 x (B) cos 2x (C) cos 3x (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 interval into 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 f 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+a

Since 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 x∈R, (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 intervals is 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.