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# Class 12-science RD SHARMA Solutions Maths Chapter 17 - Increasing and Decreasing Functions

## Increasing and Decreasing Functions Exercise MCQ

### Solution 31

Given:

Differentiating w.r.t x, we get

Take f'(x) = 0

For x < -2,

x + 2 < 0 and x + 1 < 0

Therefore, f'(x) > 0

For -2 x -1,

x + 2 > 0 and x + 1 < 0

Therefore, f'(x) < 0

Hence, f(x) is decreasing in the interval [-2, -1].

### Solution 32

Given:

Differentiating w.r.t x, we get

Take f'(x) = 0

For x < 1,

x - 3 < 0 and x - 1 < 0

Therefore, f'(x) > 0

For 1 < x < 3,

x - 3 < 0 and x - 1 > 0

Therefore, f'(x) < 0

For x > 3,

x - 3 > 0 and x - 1 > 0

So, f'(x) > 0

Hence, f(x) is decreasing for 1 < x < 3.

### Solution 33

Given:

Differentiating w.r.t x, we get

Now,

So, f'(x) > 0 when cos x > 0

i.e.

Therefore, f(x) is increasing when

Also, f'(x) < 0 when cos x < 0

i.e.

Therefore, f(x) is decreasing when

Hence, f(x) is decreasing in the interval

### Solution 34

a. Let f(x) = sin 2x

Differentiating w.r.t x, we get

f'(x) = 2cos 2x

When

Now,

b. Let g(x) = tan x

Differentiating w.r.t x, we get

f'(x) = sec2x

Now, sec2x > 0 for

Therefore, f(x) is increasing in the interval

c. Let h(x) = cos x

Differentiating w.r.t x, we get

f'(x) = -sin x

Now, sin x > 0 for

So, -sin x < 0 for

Therefore, f(x) is decreasing in the interval

d. Let h(x) = cos 3x

Differentiating w.r.t x, we get

f'(x) = -3 sin 3x

When

Now, sin 3x > 0 for

Also, sin 3x < 0 for

Thus, cos x is decreasing in the interval

### Solution 35

Given: f(x) = tan x - x

Differentiating w.r.t x, we get

f'(x) = sec2x - 1 = tan2x

Now, tan2x 0

Hence, f(x) always increases.

### Solution 1

Correct option: (b)

### Solution 2

Correct option: (c)

### Solution 3

Correct option: (c)

### Solution 4

Correct option:(b)

### Solution 5

Correct option: (a)

### Solution 6

Correct option: (c)

### Solution 7

Correct option: (b)

### Solution 8

Correct option: (c)

### Solution 9

Correct option: (a)

### Solution 10

Correct option: (c)

### Solution 11

Correct option: (b)

### Solution 12

Correct option: (a)

### Solution 13

Correct option: (b)

### Solution 14

Correct option: (d)

### Solution 15

Correct option: (d)

### Solution 16

Correct option: (c)

### Solution 17

Correct option: (c)

### Solution 18

Correct option: (d)

### Solution 19

Correct option: (a)

Every invertible function is always monotonic function.

### Solution 20

Correct option: (b)

### Solution 21

Correct option: (c)

### Solution 22

Correct option: (a)

### Solution 23

Correct option: (d)

### Solution 24

Correct option: (d)

### Solution 25

Correct option: (b)

### Solution 26

Correct option: (b)

### Solution 27

Correct option: (d)

### Solution 28

Correct option: (a)

### Solution 29

Correct option: (a)

NOTE: Option (a) should be -1 < k < 1.

### Solution 30

Correct option: (a)

## Increasing and Decreasing Functions Exercise Ex. 17.2

### Solution 1(x)

Given:

Differentiating w.r.t x, we get

Take f'(x) = 0

Clearly, f'(x) > 0 if x < -2 or x > -1

And, f'(x) < 0 if -2 < x < -1

Thus, f(x) increases on  and decreases on

### Solution 1(xxix)

Given:

Differentiating w.r.t x, we get

Take f'(x) = 0

The points x = 2, 4 and -3 divide the number line into four disjoint intervals namely

Consider the interval

In this case, x - 2 < 0, x - 4 < 0 and x + 3 < 0

Therefore, f'(x) < 0 when

Thus the function is decreasing in

Consider the interval

In this case, x - 2 < 0, x - 4 < 0 and x + 3 > 0

Therefore, f'(x) > 0 when

Thus the function is increasing in

Now, consider the interval

In this case, x - 2 > 0, x - 4 < 0 and x + 3 > 0

Therefore, f'(x) < 0 when

Thus the function is decreasing in

And now, consider the interval

In this case, x - 2 > 0, x - 4 > 0 and x + 3 > 0

Therefore, f'(x) < 0 when

Thus the function is increasing in

### Solution 30(ii)

Given:

Differentiating w.r.t x, we get

Now,

Hence, f(x) is an increasing function for all x.