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# Class 12 RD Sharma Solutions – Chapter 17 Increasing and Decreasing Functions – Exercise 17.2 | Set 3

• Last Updated : 02 Jun, 2021

### Question 26. Find the intervals in which f(x) = log (1 + x) – x/(1 + x) is increasing or decreasing.

Solution:

We have,

f(x) = log (1 + x) – x/(1 + x)

On differentiating both sides with respect to x, we get

f'(x) = f'(x) = f'(x) = f'(x) = f'(x) = f'(x) = For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> = 0

=> x = 0

Clearly, f'(x) > 0 if x > 0.

Also, f'(x) < 0 if –1 < x < 0 or x < –1.

Thus f(x) is increasing in (0, ∞) and decreasing in (–∞, –1) ∪ (–1, 0).

### Question 27. Find the intervals in which f(x) = (x + 2)e–x is increasing or decreasing.

Solution:

We have,

f(x) = (x + 2)e–x

On differentiating both sides with respect to x, we get

f'(x) = f'(x) = e–x – e–x (x + 2)

f'(x) = e–x (1 – x – 2)

f'(x) = e–x (x + 1)

For f(x), we need to find the critical point, so we get,

=> f'(x) = 0

=> e–x (x + 1) = 0

=> x = –1

Clearly, f'(x) > 0 if x < –1.

Also, f'(x) < 0 if x > –1.

Thus f(x) is increasing in (–∞, –1) and decreasing in (–1, ∞).

### Question 28. Show that the function f given by f(x) = 10x is increasing for all x.

Solution:

We have,

f(x) = 10x

On differentiating both sides with respect to x, we get

f'(x) = f'(x) = 10x log 10

Now we have, x ∈ R, we get

=> 10x > 0

=> 10x log 10 > 0

=> f'(x) > 0

Thus, f(x) is increasing for all x.

Hence proved.

### Question 29. Prove that the function f given by f(x) = x – [x] is increasing in (0, 1).

Solution:

We have,

f(x) = x – [x]

On differentiating both sides with respect to x, we get

f'(x) = f'(x) = 1

Now we have,

=> 1 > 0

=> f'(x) > 0

Thus, f(x) is increasing in the interval (0, 1).

Hence proved.

### Question 30. Prove that the function f(x) = 3x5 + 40x3 + 240x is increasing on R.

Solution:

We have,

f(x) = 3x5 + 40x3 + 240x

On differentiating both sides with respect to x, we get

f'(x) = f'(x) = 15x4 + 120x2 + 240

f'(x) = 15 (x4 + 8x2 + 16)

f'(x) = 15 (x2 + 4)2

Now we know,

=> (x2 + 4)2 > 0

=> 15 (x2 + 4)2 > 0

=> f'(x) > 0

Thus, the given f(x) is increasing on R.

Hence proved.

### Question 31. Prove that the function f given by f(x) = log cos x is strictly increasing on (–π/2, 0) and strictly decreasing on (0, π/2).

Solution:

We have,

f(x) = log cos x

On differentiating both sides with respect to x, we get

f'(x) = f'(x) = f'(x) = f'(x) = – tan x

Now for x ∈ (0, π/2), we get

=> 0 < x < π/2

=> tan 0 < tan x < tan π/2

=> 0 < tan x < 1

=> tan x > 0

=> – tan x < 0

=> f'(x) < 0

Also for x ∈ (–π/2, 0), we have,

=> –π/2 < x < 0

=> tan (–π/2) < tan x <  tan 0

=> –1 < tan x < 0

=> tan x < 0

=> – tan x > 0

=> f'(x) > 0

Thus, f(x) is strictly increasing on the interval (–π/2, 0) and strictly decreasing on the interval (0, π/2).

Hence proved.

### Question 32. Show that the function f given by f(x) = x3 – 3x2 + 4x is strictly increasing on R.

Solution:

We have,

f(x) = x3 – 3x2 + 4x

On differentiating both sides with respect to x, we get

f'(x) = f'(x) = 3x2 – 6x + 4

f'(x) = 3 (x2 – 2x + 1) + 1

f'(x) = 3 (x – 1)2 + 1

Now, we know,

=> (x – 1)2 > 0

=> 3 (x – 1)2 > 0

=> 3 (x – 1)2 + 1 > 0

=> f'(x) > 0

Thus, f(x) is strictly increasing on R.

Hence proved.

### Question 33. Show that the function f given by f(x) = cos x is strictly decreasing in (0, π), increasing in (π, 2π) and neither increasing or decreasing in (0, 2π).

Solution:

We have,

f(x) = cos x

On differentiating both sides with respect to x, we get

f'(x) = f'(x) = – sin x

Now for x ∈ (0, π), we get

=> 0 < x < π

=> sin 0 < sin x < sin π

=> 0 < sin x < 0

=> sin x > 0

=> – sin x < 0

=> f'(x) < 0

Also for x ∈ (π, 2π), we get

=> π < x < 2π

=> sin 0 < sin x < sin π

=> 0 < sin x < 0

=> sin x < 0

=> – sin x > 0

=> f'(x) > 0

Thus, f(x) is strictly increasing on the interval (π, 2π) and strictly decreasing on the interval (0, π).

So, the function is neither increasing or decreasing in (0, 2π).

Hence proved.

### Question 34. Show that f(x) = x2 – x sin x is an increasing function on (0, π/2).

Solution:

We have,

f(x) = x2 – x sin x

On differentiating both sides with respect to x, we get

f'(x) = f'(x) = 2x – (x cos x + sin x)

f'(x) = 2x – x cos x – sin x

Now for x ∈ (0, π/2), we have

=> 0 ≤ sin x ≤ 1

=> 0 ≤ cos x ≤ 1

So, this implies,

=> 2x – x cos x – sin x > 0

=> f'(x) > 0

Thus, f(x) is an increasing function on the interval (0, π/2).

Hence proved.

### Question 35. Find the value(s) of a for which f(x) = x3 – ax is an increasing function on R.

Solution:

We have,

f(x) = x3 – ax

On differentiating both sides with respect to x, we get

f'(x) = f'(x) = 3x2 – a

Now we are given that f(x) = x3 – ax is an increasing function on R, we get

=> f'(x) > 0

=> 3x2 – a > 0

=> a < 3x2

The critical point for 3x2 = 0 will be 0.

So, we get a ≤ 0.

Therefore, the values of a must be less than or equal to 0.

### Question 36. Find the value of b for which the function f(x) = sin x – bx + c is a decreasing function on R.

Solution:

We have,

f(x) = sin x – bx + c

On differentiating both sides with respect to x, we get

f'(x) = f'(x) = cos x – b + 0

f'(x) = cos x – b

Now we are given that f(x) = sin x – bx + c is a decreasing function on R, we get

=> f'(x) < 0

=> cos x – b < 0

=> b > cos x

The critical point for cos x = 0 will be 1.

So, we get b ≥ 1.

Therefore, the values of b must be greater than or equal to 1.

### Question 37. Show that f(x) = x + cos x – a is an increasing function on R for all values of a.

Solution:

We have,

f(x) = x + cos x – a

On differentiating both sides with respect to x, we get

f'(x) = f'(x) = 1 – sin x

f'(x) = f'(x) = Now for x ∈ R, we have

=> > 0

=> f'(x) > 0

Thus, the f(x) is an increasing function on R for all values of a.

Hence proved.

### Question 38. Let f defined on [0, 1] be twice differentiable such that |f”(x)| ≤ 1 for all x ∈ [0, 1]. If f(0) = f(1), then show that f'(x) < 1 for all x ∈ [0, 1].

Solution:

As f(0) = f(1) and f is differentiable, we can apply Rolle’s theorem here. So, we get

f(c) = 0 for some c ∈ [0, 1].

On applying Lagrange’s mean value theorem, we get,

For point c and x ∈ [0, 1], so we have

=> => => As we are given that |f”(d)| ≤ 1 for x ∈ [0, 1], we get

=> ≤ 1

=> |f'(x)| ≤ x – c

Now as both x and c lie in [0, 1], therefore x  – c ∈ (0, 1).

This gives us, |f'(x)| < 1 for all x ∈ (0, 1).

Hence proved.

### (i) f(x) = x |x|, x ∈ R

Solution:

We have,

f(x) = x |x|, x ∈ R

=> => => f'(x) > 0 for all values of x

Therefore, f(x) is an increasing function for all real values.

### (ii) f(x) = sin x + |sin x|, 0 < x ≤ 2π

Solution:

We have,

f(x) = sin x + |sin x|, 0 < x ≤ 2π

=> => The function cos x is positive between the interval (0, π/2).

Therefore, the function is increasing in the interval (0, π/2).

Also, the function cos x is negative between the interval (π/2, π).

Therefore, the function is decreasing in the interval (0, π/2).

Now for π ≤ x ≤ 2π, value of f'(x) is 0.

Hence the function is neither increasing nor decreasing in the interval (π, 2π).

### (iii) f(x) = sin x (1 + cos x), 0 < x ≤ π/2

Solution:

We have,

f(x) = sin x (1 + cos x)

On differentiating both sides with respect to x, we get

f'(x) = f'(x) = f'(x) = –sin2 x + cos x + cos2

f'(x) = cos2 x – sin2 x + cos x

f'(x) = cos2 x – (1 – cos2 x) + cos x

f'(x) = cos2 x – 1 + cos2 x + cos x

f'(x) = 2 cos2 x + cos x – 1

f'(x) = 2 cos2 x + 2 cos x – cos x – 1

f'(x) = 2 cos x (cos x + 1) – 1 (cos x + 1)

f'(x) = (2 cos x – 1) (cos x + 1)

For f(x) to be increasing, we must have,

=> f'(x) > 0

=> (2 cos x – 1) (cos x + 1) > 0

=> 0 < x < π/3

=> x ∈ (0, π/3)

For f(x) to be decreasing, we must have,

=> f'(x) < 0

=> (2 cos x – 1) (cos x + 1) > 0

=> π/3 < x < π/2

=> x ∈ (π/3, π/2)

Thus, f(x) is increasing on the interval x ∈ (0, π/3) and decreasing on the interval x ∈ (π/3, π/2).

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