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

### 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) = 10^{x} is increasing for all x.

**Solution:**

We have,

f(x) = 10

^{x}On differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 10

^{x}log 10Now we have, x ∈ R, we get

=> 10

^{x}> 0=> 10

^{x}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) = 3x^{5} + 40x^{3} + 240x is increasing on R.

**Solution:**

We have,

f(x) = 3x

^{5}+ 40x^{3}+ 240xOn differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 15x

^{4}+ 120x^{2}+ 240f'(x) = 15 (x

^{4}+ 8x^{2}+ 16)f'(x) = 15 (x

^{2}+ 4)^{2}Now we know,

=> (x

^{2}+ 4)^{2}> 0=> 15 (x

^{2}+ 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) = x^{3} – 3x^{2} + 4x is strictly increasing on R.

**Solution:**

We have,

f(x) = x

^{3}– 3x^{2}+ 4xOn differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 3x

^{2}– 6x + 4f'(x) = 3 (x

^{2}– 2x + 1) + 1f'(x) = 3 (x – 1)

^{2}+ 1Now, 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) = x^{2} – x sin x is an increasing function on (0, π/2).

**Solution:**

We have,

f(x) = x

^{2}– x sin xOn 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) = x^{3} – ax is an increasing function on R.

**Solution:**

We have,

f(x) = x

^{3}– axOn differentiating both sides with respect to x, we get

f'(x) =

f'(x) = 3x

^{2}– aNow we are given that f(x) = x

^{3}– ax is an increasing function on R, we get=> f'(x) > 0

=> 3x

^{2}– a > 0=> a < 3x

^{2}The critical point for 3x

^{2}= 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.

### Question 39. Find the interval in which f(x) is increasing or decreasing:

### (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) = –sin

^{2}x + cos x + cos^{2}xf'(x) = cos

^{2 }x – sin^{2}x + cos xf'(x) = cos

^{2}x – (1 – cos^{2}x) + cos xf'(x) = cos

^{2}x – 1 + cos^{2}x + cos xf'(x) = 2 cos

^{2}x + cos x – 1f'(x) = 2 cos

^{2}x + 2 cos x – cos x – 1f'(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).