# Class 12 NCERT Solutions- Mathematics Part I – Application of Derivatives – Exercise 6.2 | Set 1

### Question 1. Show that the function given by f (x) = 3x + 17 is increasing on R.

**Solution:**

If for a function f(x), f'(x) > 0 for all x, then the function is a strictly increasing function. (vice-versa is not true)

Given: f(x) = 3x + 17

f'(x) = 3 > 0 -(Always greater than zero)

Hence, 3x + 17 is strictly increasing on R.

### Question 2. Show that the function is given by f (x) = e^{2x} is increasing on R.

**Solution:**

If for a function f(x), f'(x) > 0 for all x, then the function is a strictly increasing function. (vice-versa is not true)

Given: f(x) = e

^{2x}f’(x) = 2e

^{2x }> 0Hence, f(x) = e

^{2x}is strictly increasing on ∞

### Question 3. Show that the function given by f (x) = sin x is

### (i) increasing in(0, π/2)

### (ii) decreeing in(π/2, π)

### (iii) neither increasing nor decreasing in (0, π)

**Solution:**

Given: f(x) = sin x

So, f’(x) = d/dx(sin x) = cos x

(i)Now in (0, π/2), f’(x) = cos x > 0 (positive in first quadrant)Hence, f(x) = sin x is strictly increasing in (0, π/2).

(ii)In (π/2, π), f’(x) = cos x < 0 -(negative in second quadrant)Hence, f(x) = sin x is strictly decreasing in (π/2,π)

(iii)As we know that f’(x) = cos x is positive in interval(0, π/2)and f’(x) = cos x is negative in interval (π/2, π)

So, it is neither increasing nor decreasing.

### Question 4. Find the intervals in which the function f given by f(x) = 2x^{2} – 3x is

### (i) increasing

### (ii) decreasing

**Solution:**

Given: f(x) = 2x

^{2 }– 3xf'(x) = = 4x – 3 -(1)

= x = 3/4

So the intervals are (-∞, 3/4) and (3/4, ∞)

(i)Interval (3/4, ∞) let take x = 1So, from eq(1) f'(x) > 0

Hence, f is strictly increasing in interval (3/4, ∞)

(ii)Interval (-∞, 3/4) let take x = 0.5So, from eq(1) f'(x) < 0

Hence, f is strictly decreasing in interval (-∞, 3/4)

### Question 5. Find the intervals in which the function f given by f(x) = 2x^{3} – 3x^{2} – 36x + 7 is

### (i) increasing

### (ii) decreasing

**Solution:**

Given: f(x) = 2x

^{3 }– 3x^{2 }– 36x + 7f'(x) = = 6x

^{2 }– 6x – 36 -(1)f'(x) = 6(x

^{2}– x – 6)On putting f'(x) = 0, we get

6(x

^{2}– x – 6) = 0(x

^{2}– x – 6) = 0x = -2, x = 3

So, the intervals are (-∞, -2), (-2, 3), and (3, ∞)

For (-∞, -2) interval, take x = -3

From eq(1), we get

f'(x) = (+)(-)(-) = (+) > 0

So, f is strictly increasing in interval (-∞, -2)

For (-2, 3) interval, take x = 2

From eq(1), we get

f'(x) = (+)(+)(-) = (-) < 0

So, f is strictly decreasing in interval (-2, 3)

For (3, ∞)interval, take x = 4

From eq(1), we get

f'(x) = (+)(+)(+) = (+) > 0

So, f is strictly increasing in interval (3, ∞)

(i)f is strictly increasing in interval (-∞, -2) and (3, ∞)

(ii)f is strictly decreasing in interval (-2, 3)

### Question 6. Find the intervals in which the following functions are strictly increasing or decreasing:

### (i) x^{2 }+ 2x – 5

### (ii) 10 – 6x – 2x^{2}

### (iii) -2x^{3} – 9x^{2} – 12x + 1

### (iv) 6 – 9x – x^{2 }

### (v) (x + 1)^{3} (x – 3)^{3}

**Solution:**

(i)f(x) = x^{2 }+ 2x – 5f'(x) = 2x + 2 -(1)

On putting f'(x) = 0, we get

2x + 2 = 0

x = -1

So, the intervals are (-∞, -1) and (-1, ∞)

For (-∞, -1) interval take x = -2

From eq(1), f'(x) = (-) < 0

So, f is strictly decreasing

For (-1, ∞) interval take x = 0

From eq(1), f'(x) = (+) > 0

So, f is strictly increasing

(ii)f(x) = 10 – 6x – 2x^{2}f'(x) = -6 – 4x

On putting f'(x) = 0, we get

-6 – 4x = 0

x = -3/2

So, the intervals are (-∞, -3/2) and (-3/2, ∞)

For (-∞, -3/2) interval take x = -2

From eq(1), f'(x) = (-)(-) = (+) > 0

So, f is strictly increasing

For (-3/2, ∞) interval take x = -1

From eq(1), f'(x) = (-)(+) = (-) < 0

So, f is strictly decreasing

(iii)f(x) = -2x^{3}– 9x^{2}– 12x + 1f'(x) = -6x

^{2 }– 8x – 12On putting f'(x) = 0, we get

-6x

^{2 }– 8x – 12 = 0-6(x + 1)(x + 2) = 0

x = -1, x = -2

So, the intervals are (-∞, -2), (-2, -1), and (-1, ∞)

For (-∞, -2) interval take x = -3

From eq(1), f'(x) = (-)(-)(-) = (-) < 0

So, f is strictly decreasing

For (-2, -1) interval take x = -1.5

From eq(1), f'(x) = (-)(-)(+) = (+) > 0

So, f is strictly increasing

For (-1, ∞) interval take x = 0

From eq(1), f'(x) = (-)(+)(+) = (-) < 0

So, f is strictly decreasing

(iv)f(x) = 6 – 9x – x^{2 }f'(x) = -9 – 2x

On putting f'(x) = 0, we get

-9 – 2x = 0

x = -9/2

So, the intervals are (-∞, -9/2) and (-9/2, ∞)

For f to be strictly increasing, f'(x) > 0

– 9 – 2x > 0

x > -9/2

So f is strictly increasing in interval (-∞, -9/2)

For f to be strictly decreasing, f'(x) < 0

-9 – 2x < 0

x < -9/2

So f is strictly decreasing in interval (-9/2, ∞)

(v)f(x) = (x + 1)^{3}(x – 3)^{3}f'(x) = (x + 3)

^{3}.3(x – 3)^{3 }+ (x – 3)^{3}.3(x + 1)^{2}f'(x) = 6(x – 3)

^{2}(x + 1)^{2}(x – 1)Now, the factor of (x – 3)

^{2 }and (x + 1)^{2}are non-negative for all xFor f to be strictly increasing, f'(x) > 0

(x – 1) > 0

x > 1

So, f is strictly increasing in interval (1, ∞)

For f to be strictly decreasing, f'(x) < 0

(x – 1) < 0

x < 1

So, f is strictly decreasing in interval (-∞, 1)

### Question 7. Show that y = log(1 + x) – , is an increasing function of x throughout its domain.

**Solution:**

f(x) = log(1+x)

f'(x)=

So, the domain of the given function is x > -1

Now, x

^{2 }> 0, (x + 2)^{2 }≥ 0, x + 1 > 0From the above equation f'(x) ≥ 0 ∀ x in the domain(x > -1) and f is an increasing function.

### Question 8. Find the values of x for which y = [x(x – 2)]^{2} is an increasing function.

**Solution:**

Given: y = f(x) = [x(x – 2)]

^{2 }= x^{2}(x^{ }– 2x)^{2}= x

^{4}– 4x^{3}+ 4x^{2}f'(x) = 4x

^{3}– 12x^{2}+ 8xf'(x) = 4x(x – 2)(x – 1)

x = 0, x = 1, x = 2

So, (∞, 0], [0, 1], [1, 2], [2,∞)

For (∞, 0], let x = -1

So, f'(x) = (-)(-)(-) = (-) ≤ 0

f(x) is decreasing

For [0, 1], let x = 1/2

So, f'(x) = (+)(-)(-) = (+) ≥ 0

f(x) is increasing

Similarly, for [1, 2], f(x) is decreasing

For [2,∞), f(x) is increasing

So, f(x) is increasing in interval [0, 1] and [2,∞)

### Question 9. Prove that y = is an increasing function of θ in[0, π/2].

**Solution:**

y = f(θ) =

Now 0 ≤ θ ≤ π/2, and we have 0 ≤ cosθ ≤ 1,

So, 4 – cosθ > 0

Therefore f'(θ) ≥ 0 for 0 ≤ θ ≤ π/2

Hence, f'(x) = is a strictly increasing in the interval (θ, π/2).

### Question 10. Prove that the logarithmic function is increasing on (0, ∞).

**Solution:**

Given: f(x) = log(x) -(logarithmic function)

f'(x) = 1/x ∀ x in (0, ∞)

Therefore, x > 0, so, 1/x > 0

Hence, the logarithmic function is strictly increasing in interval (0, ∞)