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

• Last Updated : 06 Apr, 2021

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) = e2x 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) = e2x

f’(x) = 2e2x > 0

Hence, f(x) = e2x is strictly increasing on ∞

(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.

(ii) decreasing

Solution:

Given: f(x) = 2x2 – 3x

f'(x) =  = 4x – 3           -(1)

= x = 3/4

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

(i) Interval (3/4, ∞) let take x = 1

So, from eq(1) f'(x) > 0

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

(ii) Interval (-∞, 3/4) let take x = 0.5

So, from eq(1) f'(x) < 0

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

(ii) decreasing

Solution:

Given: f(x) = 2x3 – 3x2 – 36x + 7

f'(x) =  = 6x2 – 6x – 36         -(1)

f'(x) = 6(x2 – x – 6)

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

6(x2 – x – 6) = 0

(x2 – x – 6) = 0

x = -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)

(v) (x + 1)3 (x – 3)3

Solution:

(i) f(x) = x2 + 2x – 5

f'(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 – 2x2

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) = -2x3 – 9x2 – 12x + 1

f'(x) = -6x2 – 8x – 12

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

-6x2 – 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

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 x

For 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, x2 > 0, (x + 2)2 ≥ 0, x + 1 > 0

From 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 = x2(x – 2x)2

= x4 – 4x3 + 4x2

f'(x) = 4x3 – 12x2 + 8x

f'(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, ∞)

Chapter 6 Application of Derivatives – Exercise 6.2 | Set 2

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