Class 12 NCERT Solutions- Mathematics Part I – Application of Derivatives – Exercise 6.2
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 ∞
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) = 2x2 – 3x is
(i) increasing
(ii) decreasing
Solution:
Given: f(x) = 2x2 – 3x
f'(x) = [Tex]\frac{d(2x^2-3x)}{dx} [/Tex] = 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)
Question 5. Find the intervals in which the function f given by f(x) = 2x3 – 3x2 – 36x + 7 is
(i) increasing
(ii) decreasing
Solution:
Given: f(x) = 2x3 – 3x2 – 36x + 7
f'(x) = [Tex]\frac{d(2x^3-3x^2-36x+7)}{dx} [/Tex] = 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)
Question 6. Find the intervals in which the following functions are strictly increasing or decreasing:
(i) x2 + 2x – 5
(ii) 10 – 6x – 2x2
(iii) -2x3 – 9x2 – 12x + 1
(iv) 6 – 9x – x2
(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 – x2
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) – [Tex]\frac{2x}{2+x} [/Tex], is an increasing function of x throughout its domain.
Solution:
f(x) = log(1+x)[Tex]\frac{-2x}{2+x}, x>-1[/Tex]
[Tex]\frac{dy}{dx}=f'(x)=\frac{1}{1+x}-\frac{(2x+1).2-2x(1)}{(2x+1)^2} [/Tex]
f'(x)=[Tex]\frac{1}{1+x}-\frac{4+2x-2x}{(2x+1)^2}[/Tex]
[Tex]f'(x)=\frac{1}{1+x}-\frac{4}{(2+x)^2}[/Tex]
[Tex]f'(x)=\frac{(x+2)^2-4(x+1)}{(x+1)(x+2)}[/Tex]
[Tex]f'(x)=\frac{x^2}{(x+1)(x+2)^2}[/Tex]
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 = [Tex]\frac{4sin\theta}{2+cos\theta}-\theta [/Tex] is an increasing function of θ in[0, π/2].
Solution:
y = f(θ) = [Tex]\frac{4\sinθ}{(2+\cosθ)^2}-θ[/Tex]
[Tex]f'(θ)=\frac{(2+\cosθ).4\cosθ-4\sinθ(-\sinθ)}{(2+\cosθ)^2} [/Tex]
[Tex]f'(θ)=\frac{4.(2\cosθ+\cos^2θ+\sin^2θ)}{(2+\cosθ)^2}-1[/Tex]
[Tex]f'(θ)=\frac{8\cosθ+4}{(\cosθ+2)^2}[/Tex]
[Tex]f'(θ)=\frac{8\cosθ+4-(2+\cosθ)^2}{(2+\cosθ)^2}[/Tex]
[Tex]f'(θ)=\frac{4\cosθ-\cos^2θ}{(2+\cosθ)^2}[/Tex]
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) = [Tex]\frac{4\sinθ}{2+\cosθ}-θ [/Tex] 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, ∞)
Question 11. Prove that the function f given by f(x) = x2 – x + 1 is neither strictly increasing nor decreasing on (– 1, 1).
Solution:
Given: f(x) = x2 – x + 1
f'(x) = 2x – 1
For strictly increasing, f'(x) > 0
2x – 1 > 0
x > 1/2
So, f(x) function is increasing for x > 1/2 in the interval (1/2, 1) -(Given interval is (-1, 1)
Similarly, for decreasing f'(x) < 0
2x – 1 < 0
x < 1/2
So, f(x) function is increasing for x < 1/2 in the interval (-1, 1/2) -(Given interval is (-1, 1)
Hence, the function f(x) = x2 – x + 1 is neither strictly increasing nor decreasing.
Question 12. Which of the following functions are decreasing on (0, π/2).
(A) cos x (B) cos 2x (C) cos 3x (D) tan x
Solution:
(A) f(x) = cos x
f'(x) = -sin x
Now in (0, π/2) interval, sin x is positive(because it is second quadrant)
So, -sin x < 0
∴ f'(x) < 0
f(x) = cos x is strictly decreasing on(0, π/2).
(B) f(x) = cos 2x
f'(x) = -2 sin 2x
Now in (0, π/2) interval, sin x is positive(because it is second quadrant)
-sin 2x < 0
∴ f'(x) < 0,
f(x) = cos 2x is strictly decreasing on(0, π/2).
(C) f(x) = cos 3x
f'(x) = -3sin 3x
Let 3x = t
So in sin 3x = sin t
When t ∈(0, π), sin t + >0 or 3x ∈ (0, π)
But when π/3 < x < π/2
π < 3x < 3π/2
Here sin 3x < 0
So, in x ∈ (0, π/3),
f'(x) = -3sin 3x < 0 & in x∈(π/3, π/2), f'(x) = -3sin 3x > 0
f'(x) is changing signs, hence f(x) is not strictly decreasing.
(D) f(x) = tan x
f'(x) = sec2x
Now in x ∈ (0, π/2), sec2x > 0
Hence, f(x) is strictly increasing on(0, π/2).
So, option (A) and (B) are decreasing on (0, π/2).
Question 13. On which of the following intervals is the function f given by f(x) = x100 + sin x – 1 decreasing ?
(A) (0, 1) (B) π/2, π (C) 0, π/2 (D) None of these
Solution:
f(x) = x100 + sin x – 1
f'(x) = 100x99 + cos x
(A) In (0, 1) interval, x > 0, so 100x99 > 0
and for cos x: (0, 1°) = (0, 0.57°) > 0
Hence, f(x)is strictly increasing in interval(0, 1)
(B) In (π/2, π) interval,
For 100x99: x ∈ (π/2, π) = (11/7, 22/7) = (1.5, 3.1) > 1
So, x99 > 1. Hence 100x99 > 100
For Cos x: (π/2, π) in second quadrant and in second quadrant cos x is negative, so the value is in be -1 and 0.
Hence, f(x)is strictly increasing in interval (π/2, π)
(C) In (0, π/2) interval, both cos x > 0 and 100x99 > 0
So f'(x) > 0
Hence, f(x)is strictly increasing in interval (0, π/2)
So, the correct option is (D).
Question 14. For what values of a the function f given by f(x) = x2 + ax + 1 is increasing on (1, 2)?
Solution:
Given: f(x) = x2 + ax + 1
f'(x) = 2x + a
Now, x ∈ (1, 2), 2x ∈ (2, 4)
2x + a ∈ (2 + a, 4 + a)
For f(x) to be strictly increasing, f'(x) > 0
If the minimum value of f'(x) > 0 then
f'(x) on its entire domain will be > 0.
f'(x)min > 0
2 + a > 0
a > -2
Question 15. Let I be any interval disjoint from [–1, 1]. Prove that the function f given by [Tex]f(x)=x+\frac{1}{x} [/Tex] is increasing on I.
Solution:
Clearly the maximum interval I is R-(-1,1)
Now, f(x) = [Tex]x+\frac{x}{1}[/Tex]
f'(x) = [Tex]1-\frac{1}{x^2}=\frac{x^2-1}{x^2}[/Tex]
It is given that I be any interval disjoint from [–1, 1]
So, for every x ∈ I either x < -1 or x > 1
So, for x < -1, f'(x) is positive.
So, for x < 1, f'(x) is positive.
Hence, f'(x) > 0 ∀ x ∈ I, so, f(x) is strictly increasing on I.
Question 16. Prove that the function f given by f(x) = log sin x is increasing on (0, π/2) and decreasing on (π/2, π).
Solution:
f(x) = log sin x
f'(x) = [Tex]\frac{1}{\sin x}.\cos x=\cot x [/Tex]
Interval (0, π/2), it is first quadrant, here cot x is positive.
So, f'(x) = cot x is positive (i.e., cot x > 0)
Hence, f(x) is strictly increasing in interval (0, π/2)
Interval (π/2, π), it is second quadrant, here cot x is negative.
So, f'(x) = cot x is negative (i.e., cot x < 0)
Hence, f(x) is strictly decreasing in interval (π/2, π)
Question 17. Prove that the function f given by f(x) = log|cos x| is decreasing on (0, π/2) and increasing on (π/2, π).
Solution:
f(x) = log cos x
f'(x) = 1/cos x (-sin x) = -tan x
Interval (0, π/2), it is first quadrant, here tan x is positive.
So, f'(x) = -tan x is negative(i.e., tan x < 0)
Hence, f(x) is strictly decreasing in interval (0, π/2)
Interval (π/2, π), it is second quadrant, here tan x is negative.
So, f'(x) = -tan x is positive (i.e., tan x > 0)
Hence, f(x) is strictly increasing in interval (π/2, π)
Question 18. Prove that the function given by f(x) = x3 – 3x2 + 3x – 100 is increasing in R.
Solution:
f(x) = x3 – 3x2 + 3x – 100
f'(x) = 3x2 – 6x + 3
f'(x) = 3(x2 – 2x + 1)
f'(x) = 3(x – 1)2 ≥ 0 ∀ x in R
So f(x) is strictly increasing in R.
Question 19. The interval in which y = x2 e-x is increasing is
(A) (– ∞, ∞) (B) (– 2, 0) (C) (2, ∞) (D) (0, 2)
Solution:
Given, f(x) = x2e-x
f'(x) = x2(-e-x) + e-x.2x
f'(x) = e-x(2x – x2)
f'(x) = e-x.x(2 – x)
For f(x) to be increasing, f'(x) ≥ 0
So, f'(x) ≥ 0
e-x.x.(2 – x) ≥ 0
x.(2 – x) ≥ 0
x(x – 2) ≥ 0
x ∈ [0, 2]
So, the f(x) is strictly increasing in interval (0, 2). Correct option in D.
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