ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.5

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.5

Question 1.
Show that the following functions are strictly increasing on R :
(i) f(x) = 3x+ 17 (NCERT)
(ii) f(x) = 7x – 3. (NCERT)
Solution:
(i) Given f(x) = 3x + 17
Diff. both sides w.r.t. x, we have
f’(x) = 3 > 0 ∀ x ∈ R
Thus the functionf(x) is strictly increasing on R.

Aliter:
Given f(x) = 3x + 17 and let x₁, x₂ ∈ R
s.t. x₁ > x₂
⇒ 3x₁ > 3x₂
⇒ 3x₁ + 7 > 3x₂ + 7
⇒ f(x₁) >f(x₂)
∴ f(x) is strictly increasing on R.

(ii) Given f(x) = 7x – 3
∴ f’(x) = 7 > 0 ∀ x ∈ R
Hence, the function f (x) is strictly increasing on R.

Question 2.
(i) Show that the function f given by f(x) = 3 – 7x is strictly decreasing.
(ii) Show that the function f(x) = 1 – \(\frac{1}{x}\) is strictly increasing.
Solution:
(i) Given f(x) = 3 – 7x
∴ f’(x) = – 7 < 0 ∀ x ∈ R Thus, the function f(x) is strictly decreasing on R.

(ii) Given f(x) = 1 – \(\frac{1}{x}\)
∴ f'(x) = \(\frac{1}{x^2}\) > 0 ∀ x ∈ R
[∵ x² ≥ 0 ∀ x ∈ R]
Thus f(x) is strictly increasing on R.

Question 3.
(i) Prove that the function f(x) = 3 + \(\frac{1}{x}\) is strictly decreasing.
(ii) Show that the function f(x) = x – \(\frac{1}{x}\) is strictly increasing.
Solution:
(i) Given f(x) = 3 + \(\frac{1}{x}\)
∴ f'(x) = – \(\frac{1}{x^2}\) < 0
[∵ x² ≥ 0 ∀ x ∈ R
⇒ \(\frac{1}{x^2}\) > 0 ∀ x ∈ R]
Thus, f(x) is strictly decreasing on R.

(ii) Given f(x) = x – \(\frac{1}{x}\)
∴ f'(x) = 1 + \(\frac{1}{x^2}\) ∀ x ∈ R
[∵ x² ≥ 0 ∀ x ∈ R
⇒ \(\frac{1}{x^2}\) > 0
⇒ 1 + \(\frac{1}{x^2}\) > 1 ∀ x ∈ R]
Thus f(x) is strictly increasing on R.

Question 4.
(i) Show that the function f (x) = e^{– 2x} is strictly decreasing on R.
(ii) Show that the function f (x) = 5^x is strictly increasing.
(iii) Show that the function f (x) = log x is strictly increasing. (NCERT)
Solution:
(i) Given f(x) = e^{– 2x}
f (x) = – 2 e^-2x < 0
[∵ e^{– 2x} > 0 ∀ x ∈ R]
Hence f(x) is strictly decreasing on R.

(ii) Given f(x) = 5^x
f'(x) = 5^x log 5 ∀ x ∈ R
[∵ 5^x > 0 ∀ x ∈ R]
Hence f(x) is strictly increasing on R.

(iii) Let x₁, x₂ ∈ (0, ∞) s.t. x₁ < x₂
⇒ log_e x₁ < log_e x₂
⇒ f(x₁) < f(x₂)
Thus x₁ < x₂
⇒ f(x₁) < f(x₂) ∀ x₁, x₂ ∈ (0, ∞)
Hence f(x) is increasing on (0, ∞).

Aliter :
f(x) = log x
∴ f'(x) = \(\frac{1}{x}\) > 0 ∀ x ∈ R (0, ∞)
[∵ as x > 0 ⇒ \(\frac{1}{x}\) > 0]
f(x) is strictly increasing in (0, ∞).

Question 5.
Show that the function f(x) = – x + cos x is decreasing.
Solution:
Given, f(x) = – x + cos x
∴ f'(x) = – 1 – sin x = – (1 + sin x)
Since, – 1 ≤ sin x ≤ 1
⇒ 1 + sin x ≥ 0
⇒ – (1 + sin x) ≤ 0
∴ f'(x) ≤ 0
Hence f(x) is decreasing on R.

Question 6.
For what value of k, the function f(x) = kx³ + 5 is decreasing ?
Solution:
Given f(x) = kx³ + 5 ;
f'(x) = 3kx²
Now f(x) is decreasing if f'(x) ≤ 0
if 3kx² ≤ 0 if 3k < 0
[∵ x² > 0 ∀ x ∈ R] if k< 0.

Question 7.
(i) Prove that the function f(x) = ax + b is strictly decreasing iff a < 0.
(ii) Prove that the function/(x) = [x] – x is strictly decreasing on [0, 1).
Solution:
(i) Given f(x) = ax + b
∴ f'(x) = a
f (x) is strictly decreasing iff f'(x) < 0 iff a < 0.

(ii) Given f(x) = [x] – x, x ∈ [0, 1)
Since x ∈ [0, 1)
∴ 0 < x < 1
⇒ [x] = 0
f(x) = 0 – x = – x, x ∈ [0, 1)
⇒ f'(x) = – 1 < 0
Thus f(x) is strictly decreasing in [0, 1).

Question 8.
Prove that the function f(x) = x³ – 6x² + 12x + 5 is increasing on R. (ISC 2019)
Solution:
Given f(x) = x³ – 6x² + 12x + 5
∴ f'(x) = 3x² – 12x + 12
= 3 (x² – 4x + 4)
= 3 (x – 2)² ≥ 0 ∀ x ∈ R.
Thus f(x) is increasing on R.

Question 8 (old).
Show that the function f given by f(x) = x³ – 3x² + 4x, x ∈ R, is strictly increasing on R. (NCERT)
Solution:
Given f(x) = x³ – 3x² + 4x
∴ f'(x) = 3x² – 6x + 4
= 3 (x² – 2x + \(\frac{4}{3}\))
= 3 [x² – 2x + 1 + \(\frac{1}{3}\)]
= 3 [(x – 1)² + \(\frac{1}{3}\)]
[∵ (x – 1)² ≥ 0
⇒ (x – 1)² + \(\frac{1}{3}\) ≥ \(\frac{1}{3}\) > 0
⇒ 3 [(x – 1)² + \(\frac{1}{3}\)] > 0
Thus, f(x) is strictly increasing on R.

Question 9.
Show that the function f(x) = 4x³ – 18x² + 27x – 7 is strictly increasing on R.
Solution:
Given f(x) = 4x³ – 18x² + 27x – 7
∴ f'(x) = 12x² – 36x + 27
= 3 (4x² – 12x + 9)
= 3 (2x – 3)² ≥ 0 ∀ x ∈ R.

Question 9 (old).
Prove that the function f(x) = x³ – 3x² + 3x – 100 is strictly increasing on R. (NCERT)
Solution:
Given f(x) = x³ – 3x² + 3x – 100
∴ f'(x) = 3x² – 6x + 3
= 3 (x² – 2x + 1)
= 3 (x- 1)² > 0
∴ f(x) is increasing function on R.

Question 10.
Prove that the function f(x) = 5 – 3x + 3x² – x³ is decreasing on R.
Solution:
Given f(x) = 5 – 3x + 3x² – x³ …………(1)
Diff both sides eqn. (1) w.r.t. x, we have
f(x) = – 3 + 6x – 3x²
= – 3 (x² – 2x + 1)
= – 3 (x – 1)²
since, (x – 1)² > 0 ∀ x ∈ R
⇒ – 3 (x – 1)² < 0, ∀ x ∈ R
⇒ f'(x) < 0, ∀ x ∈ R
Thus, f(x) is decreasing on R.

Question 11.
Show that f(x) = 2x – sin 2x is an increasing function.
Solution:
Given f(x) = 2x – sin 2x
Diff. both sides w.r.t. x, we have
f'(x) = 2 – 2 cos 2x
= 2 (1 – cos 2x)
= 2 × 2 sin² x = 4 sin² x
Since 0 ≤ sin² x ≤ 1 ∀ x ∈ R
⇒ 0 ≤ 4 sin² x ≤ 4
⇒ 0 ≤ f'(x) ≤ 4
Thus, f'(x) ≥ 0 ∀ x ∈ R
Hence, f(x) is increasing on R.

Question 12.
Prove that the function f given by f(x) = log cos x is strictly decreasing on (0, \(\frac{\pi}{2}\)). (NCERT)
Solution:
Given f(x) = log (cos x)
∴ f'(x) = – tan x
when x ∈ (0, \(\frac{\pi}{2}\))
⇒ tan x > 0
⇒ f'(x) < 0
∴ f(x) is strictly decreasing in (0, \(\frac{\pi}{2}\)).

Question 13.
Determine whether the following functions are increasing or decreasing for the stated values of x :
(i) f(x) = \(\frac{1}{x}\), x < 0
(ii) f(x) = \(\frac{1}{1+x^2}\), x ≥ 0
(iii) f(x) = cos² x in [0, \(\frac{\pi}{2}\)]
(iv) f(x) = – \(\frac{x}{2}\) + sin x, – \(\frac{\pi}{3}\) < x < \(\frac{\pi}{3}\)
Solution:
(i) Given f(x) = \(\frac{1}{x}\)
∴ f'(x) = – \(\frac{1}{x^2}\) < 0
[When x < 0
⇒ x² > 0
⇒ \(\frac{1}{x^2}\) > 0
⇒ – \(\frac{1}{x^2}\) < 0]
Thus f(x) is strictly decreasing.

(ii) Given f(x) = \(\frac{1}{1+x^2}\), x ≥ 0
Diff. both sides w.r.t. x, we have
f'(x) = \(\frac{-2 x}{\left(1+x^2\right)^2}\)
Since (1 + x²)² > 0 ∀ x ∈ R
⇒ \(\frac{1}{\left(1+x^2\right)^2}\) ≥ 0
⇒ \(\frac{-2 x}{\left(1+x^2\right)^2}\) ≤ 0 ∀ x ≥ 0
Thus, f'(x) ≤ 0 ∀ x ≥ 0
Hence, f(x) is decreasing for all x ≥ 0.

(iii) f(x) = cos² x
∴ f'(x) = 2 cos x (- sin x) = – sin 2x
Now f'(x) < 0
⇒ – sin 2x < 0 ⇒ sin 2x > 0
⇒ 2x ∈ (0, π)
⇒ x ∈ (0, π/2)
∴ f(x) is strictly decreasing in [0, \(\frac{\pi}{2}\)].

(iv) Given f(x) = – \(\frac{x}{2}\) + sin x
∴ f'(x) = – \(\frac{1}{2}\) + cos x
Now for x ∈ (- \(\frac{\pi}{3}\), \(\frac{\pi}{3}\))
⇒ – \(\frac{\pi}{3}\) < x < \(\frac{\pi}{3}\)
⇒ \(\frac{1}{2}\) < cos x < 1
⇒ – \(\frac{1}{2}\) + cos x > 0
⇒ f'(x) > 0
Hence f'(x) is strictly increasing function on (- \(\frac{\pi}{3}\), \(\frac{\pi}{3}\)).

Question 14.
Which of the following functions are strictly decreasing on ?
(i) cos x
(ii) cos 2x
(iii) sin2 x
(iv) tan x (NCERT)
Solution:
(i) Given f(x) = cos x ;
Diff. both sides w.r.t. x, we have
⇒ f’(x) = – sin x ∀ x ∈ (0, \(\frac{\pi}{2}\)) ;
sin x > 0
⇒ – sin x < 0
⇒ f'(x) < 0
Thus f(x) is strictly decreasing on (0, \(\frac{\pi}{2}\)).

(ii) Given f(x) = cos 2x ;
Diff. both sides w.r.t. x, we get
f'(x) = – 2 sin 2x
When 0 < x < \(\frac{\pi}{2}\) ⇒ 0 < 2x < π ∴ sin 2x > 0
⇒ – 2 sin 2x < 0
⇒ f'(x) < 0
Hence, f(x) is strictly decreasing on (0, \(\frac{\pi}{2}\)).

(iii) Given f(x) = sin² x ;
Diff. both sides w.r.t. x, we have
f'(x) = 2 sin x cos x = sin 2x
when x ∈ (0, \(\frac{\pi}{2}\))
i.e. 0 < x < \(\frac{\pi}{2}\) ⇒ 0 < 2x < π ⇒ sin 2x > 0
⇒ f'(x) > 0
Hence, f(x) is strictly increasing on (0, \(\frac{\pi}{2}\)).

(iv) Given f(x) = tan x;
Diff. both sides w.r.t. x, we have
f'(x) = sec² x > 0 ∀ x ∈ (0, \(\frac{\pi}{2}\))
Thus f(x) is strictly increasing on (0, \(\frac{\pi}{2}\)).

Question 15.
Prove that the function f(x) = sin (2x + \(\frac{\pi}{4}\)) is decreasing for \(\frac{3 \pi}{8}\) ≤ x ≤ \(\frac{5 \pi}{8}\).
Solution:
Given f(x) = sin (2x + \(\frac{\pi}{4}\))
∴ f'(x) = + 2 cos (2x + \(\frac{\pi}{4}\))
Now for x ∈ [\(\frac{3 \pi}{8}\), \(\frac{5 \pi}{8}\)]
⇒ \(\frac{3 \pi}{8}\) ≤ x ≤ \(\frac{5 \pi}{8}\)
⇒ \(\frac{3 \pi}{4}\) ≤ 2x ≤ \(\frac{5 \pi}{4}\)
⇒ \(\frac{3 \pi}{4}+\frac{\pi}{4} \leq 2 x+\frac{\pi}{4} \leq \frac{5 \pi}{4}+\frac{\pi}{4}\)
⇒ π ≤ 2x + \(\frac{\pi}{4}\) ≤ \(\frac{3 \pi}{2}\)
∴ 2x + \(\frac{\pi}{4}\) lies in 3rd quadrant.
∴ cos (2x + \(\frac{\pi}{4}\)) ≤ 0
∴ f'(x) = 2 cos (2x + \(\frac{\pi}{4}\)) ≤ 0
Hence f(x) is decreasing function on \(\left[\frac{3 \pi}{8}, \frac{5 \pi}{8}\right]\).

Question 16.
Prove that the function f(x) = sin x is
(i) strictly increasing in (0, \(\frac{\pi}{2}\))
(ii) strictly decreasing in (\(\frac{\pi}{2}\), π)
(iii) neither increasing nor decreasing in (0, π). (NCERT)
Solution:
(i) Given f(x) = sin x
∴ f’(x) = cos x > 0 ∀ x ∈ (0, \(\frac{\pi}{2}\))
[since in first quadrant, cos x > 0]
Thus f(x) is strictly increasing on (0, \(\frac{\pi}{2}\)).

(ii) Also f’(x) = cos x < 0 ∀ x ∈ (\(\frac{\pi}{2}\), π)
since in 2nd quadrant, cos x is negative.
Thusf(x) is strictly decreasing on (\(\frac{\pi}{2}\), π)

(iii) Hence f(x) is increasing on (0, \(\frac{\pi}{2}\)) and decreasing on (\(\frac{\pi}{2}\), π) Therefore, f (x) is neither increasing nor decreasing on (0, π).

Question 17.
Prove that the function f defined by f(x) = cos x is
(i) strictly incresing in (π, 2π)
(ii) strictly decreasing in (0, 2π)
(iii) neither increasing nor decreasing in (0, 2π). (NCERT)
Solution:
(i) Given f(x) = cos x
∴ f’(x) = – sin x
Now f(x) is strictly increasing if f’(x) > 0
i.e. if – sin x > 0
i.e. sin x < 0
⇒ x ∈ (π, 2π)

(ii) f(x) is strictly decreasing if f’(x) < 0
i.e. – sin x < 0 i.e. sin x > 0
⇒ x ∈ (0, π)

(iii) So f(x) is neither increasing nor decreasing in (0, 2π).
Since f(x) is strictly increasing in (π, 2π) and strictly decreasing in (0, π).

Question 18.
Prove that the following functions are neither increasing nor decreasing :
(i) f(x) = x² – x + 1 on (- 1, 1) (NCERT)
(ii) f(x) = \(\frac{1}{x^2+1}\) on R.
Solution:
(i) Given f(x) = x² – x + 1
∴ f'(x) = 2x – 1
Now f’(x) > 0
⇒ 2x – 1 > 0
⇒ x > \(\frac{1}{2}\)
∴ f(x) is increasing in (\(\frac{1}{2}\), 1).
Now f'(x) < 0
⇒ 2x – 1 < 0
⇒ x < \(\frac{1}{2}\)
∴ f(x) is decreasing on (- 1, \(\frac{1}{2}\)).

(ii) Given f(x) = \(\frac{1}{x^2+1}\) ;
Diff. both sides w.r.t. x ; we have
∴ f'(x) = \(\frac{-2 x}{\left(x^2+1\right)^2}\)
Now f(x) is increasing iff f'(x) ≥ 0
iff \(\frac{-2 x}{\left(x^2+1\right)^2}\) ≥ 0 iff – 2x ≥ 0
[∵ (x² + 1)² ≥ 0, ∀ x ∈ R]
iff x ≤ 0
and f(x) is decreasing iff f'(x) ≤ 0
iff \(\frac{-2 x}{\left(x^2+1\right)^2}\) ≤ 0 iff – 2x ≤ 0
[∵ (x² + 1)² ≥ 0, ∀ x ∈ R]
iff x ≥ 0
Hence, the function f(x) is neither increasing nor decreasing on R.

Question 19.
Prove that the function f(x) = x is
(i) strictly increasing in (0, ∞)
(ii) strictly decreasing in (- ∞, 0).
Solution:
Given f(x) = (x)
= \(\left\{\begin{array}{cc}
x & ; x>0 \\
-x & ; x<0 \\ 0 & ; x=0 \end{array}\right.\) (i) When x > 0
f(x) = x
⇒ f’(x) = 1
∴ f(x) is strictly increasing in (0, ∞)

(ii) When x < 0 f(x) = – x ⇒ f’(x) = – 1
∴ f(x) is strictly decreasing in (∞, 0).

Question 20.
Prove that the function f given by f(x) = \(\frac{4 \sin x}{2+\cos x}\) – x is increasing in [o, \(\frac{\pi}{2}\)].
Solution:
Given f(x) = \(\frac{4 \sin x}{2+\cos x}\) – x
Diff. both sides w.r.t. x, we have
f'(x) = \(\frac{(2+\cos x)(4 \cos x)-4 \sin x(-\sin x)}{(2+\cos x)^2}\) – 1
= \(\frac{8 \cos x+4}{(2+\cos x)^2}\) – 1
= \(\frac{8 \cos x+4-(2+\cos x)^2}{(2+\cos x)^2}\)
= \(\frac{-\cos ^2 x+4 \cos x}{(2+\cos x)^2}\)
= \(\frac{\cos x(4-\cos x)}{(2+\cos x)^2}\)
Since cos x > 0,
4 – cos x > 0 ∀ x ∈ (o, \(\frac{\pi}{2}\))
∴ cos x (4 – cos x) > 0 ∀ x ∈ (o, \(\frac{\pi}{2}\))
Also (2 + cos x)² > 0
∴ f'(x) > 0 ∀ x ∈ (o, \(\frac{\pi}{2}\))
Thus f is strictly increasing in [o, \(\frac{\pi}{2}\)].
[if f'(x) > 0 ∀ x ∈ (a, b) Then f is strictly increasing in [a, b]]

Question 21.
Find the intervals in which the following functions are strictly increasing or strictly decreasing :
(i) f(x) = 2x³ – 3x² – 36x + 7 (NCERT)
(ii) f(x) = 4x³ – 6x² – 72x + 30 (NCERT)
(iii) f(x) = 2x³ – 9x² + 12x + 15
(iv) f(x) = – 2x³ – 9x² – 12x + 1
(v) f(x) = x + \(\frac{1}{x}\), x ≠ 0
(vi) f(x) = \(\frac{4 x^2+1}{x}\), x ≠ 0
(vii) f(x) = x⁴ – 8x³ + 22x² – 24x + 21
(viii) f(x) = \(\frac{1}{4}\) x⁴ – x³ – 5x² + 24x + 12
(ix) f(x) = x⁴ – 4x
(x) f(x) = (x – 1)³ (x – 2)².
Solution:
(i) Given f(x) = 2x³ – 3x² – 36x + 7
∴ f'(x) = 6x² – 6x – 36
= 6 (x² – x – 6)
= 6 (x + 2) (x – 3)
Now f'(x) > 0
⇒ (x + 2) (x – 3) > 0
⇒ x > 3 or x < – 2
∴ f(x) is strictly increasing in (- ∞, – 2] ∪ [3, ∞)
Now f'(x) < 0
⇒ (x + 2) (x – 3) < 0
⇒ – 2 < x < 3
∴ f(x) is strictly decreasing in [- 2, 3].

(ii) Given f(x) = 4x³ – 6x² – 72x + 30
∴ f'(x) = 12x² – 12x – 72
= 12 (x² – x – 6)
= 12 (x + 2) (x – 3)
Now f'(x) > 0
⇒ (x + 2) (x – 3) > 0
⇒ x > 3 or x < – 2
∴ f(x) is stictly increasing in (- ∞, – 2] ∪ [3, ∞)
Now f'(x) < 0
⇒ (x + 2) (x – 3) < 0
⇒ – 2 < x < 3.
∴ f(x) is strictly decreasing in [- 2, 3].

(iii) Given f(x) = 2x³ – 9x² + 12x + 15
∴ f’(x) = 6x² – 18x + 12
= 6 (x² – 3x + 2)
= 6 (x – 1) (x – 2)
Now f'(x) > 0
⇒ 6 (x – 1) (x – 2) > 0
⇒ x > 2 or x < 1
∴ f(x) is stictly increasing in (- ∞, 1] ∪ [2, ∞)
Now f'(x) < 0
⇒ 6 (x – 1) (x – 2) < 0
⇒ (x – 1) (x – 2) < 0
⇒ 1 < x < 2
∴ f(x) is increasing in [1, 2].

(iv) Given f(x) = – 2x³ – 9x² – 12x + 1
∴ f’(x) = – 6x² – 18x – 12
= – 6 (x² + 3x + 2)
= – 6 (x + 1) (x + 2)
For f(x) to be increasing, we must have f'(x) > 0
∴ – 6 (x + 1) (x + 2) > 0
⇒ (x + 1) (x + 2) < 0
⇒ – 2 < x – 1
⇒ x ∈ (- 2, – 1)

signs of f’ (x) for different values of x
Thus, f(x) is decreasing on [- 2, – 1]
For f (x) to be decreasing we must have f’(x) < 0
– 6 (x + 1) (x + 2) < 0 (x + 1) (x + 2) > 0
x < – 2 or x > – 1
i.e. x ∈ (- ∞, – 2) ∪ (- 1, ∞)
Thus,f(x) is decreasing on (- ∞, – 2) ∪ (- 1, ∞).

(v) Given y = x + \(\frac{1}{x}\), x ≠ 0
∴ f'(x) = \(\frac{d y}{d x}\)
= 1 – \(\frac{1}{x^2}\)
= \(\frac{x^2-1}{x^2}\)
Now \(\frac{d y}{d x}\) > 0
⇒ \(\frac{x^2-1}{x^2}\) > 0
⇒ x² > 1
⇒ |x| > 1
⇒ x > 1 or x < – 1
⇒ f(x) is strictly increasing in (- ∞, – 1] ∪ [1, ∞).
Also f'(x) < 0
⇒ \(\frac{x^2-1}{x^2}\) < 0
⇒ x² – 1 < 0
⇒ |x| < 1
⇒ – 1 < x < 1 Thus, f(x) is strictly decreasing in [- 1, 1].

(vi) Given f(x) = \(\frac{4 x^2+1}{x}\)
= 4x + \(\frac{1}{x}\)
∴ f’(x) = 4 – \(\frac{1}{x^2}\)
= \(\frac{4 x^2-1}{x^2}\)
Now f'(x) > 0
⇒ \(\frac{4 x^2-1}{x^2}\) > 0
⇒ 4x² – 1 > 0
⇒ x² > \(\frac{1}{4}\)
⇒ |x| > \(\frac{1}{2}\)
⇒ x > \(\frac{1}{2}\) or x < – \(\frac{1}{2}\)
∴ f(x) is stictly increasing in (- ∞, – \(\frac{1}{2}\)] ∪[\(\frac{1}{2}\), ∞).
Now f'(x) < 0
⇒ \(\frac{4 x^2-1}{x^2}\) < 0
⇒ 4x² – 1 < 0
⇒ |x| < \(\frac{1}{2}\)
⇒ – \(\frac{1}{2}\) < x < \(\frac{1}{2}\)
∴ f(x) is strictly decreasing in [- \(\frac{1}{2}\), \(\frac{1}{2}\)].

(vii) Given f(x) = x⁴ – 8x³ + 22x² – 24x + 21
Diff. both sides w.r.t. x, we have
f'(x) = 4x³ – 24x² + 44x – 24
= 4 (x³ – 6x² + 11x – 6)
= 4 (x – 1) (x – 2) (x – 3)
Now f'(x) > 0
iff 4 (x – 1) (x – 2) (x – 3) > 0
⇒ (x – 1) (x – 2) (x – 3) > 0
⇒ 1 < x < 2 or x > 3
⇒ x ∈ (1, 2) ∪ (3, ∞)

Thus,f(x) is strictly increasing in [1, 2] ∪ [3, ∞)
Now f’(x) < 0 iff 4 (x – 1) (x – 2)( x – 3) < 0
iff (x- 1) (x – 2) (x – 3) < 0
⇒ x< 1 or 2 < x < 3
⇒ x ∈ (- ∞, 1) ∪ (2, 3)
Thus,f(x) is strictly decreasing in (- ∞, 1] ∪ [2, 3].

(viii) Given f(x) = \(\frac{1}{4}\) x⁴ – x³ – 5x² + 24x + 12
∴ f(x) = x³ – 3x² – 10x + 24
= (x – 2) (x² – x – 12)
= (x – 2) (x – 4) (x + 3)

Now f'(x) > 0 iff (x – 2) (x – 4) (x + 3) > 0
⇒ x ∈ (4, ∞) ∪ (- 3, 2) [method of intervals)
⇒ x ∈ (- 3, 2) ∪ (4, ∞)
Thus f is strictly increasing in [- 3, 2] ∪ (4, ∞)
⇒ x ∈ (- ∞, – 3) ∪ (2, 4)
Thus f is strictly decreasing in (- ∞, – 3] ∪ [2, 4]

(ix) Given f(x) = x⁴ – 4x
∴ f'(x) = 4x³ – 4
⇒ f'(x) = 4 (x – 1) (x² + x + 1)
= 4 (x – 1) [x² + x + \(\frac{1}{4}\) + \(\frac{3}{4}\)]
= 4 (x – 1) \(\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]\)
For f(x) to be increasing, we must have f'(x) > 0
⇒ 4 (x – 1) \(\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]\) > 0
⇒ (x – 1) > 0
[∵ \(\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]\) > 0]
⇒ x > 1
⇒ x ∈ (1, ∞)
Hence f(x) is increasing on [1, ∞).
For f(x) to be decreasing we must have f'(x) < 0
⇒ 4 (x – 1) \(\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]\) < 0
⇒ (x – 1) < 0 [∵ \(\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]\) > 0]
⇒ x < 1
⇒ x ∈ (- ∞, 1)
Thus f(x) is decreasing on (- ∞, 1].

(x) Given f(x) = (x – 1)³ (x – 2)²
∴f’(x) = (x – 1)³ 2 (x – 2) + (x – 2)² 3 (x – 1)²
= (x – 1)² (x – 2) [2 (x – 1) + 3 (x – 2)]
= (x – 1)² (x – 2) (5x – 8)
Now f’(x) = 0
⇒ (x – 1)² (x – 2) (5x – 8) = 0
⇒ x = 1, 2, \(\frac{8}{5}\)
Now f’(x) > 0
iff (x – 1)² (x – 2) (5x – 8) > 0
iff (x – 2) (5x – 8) > 0
[∵ (x – 1)² > 0 ∀ x ∈ R – {1}]
iff x > 2 or x < \(\frac{8}{5}\) and x ≠ 1
iff x ∈ (2, ∞) or x ∈ (- ∞, 1) (1, \(\frac{8}{5}\))
Thus f(x) is stictly increasing in (- ∞, 1) ∪ (1, \(\frac{8}{5}\)) ∪ [2, ∞)
Now f'(x) < 0
iff (x – 1)² (x – 2) (5x – 8) < 0
iff (x – 2) (5x – 8) < 0
[∵ (x – 1)² > 0 ∀ x ∈ R – {1}]
iff \(\frac{8}{5}\) < x < 2
iff x ∈ (\(\frac{8}{5}\), 2)
Thus f(x) is strictly decreasing in [\(\frac{8}{5}\), 2].

Question 21 (old).
(i) f(x) = x² – 4x + 6 (NCERT)
(ii) f(x) = x² + 2x – 5 (NCERT)
(iii) f(x) = 2x² – 3x (NCERT)
(iv) f(x) = 6 – 9x – x² (NCERT)
(xii) f(x) = \(\frac{1}{4}\) x⁴ + \(\frac{2}{3}\) x³ + \(\frac{5}{2}\) x² – 6x + 9
Solution:
(i) Given f(x) = x² – 4x – 6
⇒ f’(x) = 2x – 4
∴ f(x) is strictly increasing if f’(x) > 0 i.e. if x> 2
∴ f(x) is stictly increasing in [2, ∞)
Now f’(x) < 0 = 2x – 4 < 0 ⇒ x < 2
∴ f(x) is strictly decreasing in (- ∞, 2].

(ii) Given f(x) = x² + 2x – 5
∴ f’(x) = 2x + 2
f(x) > 0
⇒ 2x + 2 > 0
⇒ x > – 1
∴ f (x) is strictly increasing in [- 1, ∞)
Now f'(x) < 0
⇒ 2x + 2 < 0
⇒ x < – 1
∴ f(x) is strictly decreasing in (- ∞. – 1].

(iii) f(x) = 2x² – 3x
∴ f'(x) = 4x – 3
Now f'(x) > 0
⇒ 4x – 3 > 0
⇒ x > \(\frac{3}{4}\)
∴ f(x) is strictly incresing in [\(\frac{3}{4}\), ∞)
Now f'(x) < 0
⇒ 4x – 3 < 0
⇒ x < \(\frac{3}{4}\)
∴ f(x) is strictly incresing in (- ∞, \(\frac{3}{4}\)].

(iv) Given f(x) = 6 – 9x – x²
∴ f(x) = – 9 – 2x
Now f'(x) > 0
⇒ – 9 – 2x > 0
⇒ 2x < – 9
⇒ x < – \(\frac{9}{2}\)
∴ f(x) is strictly incresing in (- ∞, – \(\frac{9}{2}\)].
Now f'(x) < 0
⇒ – 9 – 2x < 0 ⇒ x > – \(\frac{9}{2}\)
∴ f(x) is strictly incresing in [- \(\frac{9}{2}\), ∞).

(xii) Given f(x) = \(\frac{x^4}{4}+\frac{2}{3} x^3-\frac{5}{2} x^2\) – 6x + 9
∴ f(x) = x³ + 2x² – 5x – 6
= (x – 2) (x² + 4x + 3)
= (x – 2) (x + 1) (x + 3)

signs of f(x) for different values of x.
For f(x) to be increasing, we must have
f(x) > 0
⇒ (x – 2) (x + 1) (x + 3) > 0
⇒ x ∈ (- 3, – 1) ∪ (2, ∞).
Thus,f(x) be strictly increasing on [- 3, – 1] ∪ (2, ∞).
For f(x) to be decreasing, we must have f(x) < 0
⇒ (x – 2) (x + 1) (x + 3) < 0
⇒ x ∈ (- ∞, – 3) ∪ (- 1, 2).
Thus, f(x) be strictly decreasing on [- ∞, – 3] ∪ (- 1, 2).

Question 22.
(i) Find the interval in which the function f(x) = x² e^-x is increasing. (NCERT)
(ii) Find the intervals in which the function given by f (x) = x² e^x is strictly increasing or strictly decreasing.
Solution:
(i) Given f(x) = x² e^-x ;
Diff. both sides w.r.t. x, we have
f'(x) = – x² e^-x + 2x e^-x
= e^-x (2x – x²)
= e^-x (2 – x) x
Now f'(x) ≥ 0
⇒ e^-x (2 – x) ≥ 0
⇒ (2 – x) x ≥ 0
[∵ e^-x > 0]
⇒ – (x – 2) x ≥ 0
⇒ (x – 2) x ≤ 0
⇒ 0 ≤ x ≤ 2

Thus f(x) is increasing in [0, 2].

(ii) Given f(x) = x² e^x
Diff. both sides w.r.t. x, we have
f'(x) = x²e^x + 2x e^x
= e^x x (x + 2)
Now f'(x) > 0 iff e^x x (x + 2) > 0
⇒ x (x + 2) > 0
[∵ e^-x > 0 ∀ x ∈ R]
⇒ x < – 2 or x > 0
⇒ x ∈ (- ∞, – 2) ∪ (0, ∞)

Thus f(x) is strictly increasing in (- ∞, – 2] ∪ [0, ∞)
Now f’(x) < 0 iff e^x x (x+2) < 0
iff x (x + 2) < 0
[∵ e^{– x} > 0 ∀ x ∈ R]
⇒ – 2 < x < 0
⇒ x ∈ (- 2, 0)
Thus,f(x) is strictly decreasing in [- 2, 0].

Question 23.
Find the intervals in which the function f(x) = 2 log (x – 2) – x² + 4x – 5 is strictly increasing or strictly decreasing.
Solution:
Given f(x) = 2 log (x – 2) – x² + 4x – 5,
D_f = (2, ∞)
Diff. both sides w.r.t. x, we have

⇒ f'(x) = – 2 (x -1) (x – 3) (x – 2)
[∵ (x – 2)² > 0 ∀ x ∈ R ∵ x ≠ 2]
Now f'(x) > 0
iff – 2 (x – 1) (x – 3) (x – 2) > 0
iff (x – 1) (x – 3) (x – 2) < 0
⇒ 2 < x < 3 or x < 1 ⇒ x ∈ (2, 3) ∪ (- ∞, 1)

But x > 2
Thus,f(x) is strictly increasing function in [2, 3].
Now f’(x) < 0
iff – 2 (x – 1) (x – 2) (x – 3) < 0 iff (x – 1) (x – 2) (x – 3) > 0
⇒ 1 < x < 2 or x > 3
⇒ x ∈ (1, 2) ∪ (3, ∞)
But x > 2
∴ x ∈ (3, ∞)
Hence f(x) is strictly decreasing in [3, ∞).

Question 24.
Find the intervals in which the function f given by f(x) = tan x – 4x, x ∈ (0, \(\frac{\pi}{2}\)) is
(i) strictly increasing
(ii) strictly decreasing
Solution:
(i) Given f(x) = tan x – 4x
∴ f'(x) = sec² – 4
= (sec x – 2) (sec x + 2)
Now f'(x) > 0
⇒ (sec x – 2) (sec x + 2) > 0
⇒ sec x > 2
[∵ x ∈ (0, \(\frac{\pi}{2}\))
⇒ sec x > 0]
⇒ x > \(\frac{\pi}{3}\) but x ∈ (0, \(\frac{\pi}{2}\))
⇒ \(\frac{\pi}{3}\) < x < \(\frac{\pi}{2}\)
∴ f(x) is strictly increasing in [\(\frac{\pi}{3}\), \(\frac{\pi}{2}\))
Now f'(x) < 0
⇒ (sec x – 2) (sec x + 2) < 0
⇒ sec x < 2 [∵ x ∈ (0, \(\frac{\pi}{2}\)) ∴ sec x > 0]
⇒ x < \(\frac{\pi}{3}\) but x ∈ (0, \(\frac{\pi}{2}\))
⇒ 0 < x < \(\frac{\pi}{3}\)
Thus f(x) is strictly decreasing in (0, \(\frac{\pi}{3}\)].