Exercise 17.2 Set 1 in Chapter 17 of RD Sharma's Class 12 mathematics textbook delves into the concepts of increasing and decreasing functions. This section typically covers problems that require students to analyze functions, determine their intervals of increase and decrease, find critical points, and apply these concepts to solve real-world problems.
Question 1. Find the intervals in which the following functions are increasing or decreasing.
(i) f(x) = 10 – 6x – 2x2
Solution:
We are given,
f(x) = 10 – 6x – 2x2
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(10 – 6x – 2x^2) f'(x) = 0 – 6 – 4x
f'(x) = – 6 – 4x
For f(x) to be increasing, we must have,
=> f'(x) > 0
=> – 6 – 4x > 0
=> – 4x > 6
=> x < –6/4
=> x < –3/2
=> x ∈ (–∞, –3/2)
For f(x) to be decreasing, we must have,
=> f'(x) < 0
=> – 6 – 4x < 0
=> – 4x < 6
=> x > –6/4
=> x > –3/2
=> x ∈ ( –3/2, ∞)
Thus, f(x) is increasing on the interval x ∈ (–∞, –3/2) and decreasing on the interval x ∈ ( –3/2, ∞).
(ii) f(x) = x2 + 2x – 5
Solution:
We are given,
f(x) = x2 + 2x – 5
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(x^2 + 2x – 5) f'(x) = 2x + 2 – 0
f'(x) = 2x + 2
For f(x) to be increasing, we must have,
=> f'(x) > 0
=> 2x + 2 > 0
=> 2x > –2
=> x > –2/2
=> x > –1
=> x ∈ (–1, ∞)
For f(x) to be decreasing, we must have,
=> f'(x) < 0
=> 2x + 2 < 0
=> 2x < –2
=> x < –2/2
=> x < –1
=> x ∈ (–∞, –1)
Thus, f(x) is increasing on the interval x ∈ (–1, ∞) and decreasing on the interval x ∈ ( –∞, –1).
(iii) f(x) = 6 – 9x – x2
Solution:
We are given,
f(x) = 6 – 9x – x2
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(6 – 9x – x^2) f'(x) = 0 – 9 – 2x
f'(x) = – 9 – 2x
For f(x) to be increasing, we must have,
=> f'(x) > 0
=> –9 – 2x > 0
=> –2x > 9
=> x > –9/2
=> x ∈ (–9/2, ∞)
For f(x) to be decreasing, we must have,
=> f'(x) < 0
=> –9 – 2x < 0
=> –2x < 9
=> x < –9/2
=> x ∈ (–∞, –9/2)
Thus, f(x) is increasing on the interval x ∈ (–9/2, ∞) and decreasing on the interval x ∈ ( –∞, –9/2).
(iv) f(x) = 2x3 – 12x2 + 18x + 15
Solution:
We are given,
f(x) = 2x3 – 12x2 + 18x + 15
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(2x^3 – 12x^2 + 18x + 15) f'(x) = 6x2 – 24x + 18 + 0
f'(x) = 6x2 – 24x + 18
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> 6x2 – 24x + 18 = 0
=> 6 (x2 – 4x + 3) = 0
=> x2 – 4x + 3 = 0
=> x2 – 3x – x + 3 = 0
=> x (x – 3) – 1 (x – 3) = 0
=> (x – 1) (x – 3) = 0
=> x = 1, 3
Clearly, f'(x) > 0 if x < 1 and x > 3.
Also, f'(x) < 0, if 1 < x < 3.
Thus, f(x) is increasing on the interval x ∈ (–∞, 1)∪ (3, ∞) and decreasing on the interval x ∈ (1, 3).
(v) f(x) = 5 + 36x + 3x2 – 2x3
Solution:
We are given,
f(x) = 5 + 36x + 3x2 – 2x3
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(5 + 36x + 3x^2 – 2x^3) f'(x) = 0 + 36 + 6x – 6x2
f'(x) = 36 + 6x – 6x2
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> 36 + 6x – 6x2 = 0
=> 6 (– x2 + x + 6) = 0
=> 6 (–x2 + 3x – 2x + 6) = 0
=> –x2 + 3x – 2x + 6 = 0
=> x2 – 3x + 2x – 6 = 0
=> (x – 3) (x + 2) = 0
=> x = 3, – 2
Clearly, f’(x) > 0 if –2 < x < 3.
Also f’(x) < 0 if x < –2 and x > 3.
Thus, f(x) is increasing on x ∈ (–2, 3) and f(x) is decreasing on interval x ∈ (–∞, –2) ∪ (3, ∞).
(vi) f(x) = 8 + 36x + 3x2 – 2x3
Solution:
We are given,
f(x) = 8 + 36x + 3x2 – 2x3
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(8 + 36x + 3x^2 – 2x^3) f'(x) = 0 + 36 + 6x – 6x2
f'(x) = 36 + 6x – 6x2
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> 36 + 6x – 6x2 = 0
=> 6 (– x2 + x + 6) = 0
=> 6 (–x2 + 3x – 2x + 6) = 0
=> –x2 + 3x – 2x + 6 = 0
=> x2 – 3x + 2x – 6 = 0
=> (x – 3) (x + 2) = 0
=> x = 3, –2
Clearly, f’(x) > 0 if –2 < x < 3.
Also f’(x) < 0 if x < –2 and x > 3.
Thus, f(x) is increasing on x ∈ (–2, 3) and f(x) is decreasing on interval x ∈ (–∞, –2) ∪ (3, ∞).
(vii) f(x) = 5x3 – 15x2 – 120x + 3
Solution:
We are given,
f(x) = 5x3 – 15x2 – 120x + 3
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(5x^3 – 15x^2 – 120x + 3) f'(x) = 15x2 – 30x – 120 + 0
f'(x) = 15x2 – 30x – 120
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> 15x2 – 30x – 120 = 0
=> 15(x2 – 2x – 8) = 0
=> 15(x2 – 4x + 2x – 8) = 0
=> x2 – 4x + 2x – 8 = 0
=> (x – 4) (x + 2) = 0
=> x = 4, –2
Clearly, f’(x) > 0 if x < –2 and x > 4.
Also f’(x) < 0 if –2 < x < 4.
Thus, f(x) is increasing on x ∈ (–∞,–2) ∪ (4, ∞) and f(x) is decreasing on interval x ∈ (–2, 4).
(viii) f(x) = x3 – 6x2 – 36x + 2
Solution:
We are given,
f(x) = x3 – 6x2 – 36x + 2
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(x^3 – 6x^2 – 36x + 2) f'(x) = 3x2 – 12x – 36 + 0
f'(x) = 3x2 – 12x – 36
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> 3x2 – 12x – 36 = 0
=> 3(x2 – 4x – 12) = 0
=> 3(x2 – 6x + 2x – 12) = 0
=> x2 – 6x + 2x – 12 = 0
=> (x – 6) (x + 2) = 0
=> x = 6, –2
Clearly, f’(x) > 0 if x < –2 and x > 6.
Also f’(x) < 0 if –2< x < 6
Thus, f(x) is increasing on x ∈ (–∞,–2) ∪ (6, ∞) and f(x) is decreasing on interval x ∈ (–2, 6).
(ix) f(x) = 2x3 – 15x2 + 36x + 1
Solution:
We are given,
f(x) = 2x3 – 15x2 + 36x + 1
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(2x^3 – 15x^2 + 36x + 1) f'(x) = 6x2 – 30x + 36 + 0
f'(x) = 6x2 – 30x + 36
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> 6x2 – 30x + 36 = 0
=> 6 (x2 – 5x + 6) = 0
=> 6(x2 – 3x – 2x + 6) = 0
=> x2 – 3x – 2x + 6 = 0
=> (x – 3) (x – 2) = 0
=> x = 3, 2
Clearly, f’(x) > 0 if x < 2 and x > 3.
Also f’(x) < 0 if 2 < x < 3.
Thus, f(x) is increasing on x ∈ (–∞, 2) ∪ (3, ∞) and f(x) is decreasing on interval x ∈ (2, 3).
(x) f(x) = 2x3 + 9x2 + 12x + 1
Solution:
We are given,
f(x) = 2x3 + 9x2 + 12x + 1
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(2x^3 + 9x^2 + 12x + 1) f'(x) = 6x2 + 18x + 12 + 0
f'(x) = 6x2 + 18x + 12
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> 6x2 + 18x + 12 = 0
=> 6 (x2 + 3x + 2) = 0
=> 6(x2 + 2x + x + 2) = 0
=> x2 + 2x + x + 2 = 0
=> (x + 2) (x + 1) = 0
=> x = –1, –2
Clearly, f’(x) > 0 if –2 < x < –1.
Also f’(x) < 0 if x < –1 and x > –2.
Thus, f(x) is increasing on x ∈ (–2,–1) and f(x) is decreasing on interval x ∈ (–∞, –2) ∪ (–2, ∞).
(xi) f(x) = 2x3 – 9x2 + 12x – 5
Solution:
We are given,
f(x) = 2x3 – 9x2 + 12x – 5
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}( 2x^3 – 9x^2 + 12x – 5) f'(x) = 6x2 – 18x + 12 – 0
f'(x) = 6x2 – 18x + 12
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> 6x2 – 18x + 12 = 0
=> 6 (x2 – 3x + 2) = 0
=> 6(x2 – 2x – x + 2) = 0
=> x2 – 2x – x + 2 = 0
=> (x – 2) (x – 1) = 0
=> x = 1, 2
Clearly, f’(x) > 0 if x < 1 and x > 2.
Also f’(x) < 0 if 1 < x < 2.
Thus, f(x) is increasing on x ∈ (–∞, 1) ∪ (2, ∞) and f(x) is decreasing on interval x ∈ (1, 2).
(xii) f(x) = 6 + 12x + 3x2 – 2x3
Solution:
We are given,
f(x) = 6 + 12x + 3x2 – 2x3
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(6 + 12x + 3x^2 – 2x^3) f'(x) = 0 + 12 + 6x – 6x2
f'(x) = 12 + 6x – 6x2
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> 12 + 6x – 6x2 = 0
=> 6 (–x2 + x + 2) = 0
=> x2 – x – 2 = 0
=> x2 – 2x + x – 2 = 0
=> (x – 2) (x + 1) = 0
=> x = 2, –1
Clearly, f’(x) > 0 if –1 < x < 2.
Also f’(x) < 0 if x < –1 and x > 2.
Thus, f(x) is increasing on x ∈ (–1, 2) and f(x) is decreasing on interval x ∈ (–∞, –1) ∪ (2, ∞).
(xiii) f(x) = 2x3 – 24x + 107
Solution:
We are given,
f(x) = 2x3 – 24x + 107
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(2x^3 – 24x + 107) f'(x) = 6x2 – 24 + 0
f'(x) = 6x2 – 24
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> 6x2 – 24 = 0
=> 6x2 = 24
=> x2 = 4
=> x = 2, –2
Clearly, f’(x) > 0 if x < –2 and x > 2.
Also f’(x) < 0 if –2 < x < 2.
Thus, f(x) is increasing on x ∈ (–∞, –2) ∪ (2, ∞), and f(x) is decreasing on interval x ∈ (–2, 2).
(xiv) f(x) = –2x3 – 9x2 – 12x + 1
Solution:
We are given,
f(x) = –2x3 – 9x2 – 12x + 1
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(–2x^3 – 9x^2 – 12x + 1) f'(x) = –6x2 – 18x – 12 + 0
f'(x) = –6x2 – 18x – 12
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> –6x2 – 18x – 12 = 0
=> 6 (–x2 – 3x – 2) = 0
=> x2 + 3x + 2 = 0
=> x2 + 2x + x + 2 = 0
=> (x + 2) (x + 1) = 0
=> x = –2, –1
Clearly, f’(x) > 0 if x < –1 and x > –2.
Also, f’(x) < 0 if –2 < x < –1.
Thus, f(x) is increasing on x ∈ (–2, –1) and f(x) is decreasing on interval x ∈ (–∞, –2) ∪ (–1, ∞).
(xv) f(x) = (x – 1) (x – 2)2
Solution:
We are given,
f(x) = (x – 1) (x – 2)2
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}((x – 1) (x – 2)^2) f'(x) = (x – 2)2 + 2 (x – 1) (x – 2)
f'(x) = (x – 2) (x – 2 + 2x – 2)
f'(x) = (x – 2) (3x – 4)
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> (x – 2) (3x – 4) = 0
=> x = 2, 4/3
Clearly, f’(x) > 0 if x < 4/3 and x > 2.
Also, f’(x) < 0 if 4/3 < x < 2.
Thus, f(x) is increasing on x ∈ (–∞, 4/3) ∪ (2, ∞) and f(x) is decreasing on interval x ∈ (4/3, 2).
(xvi) f(x) = x3 – 12x2 + 36x + 17
Solution:
We are given,
f(x) = x3 – 12x2 + 36x + 17
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(x^3 – 12x^2 + 36x + 17) f'(x) = 3x2 – 24x + 36 + 0
f'(x) = 3x2 – 24x + 36
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> 3x2 – 24x + 36 = 0
=> 3 (x2 – 8x + 12) = 0
=> x2 – 8x + 12 = 0
=> x2 – 6x – 2x + 12 = 0
=> (x – 6) (x – 2) = 0
=> x = 6, 2
Clearly, f’(x) > 0 if x < 2 and x > 6.
Also, f’(x) < 0 if 2 < x < 6.
Thus, f(x) is increasing on x ∈ (–∞, 2) ∪ (6, ∞) and f(x) is decreasing on interval x ∈ (2, 6).
(xvii) f(x) = 2x3 – 24x + 7
Solution:
We are given,
f(x) = 2x3 – 24x + 7
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(2x^3 – 24x + 7) f'(x) = 6x2 – 24 + 0
f'(x) = 6x2 – 24
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> 6x2 – 24 = 0
=> 6x2 = 24
=> x2 = 4
=> x = 2, –2
Clearly, f’(x) > 0 if x < –2 and x > 2.
Also f’(x) < 0 if –2 < x < 2.
Thus, f(x) is increasing on x ∈ (–∞, –2) ∪ (2, ∞), and f(x) is decreasing on interval x ∈ (–2, 2).
(xviii) f(x) = 3x4/10 - 4x3/5 -3x2 + 36x/5 + 11
Solution:
We are given,
f(x) = 3x4/10 - 4x3/5 -3x2 + 36x/5 + 11
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(\frac{3}{10}x^4-\frac{4}{5}x^3-3x^2+\frac{36}{5}x+11) f'(x) = 6x3/5 - 12x2/5 -3(2x) + 36/5
f'(x) = 6/5[(x - 1)(x + 2)(x - 3)]
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> 6/5[(x - 1)(x + 2)(x - 3)] = 0
=> x = 1, –2, 3
Clearly, f’(x) > 0 if –2 < x < 1 and if x > 3
Also f’(x) < 0 if 1 < x < 3.
Thus, f(x) is increasing on x ∈ (3, ∞) and f(x) is decreasing on interval x ∈ (1, 3).
(xix) f(x) = x4 – 4x
Solution:
We are given,
f(x) = x4 – 4x
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(x^4 – 4x) f'(x) = 4x3 – 4
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> 4x3 – 4 = 0
=> 4 (x3 – 1) = 0
=> x3 – 1 = 0
=> x3 = 1
=> x = 1
Clearly, f’(x) > 0 if x > 1.
Also f’(x) < 0 if x < 1.
Thus, f(x) is increasing on x ∈ (1, ∞), and f(x) is decreasing on interval x ∈ (–∞, 1).
(xx) f(x) = x4/4 + 2/3x3 - 5/2x2 - 6x + 7
Solution:
We have,
f(x) = x4/4 + 2/3x3 - 5/2x2 - 6x + 7
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(\frac{x^4}{4}+\frac{2}{3}x^3-\frac{5}{2}x^2-6x+7) f'(x) = 4x3/4 + 6x2/3 - 10x/2 - 6 + 0
f'(x) = x3 + 2x2 – 5x – 6
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> x3 + 2x2 – 5x – 6 = 0
=> (x + 1) (x + 3) (x – 2) = 0
=> x = –1, –3, 2
Clearly f'(x) > 0 if –3 < x < –1 and x > 2.
Also f'(x) < 0 if x < –3 and –1 < x < 2.
Thus, f(x) is increasing on x ∈ (–3, –1) ∪ (2, ∞) and f(x) is decreasing on interval x ∈ (–∞, –3) ∪ (–1, 2).
(xxi) f(x) = x4 – 4x3 + 4x2 + 15
Solution:
We have,
f(x) = x4 – 4x3 + 4x2 + 15
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(x^4 – 4x^3 + 4x^2 + 15) f'(x) = 4x3 – 12x2 + 8x + 0
f'(x) = 4x3 – 12x2 + 8x
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> 4x3 – 12x2 + 8x = 0
=> 4x (x2 – 3x + 2) = 0
=> 4x (x – 2) (x – 1) = 0
=> x = 0, 1, 2
Clearly f'(x) > 0 if 0 < x < 1 and x > 2.
Also f'(x) < 0 if x < 0 and 1 < x < 2.
Thus, f(x) is increasing on x ∈ (0, 1) ∪ (2, ∞) and f(x) is decreasing on interval x ∈ (–∞, 0) ∪ (1, 2).
(xxii) f(x) = 5x^{\frac{3}{2}}-3x^\frac{5}{2} , x > 0
Solution:
We have,
f(x) =
5x^{\frac{3}{2}}-3x^\frac{5}{2} On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(5x^{\frac{3}{2}}-3x^\frac{5}{2}) f'(x) =
\frac{15}{2}x^{\frac{1}{2}}-\frac{15}{2}x^{\frac{3}{2}} For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=>
\frac{15}{2}x^{\frac{1}{2}}-\frac{15}{2}x^{\frac{3}{2}} = 0=>
x^{\frac{1}{2}}-x^{\frac{3}{2}} = 0=> x1/2(1 - x) = 0
=> x = 0, 1
Clearly f'(x) > 0 if 0 < x < 1.
Also f'(x) < 0 if x > 0.
Thus, f(x) is increasing on x ∈ (0, 1) and f(x) is decreasing on interval x ∈ (1, ∞).
(xxiii) f(x) = x8 + 6x2
Solution:
We have,
f(x) = x8 + 6x2
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(x^8 + 6x^2) f'(x) = 8x7 + 12x
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> 8x7 + 12x = 0
=> 4x (2x6 + 3) = 0
=> x = 0
Clearly f'(x) > 0 if x > 0.
Also f'(x) < 0 if x < 0.
Thus, f(x) is increasing on x ∈ (0, ∞) and f(x) is decreasing on interval x ∈ (–∞, 0).
(xxiv) f(x) = x3 – 6x2 + 9x + 15
Solution:
We are given,
f(x) = x3 – 6x2 + 9x + 15
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(x^3 – 6x^2 + 9x + 15) f'(x) = 3x2 – 12x + 9 + 0
f'(x) = 3x2 – 12x + 9
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> 3x2 – 12x + 9 = 0
=> 3 (x2 – 4x + 3) = 0
=> x2 – 4x + 3 = 0
=> x2 – 3x – x + 3 = 0
=> (x – 3) (x – 1) = 0
=> x = 3, 1
Clearly f'(x) > 0 if x < 1 and x > 3.
Also f'(x) < 0 if 1 < x < 3.
Thus, f(x) is increasing on x ∈ (–∞, 1) ∪ (3, ∞) and f(x) is decreasing on interval x ∈ (1, 3).
(xxv) f(x) = [x(x – 2)]2
Solution:
We are given,
f(x) = [x(x – 2)]2
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}[x(x – 2)]^2 f'(x) =
\frac{d}{dx}(x^2-2x)^2 f'(x) = 2 (x2 – 2x) (2x – 2)
f'(x) = 4x (x – 2) (x – 1)
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> 4x (x – 2) (x – 1) = 0
=> x = 0, 1, 2
Clearly f'(x) > 0 if 0 < x < 1 and x > 2.
Also f'(x) < 0 if x < 0 and 1< x < 2.
Thus, f(x) is increasing on x ∈ (0, 1) ∪ (2, ∞) and f(x) is decreasing on interval x ∈ (–∞, 0) ∪ (1, 2).
(xxvi) f(x) = 3x4 – 4x3 – 12x2 + 5
Solution:
We are given,
f(x) = 3x4 – 4x3 – 12x2 + 5
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(3x^4 – 4x^3 – 12x^2 + 5) f'(x) = 12x3 – 12x2 – 24x
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> 12x3 – 12x2 – 24x = 0
=> 12x (x2 – x – 2) = 0
=> 12x (x + 1) (x – 2) = 0
=> x = 0, –1, 2
Clearly f'(x) > 0 if –1 < x < 0 and x > 2.
Also f'(x) < 0 if x < –1 and 0< x < 2.
Thus, f(x) is increasing on x ∈ (–1, 0) ∪ (2, ∞) and f(x) is decreasing on interval x ∈ (–∞, –1) ∪ (0, 2).
(xxvii) f(x) = 3x4/2 - 4x3 - 45x2 + 51
Solution:
We have,
f(x) = 3x4/2 - 4x3 - 45x2 + 51
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(\frac{3}{2}x^4-4x^3-45x^2+51) f'(x) = 6x3 – 12x2 – 90x
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> 6x3 – 12x2 – 90x = 0
=> 6x (x2 – 2x – 15) = 0
=> 6x (x + 3) (x – 5) = 0
=> x = 0, –3, 5
Clearly f'(x) > 0 if –3 < x < 0 and x > 5.
Also f'(x) < 0 if x < –3 and 0< x < 5.
Thus, f(x) is increasing on x ∈ (–3, 0) ∪ (5, ∞) and f(x) is decreasing on interval x ∈ (–∞, –3) ∪ (0, 5).
(xxvii) f(x) = \log(2+x)-\frac{2x}{2+x}
Solution:
We have,
f(x) =
\log(2+x)-\frac{2x}{2+x} On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(\log(2+x)-\frac{2x}{2+x}) f'(x) =
\frac{1}{2+x}-\frac{(2+x)2-2x}{(2+x)^2} f'(x) =
\frac{1}{2+x}-\frac{4+2x-2x}{(2+x)^2} f'(x) =
\frac{1}{2+x}-\frac{4}{(2+x)^2} f'(x) =
\frac{2+x-4}{(2+x)^2} f'(x) =
\frac{x-2}{(2+x)^2} Clearly f'(x) > 0 if x > 2.
Also f'(x) < 0 if x < 2
Thus, f(x) is increasing on x ∈ (2, ∞) and f(x) is decreasing on interval x ∈ (–∞, 2).
Question 2. Determine the values of x for which the function f(x) = x2 – 6x + 9 is increasing or decreasing. Also, find the coordinates of the point on the curve y = x2 – 6x + 9 where the normal is parallel to the line y = x + 5.
Solution:
Given f(x) = x2 – 6x + 9
On differentiating both sides with respect to x, we get
=> f’(x) = 2x – 6
=> f’(x) = 2(x – 3)
For f(x), we need to find the critical point, so we get,
=> f’(x) = 0
=> 2(x – 3) = 0
=> (x – 3) = 0
=> x = 3
Clearly, f’(x) > 0 if x > 3.
Also f’(x) < 0 if x < 3.
Thus, f(x) is increasing on (3, ∞) and f(x) is decreasing on interval x ∈ (–∞, 3).
Equation of the given curve is f(x) = x2 – 6x + 9.
Slope of this curve is given by,
=> m1 = dy/dx
=> m1 = 2x – 6
And slope of the line is y = x + 5
Slope of this curve is given by,
=> m2 = dy/dx
=> m2 = 1
Now according to the question,
=> m1m2 = –1
=> 2x – 6 = –1
=> 2x = 5
=> x = 5/2
Putting x = 5/2 in the curve y = x2 – 6x + 9, we get,
=> y = (5/2)2 – 6 (5/2) + 9
=> y = 25/4 – 15 + 9
=> y = 1/4
Therefore, the required coordinates are (5/2, 1/4).
Question 3. Find the intervals in which f(x) = sin x – cos x, where 0 < x < 2π is increasing or decreasing.
Solution:
We have,
f(x) = sin x – cos x
On differentiating both sides with respect to x, we get
f'(x) = \frac{d}{dx}(sin x – cos x)
f'(x) = cos x + sin x
For f(x), we need to find the critical point, so we get,
=> f'(x) = 0
=> cos x + sin x = 0
=> 1 + tan x = 0
=> tan x = –1
=> x = 3π/4 , 7π/4
Clearly f'(x) > 0 if 0 < x < 3π/4 and 7π/4 < x < 2π.
Also f'(x) < 0 if 3π/4 < x < 7π/4.
Thus, f(x) is increasing on x ∈ (0, 3π/4) ∪ (7π/4, 2π) and f(x) is decreasing on interval x ∈ (3π/4, 7π/4).
Question 4. Show that f(x) = e2x is increasing on R.
Solution:
We have,
=> f(x) = e2x
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(e^{2x}) f'(x) = 2e2x
For f(x) to be increasing, we must have
=> f’(x) > 0
=> 2e2x > 0
=> e2x > 0
Now we know, the value of e lies between 2 and 3. Therefore, f(x) will be always greater than zero.
Thus, f(x) is increasing on interval R.
Hence proved.
Question 5. Show that f(x) = e1/x, x ≠ 0 is a decreasing function for all x ≠ 0.
Solution:
We have,
=> f(x) = e1/x
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(e^{\frac{1}{x}}) f'(x) = -ex/x2
As x ∈ R, we have,
=> ex > 0
Also, we get,
=> 1/x2 > 0
This means, ex/x2 > 0
=> -ex/x2 < 0
Thus, f(x) is a decreasing function for all x ≠ 0.
Hence proved.
Question 6. Show that f(x) = loga x, 0 < a < 1 is a decreasing function for all x > 0.
Solution:
We have,
=> f(x) = loga x
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(log_a x) f'(x) = 1/xloga
As we are given 0 < a < 1,
=> log a < 0
And for x > 0, 1/x > 0
Therefore, f'(x) is,
=> 1/xloga < 0
=> f'(x) < 0
Thus, f(x) is a decreasing function for all x > 0.
Hence proved.
Question 7. Show that f(x) = sin x is increasing on (0, π/2) and decreasing on (π/2, π) and neither increasing nor decreasing in (0, π).
Solution:
We have,
f(x) = sin x
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(\sin x) f'(x) = cos x
Now for 0 < x < π/2,
=> cos x > 0
=> f'(x) > 0
And for π/2 < x < π,
=> cos x < 0
=> f'(x) < 0
Thus, f(x) is increasing on x ∈ (0, π/2) and f(x) is decreasing on interval x ∈ (π/2, π).
Hence f(x) is neither increasing nor decreasing in (0, π).
Hence proved.
Question 8. Show that f(x) = log sin x is increasing on (0, π/2) and decreasing on (π/2, π).
Solution:
We have,
f(x) = log sin x
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(\log\sin x) f'(x) = (1/sinx)cosx
f'(x) = cot x
Now for 0 < x < π/2,
=> cot x > 0
=> f'(x) > 0
And for π/2 < x < π,
=> cos x < 0
=> f'(x) < 0
Thus, f(x) is increasing on x ∈ (0, π/2) and f(x) is decreasing on interval x ∈ (π/2, π).
Hence proved.
Question 9. Show that f(x) = x – sin x is increasing for all x ∈ R.
Solution:
We have,
f(x) = x – sin x
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(x – sin x) f'(x) = 1 – cos x
Now, we are given x ∈ R, we get
=> –1 < cos x < 1
=> –1 > cos x > 0
=> f’(x) > 0
Thus, f(x) is increasing on interval x ∈ R.
Hence proved.
Question 10. Show that f(x) = x3 – 15x2 + 75x – 50 is an increasing function for all x ∈ R.
Solution:
We have,
f(x) = x3 – 15x2 + 75x – 50
On differentiating both sides with respect to x, we get
f'(x) =
\frac{d}{dx}(x^3 – 15x^2 + 75x – 50) f'(x) = 3x2 – 30x + 75 – 0
f'(x) = 3x2 – 30x + 75
f’(x) = 3(x2 – 10x + 25)
f’(x) = 3(x – 5)2
Now, as we are given x ϵ R, we get
=> (x – 5)2 > 0
=> 3(x – 5)2 > 0
=> f’(x) > 0
Thus, f(x) is increasing on interval x ∈ R.
Hence proved.
Summary
Exercise 17.2 Set 1 of Chapter 17 in RD Sharma's Class 12 mathematics textbook focuses on the application of differential calculus to analyze the behavior of functions. The problems in this exercise require students to determine intervals of increase and decrease, identify critical points, and classify them as local maxima, minima, or neither. Students work with various types of functions including polynomial, trigonometric, exponential, logarithmic, and piecewise functions. This set of problems helps students develop a deeper understanding of function behavior and prepares them for more advanced topics in calculus and mathematical analysis.