Exercise 6.2, Question 1
Show \(f(x) = 3x + 17\) is increasing on \(\mathbb R\).
Show solution
- \(f'(x) = 3 > 0\) for all x.
Answer: Shown
NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2
Exercise 6.2: Increasing and decreasing functions. On an interval, \(f'(x) > 0\) gives increasing and \(f'(x) < 0\) gives decreasing. Find where \(f'(x) = 0\) (or is undefined), split the line at those points, and test the sign of \(f'\) on each piece.
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Exercise 6.2 questions and solutions
Exercise 6.2, Question 2
Show \(f(x) = e^{2x}\) is increasing on \(\mathbb R\).
Show solution
- \(f'(x) = 2e^{2x} > 0\) for all x.
Answer: Shown
Exercise 6.2, Question 3
Show that \(f(x) = \sin x\) is:
(a) increasing in \(\left(0, \tfrac{\pi}{2}\right)\)
Show solution
- \(f'(x) = \cos x > 0\) there.
Answer: Shown
(b) decreasing in \(\left(\tfrac{\pi}{2}, \pi\right)\)
Show solution
- \(\cos x < 0\) there.
Answer: Shown
(c) neither increasing nor decreasing in \((0, \pi)\)
Show solution
- \(\cos x\) is positive on \(\left(0, \tfrac{\pi}{2}\right)\) and negative on \(\left(\tfrac{\pi}{2}, \pi\right)\), so f rises then falls.
Answer: Shown
Exercise 6.2, Question 4
Find the intervals where \(f(x) = 2x^2 - 3x\) is (a) increasing (b) decreasing.
Show solution
- \(f'(x) = 4x - 3\), zero at \(x = \tfrac34\).
Answer: (a) \(\left(\tfrac34, \infty\right)\) (b) \(\left(-\infty, \tfrac34\right)\)
Exercise 6.2, Question 5
Find the intervals where \(f(x) = 2x^3 - 3x^2 - 36x + 7\) is (a) increasing (b) decreasing.
Show solution
- \(f'(x) = 6x^2 - 6x - 36 = 6(x - 3)(x + 2)\).
- Positive outside \([-2, 3]\), negative between.
Answer: (a) \((-\infty, -2)\) and \((3, \infty)\) (b) \((-2, 3)\)
Exercise 6.2, Question 6
Find the intervals in which each function is strictly increasing or decreasing.
(a) \(x^2 + 2x - 5\)
Show solution
- \(f'(x) = 2x + 2\).
Answer: Decreasing on \((-\infty, -1)\), increasing on \((-1, \infty)\)
(b) \(10 - 6x - 2x^2\)
Show solution
- \(f'(x) = -6 - 4x\), zero at \(-\tfrac32\).
Answer: Increasing on \(\left(-\infty, -\tfrac32\right)\), decreasing on \(\left(-\tfrac32, \infty\right)\)
(c) \(-2x^3 - 9x^2 - 12x + 1\)
Show solution
- \[\begin{aligned}f'(x) &= -6x^2 - 18x - 12 \\ &= -6(x + 1)(x + 2)\end{aligned}\]: positive between \(-2\) and \(-1\).
Answer: Increasing on \((-2, -1)\); decreasing on \((-\infty, -2)\) and \((-1, \infty)\)
(d) \(6 - 9x - x^2\)
Show solution
- \(f'(x) = -9 - 2x\), zero at \(-\tfrac92\).
Answer: Increasing on \(\left(-\infty, -\tfrac92\right)\), decreasing on \(\left(-\tfrac92, \infty\right)\)
(e) \((x + 1)^3(x - 3)^3\)
Show solution
- \[\begin{aligned}f'(x) &= 3(x + 1)^2(x - 3)^3 + 3(x + 1)^3(x - 3)^2 \\ &= 6(x + 1)^2(x - 3)^2(x - 1)\end{aligned}\]
- The squares are non-negative, so the sign is that of \(x - 1\) (zero only at \(-1\), 1, 3).
Answer: Increasing on \((1, 3)\) and \((3, \infty)\), i.e. on \((1, \infty)\); decreasing on \((-\infty, -1)\) and \((-1, 1)\), i.e. on \((-\infty, 1)\)
Exercise 6.2, Question 7
Show \(y = \log(1 + x) - \dfrac{2x}{2 + x}\), \(x > -1\), is increasing throughout its domain.
Show solution
- \[\begin{aligned}y' &= \dfrac{1}{1 + x} - \dfrac{2(2 + x) - 2x}{(2 + x)^2} \\ &= \dfrac{1}{1 + x} - \dfrac{4}{(2 + x)^2} \\ &= \dfrac{(2 + x)^2 - 4(1 + x)}{(1 + x)(2 + x)^2} \\ &= \dfrac{x^2}{(1 + x)(2 + x)^2}\end{aligned}\]
- For \(x > -1\) this is \(\ge 0\), zero only at \(x = 0\), so y is increasing.
Answer: Shown
Exercise 6.2, Question 8
For which x is \(y = [x(x - 2)]^2\) increasing?
Show solution
- \[\begin{aligned}y' &= 2x(x - 2)(2x - 2) \\ &= 4x(x - 1)(x - 2)\end{aligned}\]
- Sign chart through 0, 1, 2: positive on \((0, 1)\) and \((2, \infty)\).
Answer: \(0 < x < 1\) and \(x > 2\)
Exercise 6.2, Question 9
Prove \(y = \dfrac{4\sin\theta}{2 + \cos\theta} - \theta\) is increasing in \(\left[0, \tfrac{\pi}{2}\right]\).
Show solution
- \[\begin{aligned}y' &= \dfrac{4\cos\theta(2 + \cos\theta) + 4\sin^2\theta}{(2 + \cos\theta)^2} - 1 \\ &= \dfrac{8\cos\theta + 4 - (2 + \cos\theta)^2}{(2 + \cos\theta)^2} \\ &= \dfrac{\cos\theta(4 - \cos\theta)}{(2 + \cos\theta)^2}\end{aligned}\]
- On \(\left[0, \tfrac{\pi}{2}\right]\), \(\cos\theta \ge 0\) and \(4 - \cos\theta > 0\), so \(y' \ge 0\) (zero only at \(\tfrac{\pi}{2}\)).
Answer: Proved
Exercise 6.2, Question 10
Prove that \(\log x\) is increasing on \((0, \infty)\).
Show solution
- \((\log x)' = \dfrac1x > 0\) for \(x > 0\).
Answer: Proved
Exercise 6.2, Question 11
Prove \(f(x) = x^2 - x + 1\) is neither strictly increasing nor decreasing on \((-1, 1)\).
Show solution
- \(f'(x) = 2x - 1\): negative on \(\left(-1, \tfrac12\right)\), positive on \(\left(\tfrac12, 1\right)\).
Answer: Proved: f falls then rises on \((-1, 1)\)
Exercise 6.2, Question 12
Which are decreasing on \(\left(0, \tfrac{\pi}{2}\right)\)? (A) \(\cos x\) (B) \(\cos 2x\) (C) \(\cos 3x\) (D) \(\tan x\)
Show solution
- (A) \(-\sin x < 0\): decreasing. (B) \(-2\sin 2x < 0\) since \(2x \in (0, \pi)\): decreasing.
- (C) \(-3\sin 3x > 0\) when \(3x \in \left(\pi, \tfrac{3\pi}{2}\right)\): not decreasing throughout. (D) \(\sec^2 x > 0\): increasing.
Answer: (A) and (B)
Exercise 6.2, Question 13
On which interval is \(f(x) = x^{100} + \sin x - 1\) decreasing? (A) \((0, 1)\) (B) \(\left(\tfrac{\pi}{2}, \pi\right)\) (C) \(\left(0, \tfrac{\pi}{2}\right)\) (D) none of these
Show solution
- \(f'(x) = 100x^{99} + \cos x\).
- On \((0, 1)\) and \(\left(0, \tfrac{\pi}{2}\right)\): \(\cos x > 0\) (as \(1 < \tfrac{\pi}{2}\)), so \(f' > 0\). On \(\left(\tfrac{\pi}{2}, \pi\right)\): \(100x^{99} > 100\) while \(\cos x \ge -1\), so \(f' > 0\).
Answer: (D)
Exercise 6.2, Question 14
For what a is \(f(x) = x^2 + ax + 1\) increasing on \([1, 2]\)?
Show solution
- \(f'(x) = 2x + a\) is smallest at \(x = 1\) on the interval: need \(2 + a \ge 0\).
- If \(a > -2\), \(f' > 0\) on all of \([1, 2]\). If \(a = -2\), \(f'(x) = 2(x - 1)\) is 0 only at the end point \(x = 1\) and positive on \((1, 2]\), so f is still increasing on \([1, 2]\). If \(a < -2\), \(f' < 0\) near \(x = 1\).
Answer: \(a \ge -2\)
Exercise 6.2, Question 15
I is any interval disjoint from \([-1, 1]\). Prove \(f(x) = x + \dfrac1x\) is increasing on I.
Show solution
- \[\begin{aligned}f'(x) &= 1 - \dfrac{1}{x^2} \\ &= \dfrac{x^2 - 1}{x^2}\end{aligned}\]
- Every x in I has \(|x| > 1\), so \(x^2 - 1 > 0\) and \(f'(x) > 0\).
Answer: Proved
Exercise 6.2, Question 16
Prove \(f(x) = \log\sin x\) is increasing on \(\left(0, \tfrac{\pi}{2}\right)\) and decreasing on \(\left(\tfrac{\pi}{2}, \pi\right)\).
Show solution
- \(f'(x) = \dfrac{\cos x}{\sin x} = \cot x\): positive on \(\left(0, \tfrac{\pi}{2}\right)\), negative on \(\left(\tfrac{\pi}{2}, \pi\right)\).
Answer: Proved
Exercise 6.2, Question 17
Prove \(f(x) = \log|\cos x|\) is decreasing on \(\left(0, \tfrac{\pi}{2}\right)\) and increasing on \(\left(\tfrac{3\pi}{2}, 2\pi\right)\).
Show solution
- On both intervals \(\cos x > 0\), so \(f(x) = \log\cos x\) and \(f'(x) = -\tan x\).
- \(\tan x > 0\) on \(\left(0, \tfrac{\pi}{2}\right)\): \(f' < 0\). \(\tan x < 0\) on \(\left(\tfrac{3\pi}{2}, 2\pi\right)\): \(f' > 0\).
Answer: Proved
Exercise 6.2, Question 18
Prove \(f(x) = x^3 - 3x^2 + 3x - 100\) is increasing on \(\mathbb R\).
Show solution
- \(f'(x) = 3x^2 - 6x + 3 = 3(x - 1)^2 \ge 0\), zero only at \(x = 1\).
Answer: Proved
Exercise 6.2, Question 19
\(y = x^2e^{-x}\) is increasing on: (A) \((-\infty, \infty)\) (B) \((-2, 0)\) (C) \((2, \infty)\) (D) \((0, 2)\)
Show solution
- \(y' = e^{-x}(2x - x^2) = x(2 - x)e^{-x}\), positive exactly on \((0, 2)\).
Answer: (D)
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