Exercise 6.2 answers: Application of Derivatives

Open NCERT p. 158One way to think about it

1. f(x) = 3x + 17, defined and differentiable for every real x.
2. Differentiate: f'(x) = 3.
3. f'(x) = 3 is positive for every real x, so it never touches zero and never changes sign.
4. Since f'(x) > 0 for all x in R, f is increasing on the whole of R.

In shortf'(x) = 3 > 0 for every real x, so f is increasing on R.

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Open NCERT p. 158One way to think about it

1. f(x) = e^2x, defined and differentiable for every real x.
2. Differentiate: f'(x) = 2e^2x.
3. Since e^2x is always positive for every real x, f'(x) = 2e^2x is always positive too.
4. So f'(x) > 0 for every real x, and f is increasing on the whole of R.

In shortf'(x) = 2e^2x > 0 for every real x, so f is increasing on R.

Watch this explained “A rate that never reaches zero”, 13:38 into Using the sign of the derivative to split the line into rising and falling intervals

Open NCERT p. 158One way to think about it

(a) increasing in (0, π/2)

1. f(x) = sin x is continuous and differentiable everywhere, and f'(x) = cos x.
2. For every x in (0, π/2), cos x > 0.
3. Since f'(x) > 0 on (0, π/2), f is increasing on (0, π/2).

In shortf is increasing on (0, π/2) because f'(x) = cos x > 0 there

(b) decreasing in (π/2, π)

1. For every x in (π/2, π), cos x < 0.
2. Since f'(x) < 0 on (π/2, π), f is decreasing on (π/2, π).

In shortf is decreasing on (π/2, π) because f'(x) = cos x < 0 there

(c) neither increasing nor decreasing in (0, π)

1. Take π/6 < π/2, both in (0, π): sin(π/6) = 1/2 and sin(π/2) = 1, so the value goes up. Hence f is not decreasing on (0, π).
2. Take π/2 < 5π/6, both in (0, π): sin(π/2) = 1 and sin(5π/6) = 1/2, so the value goes down. Hence f is not increasing on (0, π).
3. This matches parts (a) and (b): f rises on (0, π/2) and falls on (π/2, π), both inside (0, π).

In shortf is neither increasing nor decreasing on (0, π): it rises from 1/2 to 1 between π/6 and π/2, and falls from 1 to 1/2 between π/2 and 5π/6

Watch this explained “Neither, on an interval”, 12:00 into What increasing and decreasing mean on an interval, before any calculus

Open NCERT p. 158Matches NCERT’s answer

(a) increasing

1. f(x) = 2x² − 3x
2. f'(x) = 4x − 3
3. For f to be increasing: f'(x) > 0 ⟹ 4x − 3 > 0 ⟹ x > 3/4
4. So f is increasing on (3/4, ∞)

Answer(3/4, ∞)

(b) decreasing

1. For f to be decreasing: f'(x) < 0 ⟹ 4x − 3 < 0 ⟹ x < 3/4
2. So f is decreasing on (−∞, 3/4)

Answer(−∞, 3/4)

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Open NCERT p. 158Matches NCERT’s answer

(a) increasing

1. Differentiate: f'(x) = 6x² − 6x − 36 = 6(x − 3)(x + 2).
2. f'(x) = 0 at x = −2 and x = 3, splitting the line into three pieces.
3. f'(x) > 0 on the outer two pieces, x < −2 or x > 3.

Answerf is increasing on (−∞, −2) and (3, ∞)

(b) decreasing

1. Using the same f'(x) = 6(x − 3)(x + 2).
2. On the middle piece, −2 < x < 3, the two factors have opposite sign, so f'(x) < 0.

Answerf is decreasing on (−2, 3)

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Open NCERT p. 158Checked by computer

(a) x² + 2x − 5

1. f(x) = x² + 2x − 5, so f'(x) = 2x + 2 = 2(x + 1).
2. f'(x) = 0 only at x = −1.
3. For x < −1, f'(x) < 0; for x > −1, f'(x) > 0.

AnswerStrictly increasing on (−1, ∞); strictly decreasing on (−∞, −1)

(b) 10 − 6x − 2x²

1. f(x) = 10 − 6x − 2x², so f'(x) = −6 − 4x = −2(2x + 3).
2. f'(x) = 0 only at x = −3/2.
3. For x < −3/2, f'(x) > 0; for x > −3/2, f'(x) < 0.

AnswerStrictly increasing on (−∞, −3/2); strictly decreasing on (−3/2, ∞)

(c) −2x³ − 9x² − 12x + 1

1. f(x) = −2x³ − 9x² − 12x + 1, so f'(x) = −6x² − 18x − 12 = −6(x + 1)(x + 2).
2. f'(x) = 0 at x = −2 and x = −1, which cut the line into three pieces.
3. For x < −2 both brackets are negative, their product is positive, so f'(x) < 0.
4. For −2 < x < −1, (x + 2) > 0 and (x + 1) < 0, so f'(x) > 0.
5. For x > −1 both brackets are positive, so f'(x) < 0.

AnswerStrictly increasing on (−2, −1); strictly decreasing on (−∞, −2) and on (−1, ∞)

(d) 6 − 9x − x²

1. f(x) = 6 − 9x − x², so f'(x) = −9 − 2x.
2. f'(x) = 0 only at x = −9/2.
3. For x < −9/2, f'(x) > 0; for x > −9/2, f'(x) < 0.

AnswerStrictly increasing on (−∞, −9/2); strictly decreasing on (−9/2, ∞)

(e) (x + 1)³ (x − 3)³

1. f(x) = (x + 1)³(x − 3)³. By the product rule, f'(x) = 3(x + 1)²(x − 3)³ + 3(x + 1)³(x − 3)².
2. Take out 3(x + 1)²(x − 3)²: f'(x) = 3(x + 1)²(x − 3)²[(x − 3) + (x + 1)] = 6(x + 1)²(x − 3)²(x − 1).
3. (x + 1)² and (x − 3)² are never negative, so f'(x) has the sign of (x − 1), except that f'(x) = 0 at x = −1 and x = 3.
4. So f'(x) < 0 for x < 1 (apart from x = −1) and f'(x) > 0 for x > 1 (apart from x = 3).
5. At x = −1 the sign of f' is negative on both sides, so f is strictly decreasing on (−∞, −1] and on [−1, 1), and these share the point −1: f is strictly decreasing on all of (−∞, 1). In the same way f is strictly increasing on all of (1, ∞), through x = 3.
6. The answer key at the back of the book splits this at x = 3 and x = -1, but the slope is 0 only at those single points and the graph keeps rising (or falling) straight through them, so the function is strictly increasing on all of (1, ∞) and strictly decreasing on all of (-∞, 1).

AnswerStrictly increasing on (1, ∞); strictly decreasing on (−∞, 1)

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Open NCERT p. 158One way to think about it

1. Given: = log(1 + x) − 2x/(2 + x), x > −.
2. y = log(1 + x) − 2x/(2 + x)
3. dy/dx = 1/(1 + x) − [2(2 + x) − 2x(1)]/(2 + x)²
4. = 1/(1 + x) − [4 + 2x − 2x]/(2 + x)²
5. = 1/(1 + x) − 4/(2 + x)²
6. = [(2 + x)² − 4(1 + x)]/[(1 + x)(2 + x)²]
7. = [4 + 4x + x² − 4 − 4x]/[(1 + x)(2 + x)²]
8. = x²/[(1 + x)(2 + x)²]
9. For x > −1: x² ≥ 0, (1 + x) > 0, and (2 + x)² > 0
10. Therefore dy/dx ≥ 0 for all x > −1
11. So y is increasing throughout its domain (−1, ∞)

In shortThe derivative dy/dx = x²/[(1 + x)(2 + x)²] is non-negative for all x > −1, so the function is increasing throughout its domain.

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Open NCERT p. 158Matches NCERT’s answer

1. y = [x(x − 2)]² = (x² − 2x)².
2. By the chain rule, dy/dx = 2(x² − 2x)(2x − 2) = 4x(x − 1)(x − 2).
3. dy/dx = 0 at x = 0, 1 and 2, which cut the line into four pieces.
4. For x < 0: (−)(−)(−) is negative, so dy/dx < 0.
5. For 0 < x < 1: (+)(−)(−) is positive, so dy/dx > 0.
6. For 1 < x < 2: (+)(+)(−) is negative, so dy/dx < 0.
7. For x > 2: (+)(+)(+) is positive, so dy/dx > 0.

Answery is an increasing function on (0, 1) and on (2, ∞)

Watch this explained “Two cuts, three pieces”, 9:39 into Using the sign of the derivative to split the line into rising and falling intervals

Open NCERT p. 158One way to think about it

1. Given: = 4 sin θ/(2 + cos θ) – θ is an increasing function of θ in [0, π/2].
2. y = 4sin θ/(2 + cos θ) − θ
3. dy/dθ = [4cos θ(2 + cos θ) − 4sin θ(−sin θ)]/(2 + cos θ)² − 1
4. = [4cos θ(2 + cos θ) + 4sin² θ]/(2 + cos θ)² − 1
5. = [8cos θ + 4cos² θ + 4sin² θ]/(2 + cos θ)² − 1
6. = [8cos θ + 4(cos² θ + sin² θ)]/(2 + cos θ)² − 1
7. = [8cos θ + 4]/(2 + cos θ)² − 1
8. = [4(2cos θ + 1) − (2 + cos θ)²]/(2 + cos θ)²
9. = [8cos θ + 4 − 4 − 4cos θ − cos² θ]/(2 + cos θ)²
10. = [4cos θ − cos² θ]/(2 + cos θ)²
11. = cos θ(4 − cos θ)/(2 + cos θ)²
12. For θ ∈ [0, π/2]: cos θ ≥ 0, and 4 − cos θ > 0
13. Therefore dy/dθ ≥ 0 on [0, π/2]
14. So y is increasing on [0, π/2]

In shortSince dy/dθ = cos θ(4 − cos θ)/(2 + cos θ)² ≥ 0 for all θ ∈ [0, π/2], the function is increasing on this interval.

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Open NCERT p. 159One way to think about it

1. Let f(x) = log x (natural logarithm)
2. f'(x) = 1/x
3. For all x ∈ (0, ∞), we have x > 0
4. Therefore f'(x) = 1/x > 0 for all x ∈ (0, ∞)
5. Since the derivative is positive throughout the domain, f is strictly increasing on (0, ∞)

In shortThe logarithmic function f(x) = log x has derivative f'(x) = 1/x > 0 for all x > 0, so it is increasing on its entire domain (0, ∞).

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Open NCERT p. 159One way to think about it

1. f'(x) = 2x − 1, which is 0 at x = 1/2, a point inside (−1, 1).
2. For −1 < x < 1/2, f'(x) < 0, so f is decreasing on (−1, 1/2); for 1/2 < x < 1, f'(x) > 0, so f is increasing on (1/2, 1).
3. Check with values: 0 < 1/2 but f(0) = 1 is greater than f(1/2) = 3/4, so f is not increasing on (−1, 1), and so not strictly increasing either.
4. Also 1/2 < 3/4 but f(1/2) = 3/4 is less than f(3/4) = 13/16, so f is not decreasing on (−1, 1).

In shortf falls on (−1, 1/2) and rises on (1/2, 1), so on (−1, 1) it is neither strictly increasing nor decreasing.

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Open NCERT p. 159Matches NCERT’s answer

1. (A) cos x: derivative −sin x, which is negative throughout (0, π/2), so cos x is decreasing.
2. (B) cos 2x: derivative −2 sin 2x; as x runs over (0, π/2), 2x runs over (0, π), where sin 2x is positive, so the derivative is negative throughout — decreasing.
3. (C) cos 3x: derivative −3 sin 3x; as x runs over (0, π/2), 3x runs over (0, 3π/2). sin 3x is positive for 3x in (0, π) — that is, x in (0, π/3) — but negative for 3x in (π, 3π/2) — that is, x in (π/3, π/2). So the derivative changes sign: cos 3x is not decreasing throughout.
4. (D) tan x: derivative sec²x, always positive, so tan x is increasing, not decreasing.

Answercos x and cos 2x are decreasing on (0, π/2)

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Open NCERT p. 159Matches NCERT’s answer

1. f'(x) = 100x⁹⁹ + cos x.
2. (A) and (C): for 0 < x < 1 and for 0 < x < π/2, x⁹⁹ > 0 and cos x > 0, so f'(x) > 0. f is increasing there, not decreasing.
3. (B): for π/2 < x < π we have x > 1, so 100x⁹⁹ > 100, while cos x ≥ −1. So f'(x) > 100 − 1 > 0, and f is increasing there too.
4. So f is decreasing on none of the three intervals.

Answer(D) None of these

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Open NCERT p. 159Checked by computer

1. f'(x) = 2x + a.
2. f is increasing on [1, 2] when f'(x) ≥ 0 for every x between 1 and 2.
3. 2x + a gets bigger as x gets bigger, so on [1, 2] its smallest value is at x = 1, namely 2 + a.
4. If a ≥ −2, then f'(x) ≥ 2 + a ≥ 0 for all x in [1, 2], so f is increasing on [1, 2].
5. If a < −2, then f'(1) = 2 + a < 0, so f'(x) < 0 for x just to the right of 1 and f falls there; then f is not increasing on [1, 2].
6. The answer key at the back of the book prints a > –2, but a = –2 works too: then f(x) = (x – 1)², which rises all the way from x = 1 to x = 2, so the answer is a ≥ –2.

Answera ≥ −2, that is, a ∈ [−2, ∞)

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Open NCERT p. 159One way to think about it

1. Differentiate: f'(x) = 1 − 1/x² = (x² − 1)/x².
2. I is disjoint from [−1, 1], so every x in I satisfies |x| > 1, which means x² > 1.
3. Then x² − 1 > 0, and x² is always positive, so f'(x) = (x² − 1)/x² > 0 for every x in I.

In shortf'(x) = (x² − 1)/x² > 0 for every x in I (since |x| > 1 there), so f is increasing on I.

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Open NCERT p. 159One way to think about it

1. Differentiate: f'(x) = cos x / sin x = cot x.
2. On (0, π/2), both sin x and cos x are positive, so cot x > 0 — f is increasing there.
3. On (π/2, π), sin x is still positive but cos x is negative, so cot x < 0 — f is decreasing there.

In shortf'(x) = cot x, which is positive on (0, π/2) and negative on (π/2, π), giving the two stated behaviours.

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Open NCERT p. 159One way to think about it

1. On (0, π/2), cos x > 0, so f(x) = log(cos x); differentiating, f'(x) = −sin x / cos x = −tan x.
2. On (0, π/2), tan x > 0, so f'(x) = −tan x < 0 — f is decreasing there.
3. On (3π/2, 2π), cos x is also positive (fourth quadrant), so again f(x) = log(cos x) and f'(x) = −tan x.
4. On (3π/2, 2π), sin x < 0 and cos x > 0, so tan x < 0, making f'(x) = −tan x > 0 — f is increasing there.

In shortf'(x) = −tan x, which is negative on (0, π/2) and positive on (3π/2, 2π), giving the two stated behaviours.

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Open NCERT p. 159One way to think about it

1. Differentiate: f'(x) = 3x² − 6x + 3 = 3(x² − 2x + 1) = 3(x − 1)².
2. A square is never negative, so f'(x) ≥ 0 for every real x, with equality only at the single point x = 1.

In shortf'(x) = 3(x − 1)² ≥ 0 for all x, so f is increasing throughout R.

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Open NCERT p. 159Matches NCERT’s answer

1. Given: = x² e⁻ˣ.
2. Differentiate using the product rule: dy/dx = 2x e⁻ˣ − x² e⁻ˣ = x e⁻ˣ (2 − x).
3. Since e⁻ˣ is always positive, the sign of dy/dx matches the sign of x(2 − x).
4. x(2 − x) is positive exactly when 0 < x < 2.

Answer(D) (0, 2)

Watch this explained “Two cuts, three pieces”, 9:39 into Using the sign of the derivative to split the line into rising and falling intervals