Miscellaneous Exercise answers: Application of Derivatives

Open NCERT p. 183One way to think about it

1. Given: = log x / x has maximum at x =.
2. f(x) = (log x)/x is defined for x > 0.
3. By the quotient rule, f'(x) = [1 − log x] / x² (the derivative of log x is 1/x, and the derivative of x is 1).
4. f'(x) = 0 when 1 − log x = 0, i.e. log x = 1, i.e. x = e.
5. Differentiating again, f''(x) = (2 log x − 3)/x³.
6. At x = e: f''(e) = (2 − 3)/e³ = −1/e³, which is negative.
7. A negative second derivative at a critical point means that point is a local maximum, so f has a maximum at x = e.
8. The maximum value is f(e) = (log e)/e = 1/e.

In shortf has a (local) maximum at x = e, and the maximum value is f(e) = 1/e — as required to prove.

Watch this explained “The rule, in three parts”, 1:00 into The quicker second-derivative test, and the case where it tells you nothing

Open NCERT p. 183Matches NCERT’s answer

1. Let each equal side have length a (changing with time) and the base be the fixed length b.
2. Drop a perpendicular from the apex to the base; it splits the base into two halves of length b/2, and by Pythagoras the height is h = √(a² − b²/4).
3. Area A = (1/2)·b·h = (b/4)·√(4a² − b²).
4. Differentiate with respect to time: dA/dt = (a·b/√(4a² − b²))·(da/dt).
5. We are given da/dt = −3 cm/s (a is decreasing), and we want dA/dt at the instant a = b.
6. At a = b: √(4a² − b²) becomes √(3b²) = b√3, so dA/dt = (b·b/(b√3))·(−3) = −√3·b.
7. The negative sign shows the area is decreasing, at the rate √3·b square units per second.

AnswerThe area is decreasing at the rate of √3·b cm² per second.

Watch this explained “An answer with a symbol in it”, 16:49 into Two quantities both changing with time, and the chain rule that links their rates

Open NCERT p. 183Checked by computerReads two ways: both answers shown

1. Since 2x + x cos x = x(2 + cos x), f(x) = 4 sin x/(2 + cos x) − x.
2. f′(x) = [4 cos x (2 + cos x) + 4 sin² x]/(2 + cos x)² − 1 = (8 cos x + 4)/(2 + cos x)² − 1.
3. = [8 cos x + 4 − (4 + 4 cos x + cos² x)]/(2 + cos x)² = cos x (4 − cos x)/(2 + cos x)².
4. 4 − cos x ≥ 3 > 0 and (2 + cos x)² > 0 for every x, so f′(x) has the same sign as cos x.
5. The question does not say which values of x to consider. NCERT's answer key works on 0 ≤ x ≤ 2π, so that reading leads.
6. On 0 ≤ x ≤ 2π: cos x > 0 for 0 ≤ x < π/2 and 3π/2 < x ≤ 2π, and cos x < 0 for π/2 < x < 3π/2. Since f′ = 0 only at the single point π/2 of the first piece, f is still increasing on the closed piece 0 ≤ x ≤ π/2.
7. For all real x: f itself does not repeat (f(x + 2π) = f(x) − 2π), but the sign of f′ follows cos x, so f increases wherever cos x > 0 and decreases wherever cos x < 0.

AnswerRead on 0 ≤ x ≤ 2π (NCERT's answer key): (i) increasing on 0 ≤ x ≤ π/2 and 3π/2 < x < 2π; (ii) decreasing on π/2 < x < 3π/2. Read for all real x: (i) increasing wherever cos x > 0, that is on (2nπ − π/2, 2nπ + π/2); (ii) decreasing wherever cos x < 0, that is on (2nπ + π/2, 2nπ + 3π/2), for every integer n.

Watch this explained “The procedure”, 7:52 into Using the sign of the derivative to split the line into rising and falling intervals

Open NCERT p. 183Checked by computer

(i) increasing

1. f(x) = x³ + x⁻³, defined for every x except 0.
2. f′(x) = 3x² − 3x⁻⁴ = 3(x⁶ − 1)/x⁴.
3. x⁴ is positive for every x ≠ 0, so f′(x) has the same sign as x⁶ − 1, which is positive exactly when x < −1 or x > 1.
4. So f′(x) > 0 on (−∞, −1) and on (1, ∞).

Answerf is increasing on (−∞, −1) and on (1, ∞).

(ii) decreasing

1. x⁶ − 1 < 0 exactly when −1 < x < 1, and x = 0 is left out because f is not defined there. So f′(x) < 0 on (−1, 0) and on (0, 1).
2. These are two separate intervals: f is decreasing on each one, but not across the gap at 0 — for example f(−1/2) = −65/8 while f(1/2) = 65/8.
3. The answer key at the back of the book prints –1 < x < 1, but the function is not defined at x = 0 and jumps from very negative to very positive there (f(–½) = –65/8 but f(½) = 65/8), so it is decreasing on –1 < x < 0 and on 0 < x < 1 separately.

Answerf is decreasing on (−1, 0) and on (0, 1).

Watch this explained “The procedure”, 7:52 into Using the sign of the derivative to split the line into rising and falling intervals

Open NCERT p. 184Matches NCERT’s answer

1. Take the major axis along the x-axis (a > b) and put the vertex at A = (a, 0).
2. The triangle is isosceles with its apex on the ellipse's line of symmetry, so its base is a vertical chord: P = (c, y) and Q = (c, −y), with y = b√(1 − c²/a²) and −a < c < a.
3. Base PQ = 2y and the height from A to the base is a − c, so the area is S(c) = ½ · 2y · (a − c) = b√(1 − c²/a²) · (a − c).
4. Differentiating and putting everything over one denominator: S′(c) = −b(a − c)(a + 2c) / (a²√(1 − c²/a²)).
5. For −a < c < a, a − c is positive, so S′(c) = 0 only at c = −a/2. S′(c) is positive for c < −a/2 and negative for c > −a/2, so the area rises and then falls: c = −a/2 gives the greatest area.
6. At c = −a/2: y = b√(1 − 1/4) = (√3/2)b, and the height is a − (−a/2) = 3a/2.
7. Greatest area = (√3/2)b · (3a/2) = (3√3/4)ab.

AnswerThe maximum area is (3√3/4)ab square units.

Watch this explained “The pattern the inscribed solids share”, 17:35 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 184Matches NCERT’s answer

1. Let the base be x metres by y metres, with depth fixed at 2 m.
2. Volume = 2·x·y = 8, so x·y = 4 — the base area is always 4 m², whatever shape the base takes.
3. Base cost = Rs 70 × 4 = Rs 280, which never changes.
4. Side area = perimeter × depth = 2(x + y) × 2 = 4(x + y), so side cost = Rs 45 × 4(x + y) = Rs 180(x + y).
5. Total cost C(x) = 280 + 180(x + 4/x), using y = 4/x.
6. dC/dx = 180(1 − 4/x²); setting it to zero gives x² = 4, so x = 2 (taking the positive length).
7. At x = 2, y = 4/2 = 2 too, so the cheapest base is a 2 m × 2 m square.
8. Total cost = 280 + 180(2 + 2) = 280 + 720 = Rs 1000.
9. The second derivative of C is positive, confirming this is the least cost, not the greatest.

AnswerThe cost of the least expensive tank is Rs 1000 (with a 2 m × 2 m square base).

Watch this explained “The ones worked out”, 18:40 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 184One way to think about it

1. Let the circle have radius r and the square have side s. The perimeters add to k: 2πr + 4s = k, so s = (k − 2πr)/4. As r grows, s changes at the rate ds/dr = −π/2.
2. Sum of the areas: A = πr² + s², which is now a function of r alone.
3. dA/dr = 2πr + 2s · (ds/dr) = 2πr + 2s · (−π/2) = 2πr − πs = π(2r − s).
4. dA/dr = 0 exactly when s = 2r. (Solving with the perimeter condition, this happens at r = k/(2(π + 4)) and s = k/(π + 4).)
5. d²A/dr² = π(2 − ds/dr) = π(2 + π/2), a positive constant, so this critical point gives the least value of A.

In shortThe sum of the areas is least exactly when s = 2r, that is, when the side of the square is double the radius of the circle — as required to prove.

Watch this explained “Where the eliminating relation comes from”, 2:27 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 184Matches NCERT’s answer

1. Let the rectangle have width 2r (so the semicircle on top has radius r) and height h.
2. The perimeter here is the two vertical sides, the bottom, and the curved top (the straight line between the rectangle and the semicircle is inside the window, not an outer edge): 2h + 2r + πr = 10.
3. So h = (10 − 2r − πr)/2 = 5 − r − (π/2)r.
4. Total area (which is what lets light in) = rectangle + semicircle = 2r·h + (π/2)r² = 10r − (2 + π/2)r².
5. Differentiate with respect to r and set it to zero: 10 − (4 + π)r = 0, so r = 10/(π + 4).
6. Then h also works out to 10/(π + 4), and the width of the window is 2r = 20/(π + 4).
7. The second derivative is a negative constant, so this gives the maximum area, i.e. the maximum light.

AnswerWidth = 20/(π + 4) m and height of the rectangular part = 10/(π + 4) m.

Watch this explained “The four moves, named once”, 1:24 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 184One way to think about it

1. The triangle has a hypotenuse, so it is right-angled. Put the right angle C at the origin with the two shorter sides along the x-axis and the y-axis. The point P on the hypotenuse is at distance a from the side on the y-axis and b from the side on the x-axis, so P = (a, b).
2. Let the hypotenuse make an angle θ with the x-axis, 0 < θ < π/2, and meet the x-axis at A and the y-axis at B.
3. Drop a perpendicular from P to the x-axis: it has length b, and in that small right triangle PA = b/sin θ. Drop a perpendicular from P to the y-axis: it has length a, and PB = a/cos θ.
4. So the hypotenuse is L(θ) = AB = a/cos θ + b/sin θ.
5. L′(θ) = a sin θ/cos²θ − b cos θ/sin²θ = (a sin³θ − b cos³θ)/(sin²θ cos²θ).
6. L′(θ) = 0 when a sin³θ = b cos³θ, i.e. tan³θ = b/a, so tan θ = b^(1/3)/a^(1/3).
7. As θ goes from 0 to π/2, sin³θ increases and cos³θ decreases, so a sin³θ − b cos³θ goes from negative to positive and crosses 0 only at this θ. So L falls and then rises: this θ gives the least hypotenuse.
8. With tan θ = b^(1/3)/a^(1/3), a right triangle with sides a^(1/3), b^(1/3) gives cos θ = a^(1/3)/√(a^(2/3) + b^(2/3)) and sin θ = b^(1/3)/√(a^(2/3) + b^(2/3)).
9. Then L = a/cos θ + b/sin θ = a^(2/3)√(a^(2/3) + b^(2/3)) + b^(2/3)√(a^(2/3) + b^(2/3)) = (a^(2/3) + b^(2/3))^(3/2).

In shortThe minimum length of the hypotenuse is (a^(2/3) + b^(2/3))^(3/2) — as required to prove.

Watch this explained “The four moves, named once”, 1:24 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 184Matches NCERT’s answer

(i) local maxima

1. f′(x) = 4(x − 2)³(x + 1)³ + 3(x − 2)⁴(x + 1)² = (x − 2)³(x + 1)²[4(x + 1) + 3(x − 2)] = (x − 2)³(x + 1)²(7x − 2).
2. f′(x) = 0 at x = −1, x = 2/7 and x = 2. These three points cut the line into four pieces, and f′ keeps one sign on each piece.
3. Signs of the factors: (x + 1)² is positive on both sides of −1; (x − 2)³ is negative for x < 2 and positive for x > 2; (7x − 2) is negative for x < 2/7 and positive for x > 2/7.
4. So f′(x) is positive for x < −1, positive for −1 < x < 2/7, negative for 2/7 < x < 2, and positive for x > 2.
5. At x = 2/7, f′ changes from positive to negative, so x = 2/7 is a point of local maxima.

AnswerLocal maximum at x = 2/7.

(ii) local minima

1. At x = 2, f′ changes from negative to positive, so x = 2 is a point of local minima.

AnswerLocal minimum at x = 2.

(iii) point of inflexion

1. At x = −1, f′ is positive just to the left and just to the right: it does not change sign.
2. By the First Derivative Test, part (iii), a critical point where f′ does not change sign is neither a local maximum nor a local minimum; the book calls such a point a point of inflexion.
3. (In the wider meaning used in later courses — a point where the curve changes the way it bends — the curve also does this at x = (2 ± 3√2)/7, where f′ is not zero. This question uses the chapter's meaning.)

AnswerPoint of inflexion at x = −1.

Watch this explained “Reading the verdict off an exponent”, 9:53 into A sign change either side of a critical point decides which kind it is

Open NCERT p. 184Matches NCERT’s answer

1. Given: f(x) = cos²x + sinx, x ∈ [0, π].
2. f'(x) = −2cosx sinx + cosx = cosx(1 − 2sinx).
3. f'(x) = 0 when cosx = 0 (x = π/2) or sinx = 1/2 (x = π/6 or x = 5π/6) — all three lie inside [0, π].
4. List every candidate: the two endpoints (0 and π) plus the three critical points (π/6, π/2, 5π/6).
5. f(0) = 1 + 0 = 1. f(π) = 1 + 0 = 1. f(π/2) = 0 + 1 = 1.
6. f(π/6) = 3/4 + 1/2 = 5/4. f(5π/6) = 3/4 + 1/2 = 5/4.
7. Comparing all five values, the largest is 5/4 and the smallest is 1.

AnswerAbsolute maximum value = 5/4 (at x = π/6 and x = 5π/6); absolute minimum value = 1 (at x = 0, π/2 and π).

Watch this explained “The method, in four steps”, 8:13 into Largest and smallest over a closed interval: candidates inside plus the two ends

Open NCERT p. 184One way to think about it

1. Given: inscribed in a sphere of radius r is 4r/3.
2. Let the cone have altitude h and base radius R, inscribed in a sphere of radius r with its apex and base circle both on the sphere.
3. With the base at distance (h − r) from the sphere's centre, Pythagoras gives R² = r² − (h − r)² = 2hr − h².
4. Volume V = (1/3)πR²h = (π/3)(2hr − h²)h = (π/3)(2rh² − h³).
5. dV/dh = (π/3)(4rh − 3h²) = (π/3)h(4r − 3h).
6. dV/dh = 0 gives h = 0 (rejected, no cone) or h = 4r/3.
7. d²V/dh² = (π/3)(4r − 6h); at h = 4r/3 this is −(4π/3)r, which is negative, confirming a maximum.

In shortThe altitude of the maximum-volume cone is h = 4r/3 — as required to prove.

Watch this explained “The pattern the inscribed solids share”, 17:35 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 184One way to think about it

1. Take any two points x₁, x₂ in (a, b) with x₁ < x₂; we must show f(x₁) < f(x₂).
2. f is differentiable (hence continuous) on [x₁, x₂], so the Mean Value Theorem applies: there is some c between x₁ and x₂ with f(x₂) − f(x₁) = f'(c)·(x₂ − x₁).
3. Since c lies in (a, b), we are given f'(c) > 0, and x₂ − x₁ > 0 because x₁ < x₂.
4. So f(x₂) − f(x₁) = f'(c)·(x₂ − x₁) is a positive number times a positive number, which is positive.
5. Hence f(x₂) > f(x₁) whenever x₂ > x₁, which is exactly what it means for f to be increasing on (a, b).

In shortf is increasing on (a, b) — as required to prove, using the Mean Value Theorem.

Watch this explained “The proof, in four lines”, 2:39 into Using the sign of the derivative to split the line into rising and falling intervals

Open NCERT p. 184Checked by computer

1. Given: a sphere of radius R is 2R/√3.
2. Let the cylinder have height h and base radius ρ, inscribed in a sphere of radius R so that (h/2)² + ρ² = R² (half the height, the radius, and the sphere's radius form a right triangle).
3. So ρ² = R² − h²/4.
4. Volume V = πρ²h = π(R² − h²/4)h = π(R²h − h³/4).
5. dV/dh = π(R² − 3h²/4); setting it to zero gives h² = 4R²/3, so h = 2R/√3 (taking the positive root).
6. d²V/dh² = π(−3h/2), negative for h > 0, confirming this is a maximum.
7. At h = 2R/√3: ρ² = R² − R²/3 = 2R²/3.
8. Vmax = πρ²h = π·(2R²/3)·(2R/√3) = 4πR³/(3√3) = (4√3/9)πR³ after rationalising the denominator.

AnswerThe height for maximum volume is h = 2R/√3, and the maximum volume is (4√3/9)πR³.

Watch this explained “The pattern the inscribed solids share”, 17:35 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 184One way to think about it

1. Let the cylinder have height x and base radius ρ, standing on the cone's base with its top circle touching the cone's slanted surface.
2. At height x up from the base, the cone remaining above has height (h − x), and by similar triangles its base radius, (h − x)tanα, equals the cylinder's radius ρ.
3. So ρ = (h − x)tanα, and volume V = πρ²x = π·tan²α·(h − x)²·x.
4. dV/dx = π tan²α · [(h−x)² − 2x(h−x)] = π tan²α · (h−x)(h − 3x).
5. dV/dx = 0 gives x = h (rejected — no cylinder left) or x = h/3.
6. d²V/dx² at x = h/3 works out negative, confirming a maximum.
7. At x = h/3: ρ = (h − h/3)tanα = (2h/3)tanα, and V = π·[(2h/3)tanα]²·(h/3) = (4/27)πh³tan²α.

In shortThe cylinder's height for greatest volume is h/3 (one-third the cone's height), and the greatest volume is (4/27)πh³tan²α — as required to prove.

Watch this explained “A shape as the relation”, 5:42 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 185Matches NCERT’s answer

1. Volume of wheat in the tank at depth h: V = πr²h = π(10)²h = 100πh.
2. Differentiate with respect to time: dV/dt = 100π·dh/dt.
3. We're told dV/dt = 314 cubic metres per hour; the number 314 is clearly π rounded to 3.14 (a common school approximation), so it's meant to be used with π ≈ 3.14.
4. dh/dt = 314/(100 × 3.14) = 314/314 = 1.

Answer(A) 1 m/h

Watch this explained “Two answers that need a rough pi”, 18:15 into Two quantities both changing with time, and the chain rule that links their rates