JEE Mathematics Application of Derivatives MCQs Set 23

Answer: (c) \( 2\pi \)

Explanation:
1. For x in [0, π], (x + sin x)/2 lies in [0, π/2], so f(x) = (x + sin x)/2, which is increasing (derivative (1 + cos x)/2 ≥ 0).
2. For x in [π, 2π]: f(x) = π − f(2π − x) = π − (2π − x − sin x)/2 = (x + sin x)/2, still increasing, with f(2π) = π.
3. For x in [2π, 4π], f(x) = f(4π − x) decreases from π to 0. So f increases on [0, 2π], giving α = 2π.

Teacher's Note:
1. Find f explicitly on [0, 2π].
2. The reflection rule on [2π, 4π] makes f decrease there.

Answer: (d) 32

Explanation:
1. The limit equals \( e^{\lim f(x)/x^4} \), so \( \lim_{x\to0}\frac{f(x)}{x^4}=2 \): f(x) = \( x^4(2+px+qx^2) \).
2. f'(x) = \( x^3(8+5px+6qx^2) \) must vanish at 1 and 2: 8 + 5p + 6q = 0 and 8 + 10p + 24q = 0, giving q = 2/3, p = −12/5.
3. Then \( f\left(\frac{18}5\right)=\left(\frac{18}5\right)^4\left(2-\frac{216}{25}+\frac{2}{3}\cdot\frac{324}{25}\right)=2\left(\frac{18}5\right)^4 \), so the required value is \( 2\cdot 2^4=32 \).

Teacher's Note:
1. The limit forces f(x) to behave like 2x⁴ near 0.
2. Put the roots 1 and 2 of the remaining quadratic factor of f' into two equations.

Answer: (c) \( \frac29 \)

Explanation:
1. After 49 minutes: V = 4500π − 49 · 72π = 972π, and \( \frac43\pi r^3=972\pi \) gives r³ = 729, r = 9.
2. dV/dt = 4πr² dr/dt: 72π = 4π · 81 · (−dr/dt) in magnitude.
3. dr/dt magnitude = 72/324 = 2/9.

Teacher's Note:
1. Find the volume remaining at t = 49.
2. Use dV/dt = 4 pi r squared times dr/dt.

Answer: (a) All of A, B and C

Explanation:
1. Rolle on eˣ f(x) gives a point where eˣ(f + f') = 0, so A has a root.
2. Rolle on e⁻ˣ f(x) gives a point where e⁻ˣ(f' − f) = 0, so B has a root.
3. Rolle on f(x)² gives 2f f' = 0, so C has a root.

Teacher's Note:
1. Multiply f by a suitable exponential before applying Rolle's theorem.
2. For the product, use f squared.

Answer: (b) \( \alpha=2,\ \beta=-\frac12 \)

Explanation:
1. f'(x) = α/x + 2βx + 1 = 0 at x = −1: −α − 2β + 1 = 0, and at x = 2: α/2 + 4β + 1 = 0.
2. From the first, α = 1 − 2β; then (1 − 2β)/2 + 4β + 1 = 0 gives β = −1/2 and α = 2.

Teacher's Note:
1. Set f' = 0 at both given points.

Answer: (a) \( \frac92 \)

Explanation:
1. At a common point (x, y), slopes are 3/y and −9x/(by).
2. Product of slopes = −27x/(by²) = −27x/(6bx) (using y² = 6x) = −9/(2b).
3. Perpendicular means the product is −1, so b = 9/2.

Teacher's Note:
1. Differentiate both curves implicitly.
2. Replace y² with 6x.

Answer: (d) (III) (i) (R)

Explanation:
1. f'(x) = 1/x − ln x and f''(x) = −1/x² − 1/x < 0 for all x > 0, so f' is always decreasing.
2. f(1) = 1, f(e) = 1, f(e²) = 2 − e² < 0: so (I) is true. f'(1) = 1 > 0 and f'(e) = 1/e − 1 < 0: (II) is true. f' > 0 on (0, 1), so (III) is false; f'' never vanishes, so (IV) is false.
3. Column 2: f → −∞ (ii true, i false); f' → −∞ (iii true); f'' → 0 (iv true). Column 3: f' > 0 on (0, 1) so (P) true; f' < 0 for x > e so (Q) true; f'' < 0 so (R) false and (S) true.

Teacher's Note:
1. Decide each of the twelve items true or false.
2. A combination is correct only if all three of its items are true.

Answer: (c) 4

Explanation:
1. Inner radius r, inner height h = V/(πr²). Material = π(r + 2)²(h + 2) − πr²h = \( \pi\left[(r+2)^2\left(\frac V{\pi r^2}+2\right)-\frac V\pi\right] \).
2. Material \( =(r+2)^2\left(\frac V{r^2}+2\pi\right)-V \), with derivative \( 2(r+2)\left(\frac V{r^2}+2\pi\right)-\frac{2V(r+2)^2}{r^3} \). At r = 10 this is 0.24V + 48π − 0.288V = 0, so V = 1000π.
3. So V/(250π) = 4.

Teacher's Note:
1. Material volume = outer volume − inner volume.
2. Set d(material)/dr = 0 at r = 10.

Answer: (b) B and C

Explanation:
1. Expanding, f(x) = 2 cos x cos 3x = cos 4x + cos 2x.
2. f(0) = 2 is the maximum value (A false, B true).
3. f'(x) = −4 sin 4x − 2 sin 2x = −2 sin 2x (4 cos 2x + 1). sin 2x = 0 gives x = 0, ±π/2; cos 2x = −1/4 gives four more points in (−π, π). So more than three zeros (C true, D false).

Teacher's Note:
1. Evaluate the determinant by expanding along the first row.
2. Factor f' using sin 4x = 2 sin 2x cos 2x.

Answer: (d) \( \frac{e^\alpha+2e^\beta}3\ge e^{\frac{\alpha+2\beta}3} \)

Explanation:
1. eˣ is concave up, so with weights 1 and 2: \( \frac{e^\alpha+2e^\beta}3\ge e^{\frac{\alpha+2\beta}3} \): the last statement is true.
2. The first statement is the reverse for 2ˣ (concave up). ln is concave down so the second statement (with ≥) is false. tan⁻¹ is concave up on the negative reals, so the average is ≥, making the third false.

Teacher's Note:
1. Decide concavity of each function using its second derivative.
2. Apply the stated inequality with weights 1 and 2.

Answer: (b) 9

Explanation:
1. P, Q = (h, ±y₀) with \( y_0=\frac{\sqrt3}2\sqrt{4-h^2} \). The tangents at P and Q meet at R = (4/h, 0).
2. Area \( \Delta=\frac12\cdot2y_0\left(\frac4h-h\right)=\frac{\sqrt3\,(4-h^2)^{3/2}}{2h} \), which is decreasing in h.
3. Δ₁ = Δ(1/2) = 45√5/8 and Δ₂ = Δ(1) = 9/2. Then \( \frac8{\sqrt5}\cdot\frac{45\sqrt5}8-8\cdot\frac92=45-36=9 \).

Teacher's Note:
1. The pole of the line x = h with respect to the ellipse is (4/h, 0).
2. Both factors of Δ decrease as h increases.

Answer: (d) B and C

Explanation:
1. A is false: f(a) ≠ f(b) does not forbid a critical point inside.
2. B: derivative \( 100x^{99}+\cos x\gt0 \) on [0, 1], so true.
3. C: let \( g(x)=ax^3-2bx^2+cx \). Since 2b = a + c, g(1) = a − 2b + c = 0 = g(0), so by Rolle g' = 3ax² − 4bx + c has a root in (0, 1): true.
4. D: 3 tan x + x³ is increasing, 0 at x = 0 and about 3.49 at π/4, so exactly one root: false.

Teacher's Note:
1. Integrate the quadratic to apply Rolle's theorem.
2. A strictly increasing function crosses a level at most once.

Answer: (b) \( \frac{81}{2} \)

Explanation:
1. With base angle \( \theta \), the height is 6 sin θ and the longer base is 6 + 12 cos θ, so area \( =\frac{6+(6+12\cos\theta)}{2}\cdot6\sin\theta=36\sin\theta(1+\cos\theta) \).
2. Its derivative is \( 36(\cos\theta+\cos^2\theta-\sin^2\theta)=36(2\cos\theta-1)(\cos\theta+1) \), zero at θ = 60°, giving \( \Delta=36\cdot\frac{\sqrt3}2\cdot\frac32=27\sqrt3 \).
3. Hence \( \frac{\sqrt3\cdot27\sqrt3}{2}=\frac{81}2 \).

Teacher's Note:
1. Take the equal slanting sides as the legs and one of the bases as 6.
2. Maximise 36 sin θ (1 + cos θ).

Answer: (c) A and D

Explanation:
1. As x increases, f decreases among non-negative values, and g increases toward 0 so |g| = −g decreases. Hence f|g| decreases and fg = −f|g| increases: A is true, B false.
2. The composite f(g(x)): g increases and f is decreasing, so f(g(x)) decreases: D is true, C false.

Teacher's Note:
1. Write fg as −f·|g| and watch both factors.
2. Composing a decreasing function with an increasing one gives a decreasing function.

Answer: (d) Both true; Statement 2 is a correct explanation of Statement 1

Explanation:
1. By AM–GM, eˣ + 2e⁻ˣ ≥ 2√2 with equality at eˣ = √2, and the sum is positive and unbounded above, so 0 < f(x) ≤ 1/(2√2) (Statement 2 true).
2. 1/3 ≈ 0.333 lies in (0, 0.3536], and f is continuous with range (0, 1/(2√2)], so f(c) = 1/3 for some c (Statement 1 true) — it follows directly from the range in Statement 2.

Teacher's Note:
1. Find the range of f using AM–GM.
2. Check whether 1/3 is in that range.

Answer: (a) P–4, Q–1, R–2, S–3

Explanation:
1. F = (x₀, 4√x₀) (y₀ = 4√x₀ ≤ 6). The tangent at F is y y₀ = 8(x + x₀), so y₁ = 8x₀/y₀ = 2√x₀.
2. Area \( =\frac12(3-y_1)x_0=\frac12(3-2\sqrt{x_0})x_0 \), maximised when \( \frac12(3-3\sqrt{x_0})=0 \), i.e. x₀ = 1.
3. Then y₀ = 4, y₁ = 2, m = (4 − 3)/1 = 1 and the area is 1/2. So P–4, Q–1, R–2, S–3.

Teacher's Note:
1. Express everything in terms of x₀.
2. Differentiate the area with respect to x₀.

Answer: (a) A, C and D

Explanation:
1. Near x = 1 the factor (x² − 1)ⁿ behaves like (x − 1)ⁿ times a positive factor, and x² + x + 1 is positive.
2. For even n, (x − 1)ⁿ does not change sign, so f has an extremum at x = 1 (a minimum, since f(1) = 0 ≤ f nearby).
3. For odd n the sign of f changes at x = 1, so there is no extremum. Hence n = 2, 4, 6.

Teacher's Note:
1. An extremum at a zero of f needs f to keep the same sign on both sides.
2. Even powers do not change sign.

Answer: (d) \( \frac1{5\pi} \)

Explanation:
1. Water of depth h has radius h/2, so \( V=\frac13\pi\frac{h^2}4h=\frac{\pi h^3}{12} \).
2. dV/dt = (πh²/4) dh/dt, so 5 = (π · 100/4) dh/dt = 25π dh/dt.
3. dh/dt = 1/(5π).

Teacher's Note:
1. Express the radius in terms of the depth using tan of the semi-vertical angle.

Answer: (d) f'(c) = 2g'(c)

Explanation:
1. Let h(x) = f(x) − 2g(x). Then h(0) = 2 − 0 = 2 and h(1) = 6 − 4 = 2.
2. By Rolle's theorem, h'(c) = 0 for some c in (0, 1), i.e. f'(c) = 2g'(c).

Teacher's Note:
1. Find a combination of f and g with equal values at 0 and 1.

Answer: (b) A and B

Explanation:
1. Treat the equation as a quadratic in x: \( 4x^2+12yx+(10y^2-4y+3)=0 \) has real roots when \( 144y^2-16(10y^2-4y+3)\ge0 \), i.e. \( -16y^2+64y-48\ge0 \), i.e. \( (y-1)(y-3)\le0 \).
2. So 1 ≤ y ≤ 3: A and B are true.
3. As a quadratic in y the discriminant condition for x gives x between -3-sqrt10/2 and -3+sqrt10/2 (about -4.58 and -1.42), so C and D are false.

Teacher's Note:
1. Use the discriminant condition on x (or on y) for the existence of a real point.
2. Check the range of x separately.

Answer: (a) A, C and D

Explanation:
1. A: Rolle's theorem. B is false: f would then have f(1) < f(2).
2. C: if f' ≤ 0 everywhere then f would be decreasing, contradicting f(1) = f(2) with f non-constant; so f' takes positive values somewhere.
3. D: |f'(x)| = |f'(x) − f'(a)| ≤ 2|x − a| < 2 because |x − a| < 1.

Teacher's Note:
1. Rolle gives a point with zero derivative.
2. Bound f' using the mean value theorem on f' and |f''| ≤ 2.

Answer: (c) \( \frac{c-a}{b-c} \)

Explanation:
1. By the mean value theorem, f(c) − f(a) = (c − a) f'(ξ₁) with ξ₁ in (a, c), and f(b) − f(c) = (b − c) f'(ξ₂) with ξ₂ in (c, b).
2. f'' < 0 means f' is decreasing, so f'(ξ₁) > f'(ξ₂) > 0.
3. The ratio equals \( \frac{(c-a)f'(\xi_1)}{(b-c)f'(\xi_2)}\gt\frac{c-a}{b-c} \).

Teacher's Note:
1. Apply the mean value theorem on [a, c] and [c, b].
2. Use that f' is decreasing.

Answer: (a) All of A, B and C

Explanation:
1. The point (2, −1) corresponds to t = 2 (t² + 3t − 10 = 0 gives t = 2 or −5, and only t = 2 gives y = −1).
2. dy/dx = (4t − 2)/(2t + 3) = 6/7 at t = 2, so B is true.
3. Subtangent = |y/m| = 7/6 (A true); tangent length = \( |y|\sqrt{1+\frac1{m^2}}=\sqrt{1+\frac{49}{36}}=\frac{\sqrt{85}}6 \) (C true).

Teacher's Note:
1. Locate the parameter value first.
2. Use the formulas for the lengths with k = −1 and m = 6/7.

Answer: (a) 25

Explanation:
1. For radius r and arc length s: 2r + s = 20, area = rs/2 = r(20 − 2r)/2 = r(10 − r).
2. This is maximum at r = 5, with area 25.

Teacher's Note:
1. Area of a sector = (1/2) r × arc length.

Answer: (c) A–q, B–r, C–p

Explanation:
1. A: for z = e^{iθ}, 2iz/(1 − z²) = 2i/(e^{−iθ} − e^{iθ}) = −1/sin θ, whose real values are (−∞, −1] ∪ [1, ∞), i.e. (q).
2. B: with t = 3^{x−1}, |8t/(3(1 − t²))| ≤ 1 gives t ≤ 1/3 or t ≥ 3, i.e. x ≤ 0 or x ≥ 2: (r).
3. C: the determinant equals 1 + tan²θ + 1 + tan²θ = 2 sec²θ ∈ [2, ∞): (p). So A–q, B–r, C–p.

Teacher's Note:
1. Write z = cos θ + i sin θ in part A.
2. Expand the determinant along the first row.