JEE Mathematics Application of Derivatives MCQs Set 24

Answer: (a) 5

Explanation:
1. For x ≥ 0: f = x + 1 − x² on [0, 1] (maximum at x = 1/2, value 5/4) and x² + x − 1 for x ≥ 1 (increasing).
2. f(0) = 1 is a local minimum (corner), and at x = 1 there is a corner minimum (f(1) = 1 < nearby values).
3. By evenness the same holds at x = −1/2 and x = −1. Points: −1, −1/2, 0, 1/2, 1, which is 5.

Teacher's Note:
1. Treat x ≥ 0 and use f(−x) = f(x).
2. Corners where the modulus changes can be extrema too.

Answer: (b) [−2, 2]

Explanation:
1. f''(x) = 12x² + 6ax + 3 ≥ 0 for all x.
2. This needs the discriminant 36a² − 144 ≤ 0, i.e. a² ≤ 4.
3. So a is in [−2, 2].

Teacher's Note:
1. A quadratic with positive leading coefficient is non-negative everywhere when its discriminant is non-positive.

Answer: (b) A and B

Explanation:
1. d/dx (x + sin x) = 1 + cos x ≥ 0, zero only at isolated points, so x + sin x is increasing: A is true and C is false.
2. sec x increases on (0, π/2) and decreases on (π/2, π) (and is undefined at π/2), so over its domain it is neither increasing nor decreasing: B is true and D is false.

Teacher's Note:
1. A function with non-negative derivative vanishing only at isolated points is increasing.
2. Look at the sign of sec x tan x.

Answer: (c) y = 3

Explanation:
1. dy/dx = 1 − 8/x³ = 0 gives x = 2.
2. y(2) = 2 + 4/4 = 3.
3. The horizontal tangent is y = 3.

Teacher's Note:
1. Parallel to the x-axis means slope 0.

Answer: (c) A, B and C

Explanation:
1. f'(x) = \( \frac1{1+x^2}-\frac1{2x}=-\frac{(x-1)^2}{2x(1+x^2)}\le0 \) on (0, ∞), so f decreases there (C true); f is not defined for x ≤ 0, so D is false.
2. Greatest value at x = 1/√3: \( \frac\pi6-\frac12\ln\frac1{\sqrt3}=\frac\pi6+\frac14\ln3 \) (A true).
3. Least value at x = √3: \( \frac\pi3-\frac14\ln3 \) (B true).

Teacher's Note:
1. Factor the numerator of f' as a perfect square.
2. A decreasing function has its extremes at the ends of the interval.

Answer: (d) B and D

Explanation:
1. f'(x) = 5(x⁴ − 1), so critical points are ±1: local maximum f(−1) = a + 4 and local minimum f(1) = a − 4.
2. Three real roots need a − 4 < 0 < a + 4, i.e. −4 < a < 4 (D true).
3. For a > 4 the local minimum is positive and there is only one real root (B true); for a < −4 also only one root (C false).

Teacher's Note:
1. Find the values of f at the local extremes.
2. Three roots need the local max above and local min below the axis.

Answer: (d) 30 and −30

Explanation:
1. The point (a, a) lies on the curve. Differentiating: \( 6y^2y'=2ax+3x^2 \), so at (a, a) the slope is \( \frac{5a^2}{6a^2}=\frac56 \).
2. Tangent: \( y-a=\frac56(x-a) \) has intercepts \( \alpha=-\frac a5 \) and \( \beta=\frac a6 \).
3. Then \( a^2\left(\frac1{25}+\frac1{36}\right)=\frac{61a^2}{900}=61 \), so a² = 900 and a = ±30.

Teacher's Note:
1. Find the slope at (a, a) by implicit differentiation.
2. Intercepts of y − a = m(x − a) are a − a/m on the x-axis and a − ma on the y-axis.

Answer: (d) B and C

Explanation:
1. Let h = f − 3g: h(−1) = 3, h(0) = 3, h(2) = 3.
2. By Rolle's theorem h' has at least one zero in (−1, 0) and one in (0, 2). Since h'' never vanishes there, h' is strictly monotone on each interval, so exactly one zero in each.
3. So B and C are true; A (three solutions) and D (two in each) are false.

Teacher's Note:
1. Compute h at the three points.
2. If h'' has a constant sign, h' can vanish only once on the interval.

Answer: (b) Both true; Statement 2 is not a correct explanation of Statement 1

Explanation:
1. For x near 0, tan x / x = 1 + x²/3 + ... ≥ 1 = f(0), so x = 0 is a minimum (Statement 1 true).
2. f'(0) = lim (f(x) − 1)/x = lim (x/3 + ...) = 0 (Statement 2 true).
3. A zero derivative alone does not make a point a minimum, so Statement 2 does not explain Statement 1.

Teacher's Note:
1. Expand tan x / x near zero.
2. Zero derivative is necessary but not sufficient for a minimum.

Answer: (d) \( \left(0,\frac{2\pi}3\right) \)

Explanation:
1. Since \( \sqrt{\frac{1+\sin x}{1-\sin x}}=\tan\left(\frac\pi4+\frac x2\right) \), f(x) = π/4 + x/2 on (0, π/2), so f' = 1/2.
2. At x = π/6: f = π/4 + π/12 = π/3, normal slope −2.
3. Normal: y − π/3 = −2(x − π/6). At x = 0, y = π/3 + π/3 = 2π/3: the point (0, 2π/3).

Teacher's Note:
1. Simplify f(x) using half-angle identities.
2. The normal slope is −1/f'.

Answer: (c) (II) (iii) (S)

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 item true or false first.
2. A combination is correct only if all three of its items are true.

Answer: (a) A and D

Explanation:
1. Both numerator and denominator → 0, so by L'Hôpital's rule the limit equals \( \frac{f(2)g'(2)}{f''(2)g'(2)}=\frac{f(2)}{f''(2)} \) = 1.
2. So f''(2) = f(2) > 0 (f takes positive values), with f'(2) = 0: a local minimum (A true). C is false since f''(2) = f(2).
3. At x = 2, f(x) − f''(x) = 0, so D is true.

Teacher's Note:
1. Apply L'Hôpital's rule once.
2. f is positive, so f''(2) = f(2) is positive.

Answer: (c) 50

Explanation:
1. The left side is strictly increasing (odd powers), so there is exactly one real root.
2. The terms pair up as \( (x-k)^{2013}+(x-(2014-k))^{2013} \), which vanishes at the midpoint \( x=\frac{1+2013}{2}=1007 \), so the root is 1007.
3. Digit squares: 1 + 0 + 0 + 49 = 50.

Teacher's Note:
1. Odd powers make the sum increasing, so the root is unique; find it by symmetry.
2. The nodes 1, 2, ..., 2013 are symmetric about 1007.

Answer: (b) \( \frac7{4\sqrt2} \)

Explanation:
1. A point of the curve is (y² + 2, y), whose distance from x − y = 0 is \( \frac{y^2-y+2}{\sqrt2} \).
2. The numerator is least at y = 1/2 where it equals 7/4.
3. Shortest distance = 7/(4√2).

Teacher's Note:
1. Parametrise the curve by y.
2. y² − y + 2 is least at y = 1/2.

Answer: (d) \( 20\sqrt2-2 \)

Explanation:
1. Put \( t=2^{x^2}+1 \), where t ≥ 2. Then \( f=t-2+\frac{200}{t} \).
2. By AM–GM, \( t+\frac{200}t\ge2\sqrt{200}=20\sqrt2 \), with equality at \( t=10\sqrt2\approx14.1\ge2 \), which is attainable.
3. Least value \( =20\sqrt2-2\approx26.28 \).

Teacher's Note:
1. Substitute t = 2^(x²) + 1 and note t ≥ 2.
2. Check that the AM–GM equality value of t lies in the allowed range.

Answer: (c) \( \pm\frac12 \)

Explanation:
1. At (0, 1): k = 1 and m = p + p = 2p.
2. Subtangent \( =\frac1{2|p|} \) and subnormal = 2|p|.
3. Equality gives 4p² = 1, so p = ±1/2.

Teacher's Note:
1. Compute the slope at x = 0 first.
2. Set |k/m| = |km| with k = 1.

Answer: (c) A, C and D

Explanation:
1. f is continuous at 0, 1 and 3 (values 1, 1, 1/3). Left of 1, f' = 2x − 1 → 1; right of 1, f' = 2x² − 8x + 7 → 1, so f' is continuous at 1, but f'' jumps from 2 to −4: f' is not differentiable at 1 (A true).
2. On x < 0, f' = 5(x + 1)⁴ − 2 is negative near x = −1, so f is not increasing on (−∞, 0): B false.
3. f is continuous, tends to −∞ as x → −∞ (odd degree) and to +∞ as x → ∞, so it is onto (C true). f' increases up to x = 1 (slope of f'(x) = 2x − 1 is 2) and then decreases (slope 4x − 8 < 0), so f' has a local maximum at x = 1 (D true).

Teacher's Note:
1. Check continuity of f and f' at the join points.
2. Use the sign of f'' on either side of x = 1.

Answer: (d) a = b = c

Explanation:
1. f' = 2x(eˣ² − e⁻ˣ²) ≥ 0 on [0, 1], so a = f(1) = e + 1/e.
2. g' = eˣ² + 2x²eˣ² − 2x e⁻ˣ² > 0 on [0, 1] (since eˣ² ≥ 1 ≥ x and e⁻ˣ² ≤ 1), so b = g(1) = e + 1/e.
3. h' = 2x eˣ² + 2x³ eˣ² − 2x e⁻ˣ² ≥ 0, so c = h(1) = e + 1/e. Hence a = b = c.

Teacher's Note:
1. Show each function is increasing on [0, 1] so the maximum is at x = 1.
2. All three equal e + 1/e at x = 1.

Answer: (a) A and B

Explanation:
1. f'(x) = a/x + 2bx + 1 = 0 at x = −1 and x = 2: −a − 2b + 1 = 0 and a/2 + 4b + 1 = 0.
2. Substituting a = 1 − 2b: (1 − 2b)/2 + 4b + 1 = 0, so b = −1/2 and a = 2.

Teacher's Note:
1. Set the derivative to zero at both points.
2. Solve the two linear equations in a and b.

Answer: (a) \( \frac12 \)

Explanation:
1. f'(x) = \( 2e^x+ae^{-x}+2a-3 \). With u = eˣ > 0 we need \( 2u^2+(2a-3)u+a\ge0 \) for all u > 0.
2. As u → 0 this requires a ≥ 0. If 2a − 3 ≥ 0 it is fine; for 0 ≤ a < 3/2 we need the discriminant \( (2a-3)^2-8a\le0 \), i.e. \( 4a^2-20a+9\le0 \), so \( \frac12\le a\le\frac92 \).
3. The smallest admissible a is 1/2.

Teacher's Note:
1. Reduce to a quadratic in u = e^x with u > 0.
2. Check small values of a such as 0 and 1/4 directly.

Answer: (c) \( \frac16\sqrt{\frac73} \)

Explanation:
1. The squared distance from (x, x^{3/2} + 7) to (1/2, 7) is \( D=\left(x-\frac12\right)^2+x^3 \).
2. D' = 2(x − 1/2) + 3x² = 3x² + 2x − 1 = (3x − 1)(x + 1) = 0 gives x = 1/3 for x ≥ 0.
3. D = 1/36 + 1/27 = 7/108, so the distance is \( \sqrt{\frac7{108}}=\frac16\sqrt{\frac73} \).

Teacher's Note:
1. Minimise the squared distance.
2. Only x ≥ 0 is allowed.

Answer: (d) \( f'(x)\lt f(x),\ 0\lt x\lt\frac14 \)

Explanation:
1. Let g(x) = e⁻ˣ f(x). Then g'(x) = e⁻ˣ(f' − f) and g'' ≥ 1 > 0, so g' is strictly increasing.
2. g has its minimum at x = 1/4, so g'(1/4) = 0, hence g' < 0 for x < 1/4 and g' > 0 for x > 1/4.
3. So f' < f on (0, 1/4) and f' > f on (1/4, 1).

Teacher's Note:
1. g' is increasing because g'' is positive.
2. Zero of g' at 1/4 fixes its sign on either side.

Answer: (d) A and C

Explanation:
1. By the mean value theorem, the difference quotient equals the derivative at some point.
2. Since 0 < 1/(1 + t²) ≤ 1, the quotient for tan⁻¹ is at most 1 in absolute value: A true, B false.
3. Since |cos t| ≤ 1, |sin x − sin y| ≤ |x − y|: C true, D false.

Teacher's Note:
1. Use the mean value theorem on each function.
2. Bound the derivative of each function by 1.

Answer: (c) \( \frac{\sin\alpha+\sin\beta}2\le\sin\frac{\alpha+\beta}2 \) for α, β in (0, π)

Explanation:
1. On (0, π), (sin x)'' = −sin x < 0, so sin is concave down.
2. For a concave down function, the average of the values is at most the value at the average: \( \frac{\sin\alpha+\sin\beta}2\le\sin\frac{\alpha+\beta}2 \).
3. So the third statement is correct (the first is the reverse; on (π, 2π) sin is concave up so the second is also reversed).

Teacher's Note:
1. Find the sign of the second derivative of sin x on each interval.
2. Concave down means the chord lies below the curve.

Answer: (a) 8

Explanation:
1. Differentiate: \( 2(y-x^5)(y'-5x^4)=(1+x^2)^2+4x^2(1+x^2) \).
2. At (1, 3): 2 · 2 · (y' − 5) = 4 + 8 = 12, so y' − 5 = 3.
3. y' = 8.

Teacher's Note:
1. Differentiate both sides implicitly.
2. Substitute x = 1, y = 3.