JEE Mathematics Maxima and Minima MCQs Set 08

Answer: (c) \( \sqrt{\frac\alpha\beta} \)

Explanation:
1. The number of batches is \( \frac Nx \), so the total time is \( T = \frac Nx(\alpha + \beta x^2) = N\left(\frac\alpha x + \beta x\right) \).
2. \( \frac{dT}{dx} = N\left(-\frac{\alpha}{x^2} + \beta\right) = 0 \Rightarrow x = \sqrt{\frac\alpha\beta} \).
3. The second derivative \( \frac{2N\alpha}{x^3} > 0 \), so it is a minimum.

Teacher's Note:
1. Total time is the number of batches times the time per batch.
2. Differentiate with respect to x and set the result to zero.

Answer: (b) \( ax^2 + 2bx + c - y(Ax^2 + 2Bx + C) = 0 \) has equal roots, that is, is a perfect square

Explanation:
1. Writing the relation as a quadratic in x, \( (a - yA)x^2 + 2(b - yB)x + (c - yC) = 0 \), real x exists only when its discriminant is non-negative.
2. The extreme values of y are the boundary values where the discriminant is zero.
3. That is exactly when the quadratic has equal roots.

Teacher's Note:
1. The range of y is where the equation has real roots.
2. The edges of that range correspond to a zero discriminant.

Answer: (b) All b if a = 0

Explanation:
1. At a relative minimum of the differentiable function f, f'(0) = 0.
2. For y' to vanish at 0 we need \( y'(0) = f'(0) + a = a = 0 \).
3. With a = 0, y = f(x) + b is just f shifted by b, so the minimum is kept for every b.

Teacher's Note:
1. Adding a linear term changes the slope unless the coefficient is zero.
2. Adding a constant never moves an extremum.

Answer: (a) \( \frac\pi6 \)

Explanation:
1. With the apex at A and base BC, the base angles are \( \frac{\pi - 2\theta}{2} \). The sides are \( AB = AC = 2a\cos\theta \) and \( BC = 2a\sin2\theta \).
2. Area \( = \frac12(AB)(AC)\sin2\theta = 2a^2\cos^2\theta\sin2\theta \). Since \( \cos^2\theta\sin2\theta = 2\cos^3\theta\sin\theta \), the derivative is \( 2\cos^2\theta(\cos^2\theta - 3\sin^2\theta) = 0 \), so \( \tan^2\theta = \frac13 \).
3. So \( \theta = \frac\pi6 \) (taking the positive root) and the vertical angle is 60 degrees: the triangle is equilateral.

Teacher's Note:
1. The most efficient shape is the equilateral triangle.
2. Its vertical angle is 60 degrees, so theta is 30 degrees.

Answer: (d) \( \sqrt{1 - 2a + 2a^2} \)

Explanation:
1. On the curve \( y^2 = 2x - 2x^2 \), with \( 0 \le x \le 1 \). The squared distance is \( D = (x - a)^2 + 2x - 2x^2 = -x^2 + 2(1 - a)x + a^2 \).
2. \( D' = -2x + 2(1 - a) = 0 \Rightarrow x = 1 - a \), which lies in [0, 1], and D'' = -2 < 0 so it is a maximum.
3. \( D = -(1 - a)^2 + 2(1 - a)^2 + a^2 = (1 - a)^2 + a^2 = 1 - 2a + 2a^2 \).

Teacher's Note:
1. Work with the square of the distance to avoid the square root.
2. Use the curve equation to eliminate y.

Answer: (a) 1

Explanation:
1. \( f'(x) = (x - 1)(x - 2)^2[2(x - 2) + 3(x - 1)] = (x - 1)(x - 2)^2(5x - 7) \).
2. The zeros are x = 1, \( \frac75 \) and 2. Sign of f': positive for x < 1, negative on \( \left(1, \frac75\right) \), positive on \( \left(\frac75, \infty\right) \) except at 2.
3. f' changes from + to - at x = 1 (maximum) and from - to + at \( \frac75 \) (minimum); there is no sign change at x = 2.

Teacher's Note:
1. Factorise the integrand and apply the fundamental theorem.
2. The squared factor at x = 2 gives no extremum.

Answer: (c) \( \left(1 + 2^{n/2}\right)^2 \)

Explanation:
1. Expanding: \( 1 + \frac{1}{\sin^n\alpha} + \frac{1}{\cos^n\alpha} + \frac{1}{(\sin\alpha\cos\alpha)^n} \).
2. With \( s = \sin\alpha\cos\alpha \le \frac12 \): by AM-GM \( \frac{1}{\sin^n\alpha} + \frac{1}{\cos^n\alpha} \ge \frac{2}{s^{n/2}} \), and every term is smallest when \( \sin\alpha = \cos\alpha \).
3. At \( \alpha = \frac\pi4 \): \( \sin^n\alpha = 2^{-n/2} \), so the value is \( \left(1 + 2^{n/2}\right)^2 \).

Teacher's Note:
1. Symmetry between sine and cosine suggests the minimum is at 45 degrees.
2. Check using AM-GM that every term is smallest there.

Answer: (b) \( \frac13 \)

Explanation:
1. \( \tan A\tan B = \frac{\cos(A - B) - \cos(A + B)}{\cos(A - B) + \cos(A + B)} = \frac{\cos(A - B) - \frac12}{\cos(A - B) + \frac12} \).
2. This increases with \( \cos(A - B) \), whose maximum is 1 at A = B.
3. Maximum \( = \frac{1/2}{3/2} = \frac13 \).

Teacher's Note:
1. Convert the product of tangents into cosines of the sum and difference.
2. The expression is largest when A equals B.

Answer: (a) 2

Explanation:
1. Parallel tangents occur at the same value of x where the slopes are equal: \( 2x - 1 = 3x^2 - 2x - 2 \).
2. \( 3x^2 - 4x - 1 = 0 \) has discriminant \( 16 + 12 = 28 > 0 \), so there are two real values of x.
3. Hence there are 2 pairs of parallel tangents (at those two values of x).

Teacher's Note:
1. Equate the two derivatives at the same x.
2. The discriminant tells the number of real solutions.

Answer: (c) \( \frac{2a}{\sqrt3} \)

Explanation:
1. Let the cylinder have radius r and height h. \( r^2 + \frac{h^2}{4} = a^2 \).
2. \( V = \pi r^2h = \pi\left(a^2 - \frac{h^2}{4}\right)h \), \( \frac{dV}{dh} = \pi\left(a^2 - \frac{3h^2}{4}\right) = 0 \Rightarrow h = \frac{2a}{\sqrt3} \).
3. The second derivative is negative, so V is maximum.

Teacher's Note:
1. Eliminate r with the right triangle in the cross-section.
2. Maximise V as a function of h.

Answer: (b) \( \left(\sqrt a + \sqrt b\right)^2 \)

Explanation:
1. Let OP = p and OQ = q. The line \( \frac xp + \frac yq = 1 \) passes through (a, b): \( \frac ap + \frac bq = 1 \).
2. \( p + q = (p + q)\left(\frac ap + \frac bq\right) = a + b + \frac{aq}{p} + \frac{bp}{q} \ge a + b + 2\sqrt{ab} \) by AM-GM.
3. The minimum value is \( \left(\sqrt a + \sqrt b\right)^2 \).

Teacher's Note:
1. Multiply the sum by 1 written as a/p + b/q, then use AM-GM.
2. Equality holds when the two cross terms are equal.

Answer: (b) -9

Explanation:
1. \( f'(x) = 2kx + \frac{2k^2 - 81}{2} = 0 \) at \( x = \frac94 \): \( \frac{9k}{2} + \frac{2k^2 - 81}{2} = 0 \Rightarrow 2k^2 + 9k - 81 = 0 \).
2. \( k = \frac{-9 \pm 27}{4} = \frac92 \) or -9.
3. For a maximum the parabola must open downward, so k < 0 and k = -9.

Teacher's Note:
1. Use f'(9/4) = 0 and then the sign of the leading coefficient.
2. The positive root gives a minimum, not a maximum.

Answer: (b) \( \alpha - \sin\alpha < \beta - \sin\beta \)

Explanation:
1. Let \( g(x) = x - \sin x \). Then \( g'(x) = 1 - \cos x \ge 0 \), zero only at isolated points, so g is strictly increasing.
2. For \( \alpha < \beta \) this gives \( g(\alpha) < g(\beta) \): \( \alpha - \sin\alpha < \beta - \sin\beta \).

Teacher's Note:
1. Define an auxiliary function and study its monotonicity.
2. A strictly increasing function preserves the order of inputs.

Answer: (a) \( 3 \times \sqrt[3]{36} \)

Explanation:
1. For roots r1, r2, r3: \( r_1r_2r_3 = 6 \) and \( \mu = r_1r_2 + r_2r_3 + r_3r_1 \).
2. By AM-GM, \( \mu \ge 3\sqrt[3]{(r_1r_2r_3)^2} = 3\sqrt[3]{36} \).
3. Equality holds when all roots equal \( \sqrt[3]{6} \), so the minimum is attained.

Teacher's Note:
1. Use Vieta's relations for the sum and product of roots.
2. AM-GM is attained when the three roots are equal.

Answer: (a) A maximum

Explanation:
1. The derivative is positive just before a and negative just after a.
2. So f increases up to a and then decreases, which means f(a) is a local maximum.

Teacher's Note:
1. This is the first derivative test.
2. Positive to negative means a maximum.

Answer: (c) \( \sqrt{a^2 - b^2} \)

Explanation:
1. \( f'(\theta) = \sec\theta(a\tan\theta - b\sec\theta) = 0 \Rightarrow \sin\theta = \frac ba \).
2. At that angle \( \cos\theta = \frac{\sqrt{a^2 - b^2}}{a} \), so \( f = \frac{a - b\sin\theta}{\cos\theta} = \frac{a - \frac{b^2}{a}}{\frac{\sqrt{a^2 - b^2}}{a}} = \sqrt{a^2 - b^2} \).
3. Since \( a > b \), this is a minimum with positive value.

Teacher's Note:
1. Write f as a ratio with cos theta in the denominator.
2. A sin theta equal to b over a is allowed because a is greater than b.

Answer: (b) A local maximum at \( x = \frac\pi2 \)

Explanation:
1. At \( x = \frac\pi2 \): \( f\left(\frac\pi2\right) = 1 + 1 = 2 \). Just to the left, \( \cos x \) is close to 0, and just to the right, \( 1 + \sin x \) is slightly less than 2.
2. So f(π/2) = 2 is greater than all nearby values: a local maximum.
3. f(-1) = -1 + 1 + 10 = 10 is the greatest value on the interval (not a minimum), and f(π) = 1 is small, so (c) and (d) are false.

Teacher's Note:
1. Evaluate the function on both sides of the break point.
2. Compare with the end-point values to test absolute extremes.

Answer: (c) 32, 32

Explanation:
1. Let the parts be x and 64 - x. \( S = x^3 + (64 - x)^3 \), \( S' = 3x^2 - 3(64 - x)^2 = 0 \Rightarrow x = 32 \).
2. \( S'' = 6x + 6(64 - x) = 384 > 0 \), so it is a minimum.

Teacher's Note:
1. With a fixed sum the sum of cubes is least when the parts are equal.
2. The second derivative is a positive constant.

Answer: (a) 2

Explanation:
1. Let \( t = 3^{x+1} > 0 \). Then \( f = t + \frac1t \ge 2 \) by AM-GM.
2. Equality holds at t = 1, i.e. x = -1, so the minimum is 2.

Teacher's Note:
1. Substitute t equals 3 to the power x + 1.
2. The sum of a positive number and its reciprocal is at least 2.

Answer: (b) \( \tan^{-1}\sqrt2 \)

Explanation:
1. Let the slant height be l and semi-vertical angle \( \alpha \). Then \( r = l\sin\alpha \), \( h = l\cos\alpha \), and \( V = \frac\pi3l^3\sin^2\alpha\cos\alpha \).
2. \( \frac{dV}{d\alpha} \propto 2\sin\alpha\cos^2\alpha - \sin^3\alpha = 0 \Rightarrow \tan^2\alpha = 2 \).
3. So \( \alpha = \tan^{-1}\sqrt2 \), and the second derivative test confirms a maximum.

Teacher's Note:
1. Express the volume in terms of the semi-vertical angle only.
2. The condition tan squared alpha = 2 is a standard result.

Answer: (d) Does not exist

Explanation:
1. Let \( t = \log_ax \). Then the expression is \( t + \frac1t \), where \( t \ne 0 \).
2. If \( 0 < a < 1 < x \), then t is negative and can take any negative value, so \( t + \frac1t \) is unbounded below.
3. Hence the expression has no minimum value.

Teacher's Note:
1. The condition 0 less than a less than x allows a to be less than 1.
2. The minimum of t + 1/t is 2 only when t is positive.