JEE Mathematics Maxima and Minima MCQs Set 12

Answer: (c) 2abc

Explanation:
1. S = b²x + c²a²/x.
2. By AM–GM, S ≥ 2√(b²x · c²a²/x) = 2abc, attained when b²x = c²a²/x.

Teacher's Note:
1. Apply AM–GM to the two terms whose product is constant.

Answer: (c) 4

Explanation:
1. f′(x) = 4x³ − 2x − 2 = 2(x − 1)(2x² + 2x + 1), and 2x² + 2x + 1 > 0, so x = 1 is the only critical point.
2. f″(1) = 12 − 2 = 10 > 0, and f → ∞ at both ends, so the global minimum is f(1) = 1 − 1 − 2 + 6 = 4.

Teacher's Note:
1. Find the single critical point and check the end behaviour.

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

Explanation:
1. a sec θ − b tan θ = (a − b sin θ)/cos θ. Let it equal k, so a − b sin θ = k cos θ, i.e. k cos θ + b sin θ = a.
2. This has a solution only if k² + b² ≥ a², so k ≥ √(a² − b²) (k > 0).
3. The minimum value is √(a² − b²).

Teacher's Note:
1. Write the expression as R cos(θ − φ) type equation in k.

Answer: (b) 1

Explanation:
1. The maximum possible value is 2, needing cos x = 1 and cos √2x = 1 together.
2. x = 2πm and √2 x = 2πn would give √2 = n/m, impossible unless m = n = 0.
3. So the maximum is attained only at x = 0: exactly 1 value.

Teacher's Note:
1. Irrationality of √2 limits the simultaneous solutions.

Answer: (a) π/2

Explanation:
1. Slope s = eˣ(sin x + cos x), s′ = 2eˣ cos x = 0 at x = π/2 and 3π/2.
2. s″ = 2eˣ(cos x − sin x) is negative at π/2 and positive at 3π/2.
3. So the local maximum is at a = π/2.

Teacher's Note:
1. Differentiate the slope again and use the second derivative test.

Answer: (a) a maximum

Explanation:
1. f′ changes sign from positive to negative at a.
2. The function increases and then decreases, so f(a) is a maximum.

Teacher's Note:
1. First derivative test.

Answer: (c) f(x) has a minimum at x = 1

Explanation:
1. f′ = 3(x + 3)(x − 1), f″ = 6x + 6.
2. f″(1) = 12 > 0, so x = 1 is a minimum; f″(−3) = −12 < 0, so x = −3 is a maximum.

Teacher's Note:
1. Second derivative test at the roots of f′.

Answer: (b) \( \sqrt[e]{e} \)

Explanation:
1. Let y = (1/x)^(2x²), ln y = −2x² ln x.
2. d/dx: −4x ln x − 2x = 0 gives ln x = −1/2, x = e^(−1/2).
3. ln y = −2e⁻¹(−1/2) = 1/e, so y = e^(1/e), the e-th root of e.

Teacher's Note:
1. Maximise the logarithm of the function.

Answer: (a) π/6

Explanation:
1. sin θ sin φ = ½[cos(θ − φ) − cos(θ + φ)] = ½[cos(θ − φ) − 1/2].
2. It is maximum when θ = φ, so θ = φ = π/6.

Teacher's Note:
1. Convert the product into a difference of cosines.

Answer: (d) neither a maximum nor a minimum

Explanation:
1. f′(x) = 3(x − 2)² ≥ 0, which vanishes at x = 2 without changing sign.
2. So f is increasing and x = 2 is a point of inflection, neither a maximum nor a minimum.

Teacher's Note:
1. The first derivative does not change sign at 2.

Answer: (a) \( \frac{p+q+r}{3} \)

Explanation:
1. f′ = 2[(x − p) + (x − q) + (x − r)] = 0 gives x = (p + q + r)/3.
2. f″ = 6 > 0, so it is a minimum.

Teacher's Note:
1. The sum of squared deviations is least at the mean.

Answer: (b) exactly one minimum

Explanation:
1. f′(x) = 4x + 16x³ + … = 2x(2 + 8x² + …) where the bracket is positive.
2. So f′ = 0 only at x = 0, with f′ negative before and positive after: exactly one minimum.

Teacher's Note:
1. All the powers are even with positive coefficients.

Answer: (c) must have neither a maximum nor a minimum

Explanation:
1. At a local extremum of a differentiable function f′ = 0.
2. Since f′(a) ≠ 0, f has neither a local maximum nor a local minimum at a.

Teacher's Note:
1. Necessary condition for an extremum: f′(a) = 0.

Answer: (a) 1

Explanation:
1. f(1) = 2. Just left of 1, f ≈ sin⁻¹λ + 1, and for x > 1, f = 2x > 2.
2. A local minimum needs sin⁻¹λ + x² ≥ 2 near x = 1 from the left, i.e. sin⁻¹λ ≥ 1, which holds for λ = 1 (π/2 ≈ 1.57) but not for λ = 0 or −1.

Teacher's Note:
1. Compare f near 1 on the left with f(1) = 2.

Answer: (a) \( \frac12 \)

Explanation:
1. Maximise x − x²: derivative 1 − 2x = 0 gives x = 1/2 and second derivative −2 < 0.

Teacher's Note:
1. Quadratic with negative leading coefficient is maximised at the vertex.

Answer: (b) \( \frac12 \)

Explanation:
1. 1/a + 1/b = 8/(ab), and ab ≤ 16 by AM–GM (a = b = 4).
2. So the minimum is 8/16 = 1/2.

Teacher's Note:
1. Minimising the sum of reciprocals means maximising the product.

Answer: (c) (2, 2)

Explanation:
1. For a point (x, y) on the curve, D² = x² + (y − 3)² = 2y + (y − 3)² = y² − 4y + 9.
2. This is least at y = 2, giving x² = 4, so x = ±2.
3. The point (2, 2) is one of the nearest points.

Teacher's Note:
1. Eliminate x using the curve and minimise D² in y.

Answer: (b) three distinct real roots

Explanation:
1. A cubic with two distinct turning points has local maximum and minimum values f(α) and f(β).
2. Opposite signs of f(α) and f(β) mean the graph crosses the x-axis on each side and between them: three distinct real roots.

Teacher's Note:
1. Count crossings using the signs of the local maximum and minimum values.

Answer: (d) 2 and 3 only

Explanation:
1. f′(x) = (1 − x²)e^(−x²/2).
2. It vanishes at x = ±1 and changes sign there (− to + at −1, + to − at 1).
3. At x = 0, f′ = 1 ≠ 0: no extremum.

Teacher's Note:
1. Fundamental theorem of calculus, then sign analysis of f′.

Answer: (d) f′(x) has a minimum

Explanation:
1. f′ = 3x² + 6x + 2, whose derivative f″ = 6x + 6 changes from negative to positive at x = −1.
2. So f′ has a minimum at x = −1 (f itself has a point of inflection).

Teacher's Note:
1. The minimum of f′ occurs where f″ = 0 and f‴ > 0.

Answer: (a) 1 and 3 only

Explanation:
1. Near x = 1, (x − 1)⁴ ≥ 0 and (x − 2)ⁿ has sign (−1)ⁿ: for odd n, f ≤ 0 = f(1), a maximum; for even n, f ≥ 0, a minimum.
2. Near x = 2, (x − 1)⁴ > 0: for even n, (x − 2)ⁿ ≥ 0 gives a minimum; for odd n the sign changes, so no extremum.

Teacher's Note:
1. Determine the sign of f near each point.