JEE Mathematics Maxima and Minima MCQs Set 06

Answer: (a) x = 2

Explanation:
1. \( f'(x) = 6x^2 - 30x + 36 = 6(x - 2)(x - 3) \).
2. \( f''(x) = 12x - 30 \): f''(2) = -6 < 0 (maximum) and f''(3) = 6 > 0 (minimum).
3. The maximum is at x = 2.

Teacher's Note:
1. Factorise the derivative to find the critical points.
2. The second derivative test tells maximum from minimum.

Answer: (c) \( \frac1e \)

Explanation:
1. \( \ln f = -x\ln x \Rightarrow \frac{f'}{f} = -(\ln x + 1) \).
2. f' = 0 gives \( x = \frac1e \). For x below it f' > 0 and above it f' < 0.
3. So the maximum is at \( x = \frac1e \).

Teacher's Note:
1. Logarithmic differentiation simplifies variable powers.
2. The sign of f' changes from positive to negative at the point.

Answer: (a) \( \pi \)

Explanation:
1. \( y' = a\sin x = 0 \Rightarrow x = n\pi \).
2. \( y'' = a\cos x \). At \( x = \pi \): \( y'' = -a < 0 \), a maximum with y = 2a.

Teacher's Note:
1. A maximum of 1 minus cos x occurs where cos x is -1.
2. The second derivative test confirms it.

Answer: (c) Neither maximum nor minimum

Explanation:
1. \( f'(x) = 1 + \cos x \ge 0 \) for all x, and it is zero only at isolated points.
2. So f is increasing on the whole real line.
3. An increasing function has no extremum.

Teacher's Note:
1. A derivative that never changes sign gives no extremum.
2. Isolated zeros of the derivative do not matter here.

Answer: (a) x = -1 is a minimum

Explanation:
1. \( y' = e^x(1 + x) = 0 \Rightarrow x = -1 \).
2. \( y'' = e^x(2 + x) \): at x = -1, \( y'' = e^{-1} > 0 \), a minimum.
3. At x = 0, y' = 1 is not zero, so there is no extremum.

Teacher's Note:
1. The factor e to the x is never zero.
2. Use the second derivative test at x = -1.

Answer: (d) None of these

Explanation:
1. Let the factors be x and \( \frac{36}{x} \). Then \( S = x + \frac{36}{x} \) and \( S' = 1 - \frac{36}{x^2} = 0 \Rightarrow x = 6 \).
2. \( S'' = \frac{72}{x^3} > 0 \), a minimum. The factors are 6 and 6, which are not in options (a), (b) or (c).

Teacher's Note:
1. Two positive numbers with a fixed product have the least sum when equal.
2. The numbers 6 and 6 are not listed.

Answer: (c) \( f'(x) = 0 \) but it is not sufficient

Explanation:
1. At an interior extremum where the derivative exists, \( f'(x) = 0 \).
2. The condition is not sufficient: for \( f(x) = x^3 \), f'(0) = 0 but there is no extremum.
3. So it is necessary but not sufficient.

Teacher's Note:
1. Necessary means it must hold, sufficient means it guarantees the result.
2. Keep the example x cubed in mind.

Answer: (a) 16, -39

Explanation:
1. \( f'(x) = 12x^3 - 24x^2 + 24x - 48 = 12(x - 2)(x^2 + 2) = 0 \Rightarrow x = 2 \).
2. f(1) = -16, f(2) = -39, f(3) = 16.
3. Maximum 16 and minimum -39.

Teacher's Note:
1. Factorise the derivative; the quadratic factor has no real root.
2. Compare with the end-point values.

Answer: (d) 4

Explanation:
1. \( f'(x) = 6(x - 2)(x + 1) \).
2. f(-2) = 1, f(-1) = 12, f(2) = -15, f(4) = 37.
3. The maximum value 37 is at x = 4.

Teacher's Note:
1. The greatest value on a closed interval can be at an end point.
2. Evaluate f at every candidate point.

Answer: (a) -2

Explanation:
1. \( x = 4 + 2y \), so \( xy = 4y + 2y^2 \).
2. The derivative \( 4 + 4y = 0 \Rightarrow y = -1 \), and the second derivative 4 > 0.
3. Minimum \( = -4 + 2 = -2 \).

Teacher's Note:
1. Eliminate x using the given relation.
2. The product is a quadratic in y with a positive leading coefficient.

Answer: (b) 4, 2

Explanation:
1. \( -1 \le \sin 4x \le 1 \), so \( 2 \le \sin 4x + 3 \le 4 \).
2. This is always positive, so the modulus does not change it.
3. The maximum is 4 and the minimum is 2.

Teacher's Note:
1. Bound the sine first.
2. The expression inside the modulus is always positive.

Answer: (b) 177

Explanation:
1. \( f'(x) = 3(x - 2)(x - 6) \).
2. f(1) = 42, f(2) = 49, f(6) = 17, f(10) = 177.
3. The maximum is 177, at the end point x = 10.

Teacher's Note:
1. Check critical points and end points.
2. Here the greatest value is at an end point.

Answer: (d) Neither a point of maximum nor a point of minimum

Explanation:
1. \( f'(x) = x(2\log x + 1) > 0 \) for every x in (1, e).
2. So the function is strictly increasing on the interval.
3. Hence it has no point of local maximum or minimum inside the interval.

Teacher's Note:
1. An increasing function has no interior extremum.
2. The end points are not in an open interval.

Answer: (b) 99, 1

Explanation:
1. Let the parts be x and y with \( x + y = 100 \). \( S = 2x + y^2 = 2(100 - y) + y^2 \).
2. \( \frac{dS}{dy} = -2 + 2y = 0 \Rightarrow y = 1 \), and the second derivative is 2 > 0.
3. So the parts are 99 and 1.

Teacher's Note:
1. Eliminate one part with the sum condition.
2. The second derivative confirms a minimum.

Answer: (a) \( (a - b)^2 \)

Explanation:
1. \( u^2 = a^2 + b^2 + 2\sqrt{P} \), where \( P = (a^2\cos^2\theta + b^2\sin^2\theta)(a^2\sin^2\theta + b^2\cos^2\theta) = a^2b^2 + \frac{(a^2 - b^2)^2}{4}\sin^22\theta \).
2. \( \sqrt P \) ranges from ab to \( \frac{a^2 + b^2}{2} \). So \( u^2 \) ranges from \( (a + b)^2 \) to \( 2(a^2 + b^2) \).
3. Difference \( = 2(a^2 + b^2) - (a + b)^2 = (a - b)^2 \).

Teacher's Note:
1. Expand the product under the root and use sin 2 theta to find its range.
2. The maximum and minimum of sin squared 2 theta are 1 and 0.

Answer: (a) 12

Explanation:
1. By AM-GM, \( 2x + 3y \ge 2\sqrt{6xy} = 2\sqrt{36} = 12 \).
2. Equality holds when 2x = 3y, i.e. x = 3 and y = 2.
3. The minimum value is 12.

Teacher's Note:
1. AM-GM gives the minimum when the two terms are equal.
2. Check that the equality case satisfies xy = 6.

Answer: (c) 1

Explanation:
1. \( f(x) = x + \frac1x \), \( f'(x) = 1 - \frac{1}{x^2} = 0 \Rightarrow x = 1 \) (for x > 0).
2. \( f''(1) = 2 > 0 \), a minimum with value 2.

Teacher's Note:
1. Restrict to positive x, since for negative x the sum has a maximum instead.
2. AM-GM also gives x + 1/x at least 2.

Answer: (a) x = 1

Explanation:
1. \( f'(x) = e^x(1 - x)^2 - 2e^x(1 - x) = e^x(1 - x)(-1 - x) = e^x(x - 1)(x + 1) \).
2. f' is negative on (-1, 1) and positive for x > 1.
3. So f has a minimum at x = 1 (and a maximum at x = -1).

Teacher's Note:
1. Factorise the derivative and study its sign.
2. The minimum occurs where f' changes from negative to positive.

Answer: (c) 54

Explanation:
1. \( f'(x) = 3(x - 3)(x - 5) \). The sign of f' changes from + to - at x = 3.
2. So the local maximum is at x = 3: \( f(3) = 27 - 108 + 135 = 54 \).

Teacher's Note:
1. A local maximum is where the derivative changes from positive to negative.
2. The local minimum is at x = 5.

Answer: (a) A local maximum

Explanation:
1. For \( 0 < |x| < 1 \), \( f(x) = |x| < 1 = f(0) \).
2. So f(0) is greater than all nearby values, which makes x = 0 a local maximum.

Teacher's Note:
1. Compare f(0) with values in a small neighbourhood.
2. The function is not continuous at 0, so the derivative tests do not apply.

Answer: (b) \( x = \frac1e \)

Explanation:
1. \( \ln y = x\ln x \Rightarrow \frac{y'}{y} = \ln x + 1 \).
2. y' = 0 gives \( \ln x = -1 \Rightarrow x = \frac1e \).

Teacher's Note:
1. Use logarithmic differentiation for a variable base and exponent.
2. The stationary point is a minimum.