JEE Mathematics Maxima and Minima MCQs Set 05

Answer: (b) 12

Explanation:
1. Slope \( g(x) = y' = -3x^2 + 6x + 9 \).
2. \( g'(x) = -6x + 6 = 0 \Rightarrow x = 1 \), and \( g'' = -6 < 0 \), so g is maximum there.
3. \( g(1) = -3 + 6 + 9 = 12 \).

Teacher's Note:
1. To maximise the slope, treat the slope as a new function and differentiate it.
2. The second derivative of the slope is negative, so the extremum is a maximum.

Answer: (b) x = 1 and x = 3

Explanation:
1. \( f'(x) = x(e^x - 1)(x - 1)(x - 2)^3(x - 3)^5 \), with zeros at 0, 1, 2, 3.
2. At x = 0, \( x(e^x - 1) \ge 0 \) on both sides, so f' does not change sign: no extremum.
3. At x = 1 the sign changes from - to +, so it is a minimum. At x = 2 it changes from + to -, a maximum. At x = 3 it changes from - to +, a minimum.

Teacher's Note:
1. Local minima occur where the derivative changes from negative to positive.
2. The factor x(e^x - 1) is non-negative, so x = 0 is not an extremum.

Answer: (c) 2

Explanation:
1. \( f'(x) = 6x^2 - 18ax + 12a^2 = 6(x - a)(x - 2a) \), \( f''(x) = 12x - 18a \).
2. f''(a) = -6a < 0 gives the maximum at p = a, and f''(2a) = 6a > 0 gives the minimum at q = 2a.
3. \( p^2 = q \Rightarrow a^2 = 2a \Rightarrow a = 2 \) (a = 0 is rejected because a > 0).

Teacher's Note:
1. Find which critical point is the maximum before using the given relation.
2. The condition a greater than 0 removes the root a = 0.

Answer: (b) x = 1 and x = -1

Explanation:
1. \( \phi'(x) = e^{-x^2/2}(1 - x^2) \).
2. \( \phi'(x) = 0 \Rightarrow 1 - x^2 = 0 \Rightarrow x = \pm1 \).
3. The factor \( 1 - x^2 \) changes sign at both points, so both are points of extremum.

Teacher's Note:
1. The derivative of an integral with a variable upper limit is the integrand.
2. A simple root of the derivative always gives a sign change.

Answer: (c) 160 and 128

Explanation:
1. \( y' = 3x^2 - 36x + 96 = 3(x - 4)(x - 8) \).
2. \( y'' = 6x - 36 \): at x = 4, y'' = -12 < 0, a local maximum \( y = 64 - 288 + 384 = 160 \). At x = 8, y'' = 12 > 0, a local minimum \( y = 512 - 1152 + 768 = 128 \).

Teacher's Note:
1. Use the second derivative test at each critical point.
2. Local extrema need not be the greatest or least values of the function.

Answer: (d) -3

Explanation:
1. Let \( c = \cos 2x \in [-1, 1] \). Then \( y = 2c - (2c^2 - 1) = -2c^2 + 2c + 1 \).
2. This downward parabola in c has its minimum on [-1, 1] at an end point: y(-1) = -2 - 2 + 1 = -3 and y(1) = 1.
3. The minimum value is -3, attained at \( x = \frac\pi2 \).

Teacher's Note:
1. Substituting c = cos 2x reduces the problem to a quadratic on a closed interval.
2. For a downward parabola the minimum lies at an end point.

Answer: (a) 2

Explanation:
1. \( y = \frac1x \), so \( z = x + \frac1x \) and \( z' = 1 - \frac{1}{x^2} = 0 \Rightarrow x = 1 \) (since x > 0).
2. \( z'' = \frac{2}{x^3} > 0 \) at x = 1, so z is a minimum.
3. The minimum value is 1 + 1 = 2.

Teacher's Note:
1. Eliminate one variable using the given product.
2. The AM-GM inequality gives the same result.

Answer: (c) 1

Explanation:
1. Let the numbers be x and 4 - x with 0 < x < 4. The sum of reciprocals is \( \frac{4}{x(4 - x)} \).
2. This is minimum when \( x(4 - x) \) is maximum, which happens at x = 2.
3. The minimum value is \( \frac{4}{4} = 1 \).

Teacher's Note:
1. A fraction with a fixed numerator is smallest when the denominator is largest.
2. The product of two positive numbers with a fixed sum is greatest when they are equal.

Answer: (b) \( \frac{1}{\sqrt3} \)

Explanation:
1. Maximise \( f(x) = x - x^3 \). \( f'(x) = 1 - 3x^2 = 0 \Rightarrow x = \pm\frac{1}{\sqrt3} \).
2. \( f''(x) = -6x < 0 \) at \( x = +\frac{1}{\sqrt3} \), so this is the maximum.
3. The number is \( \frac{1}{\sqrt3} \).

Teacher's Note:
1. Excess of a number over its cube is x minus x cubed.
2. The positive root is the maximum, the negative root is the minimum.

Answer: (b) \( \sqrt{\frac23}R \)

Explanation:
1. Let r and h be the radius and height. \( r^2 + \frac{h^2}{4} = R^2 \Rightarrow h = 2\sqrt{R^2 - r^2} \).
2. \( V = \pi r^2h = 2\pi r^2\sqrt{R^2 - r^2} \). Setting \( \frac{dV}{dr} = 0 \): \( 2(R^2 - r^2) = r^2 \Rightarrow r^2 = \frac23R^2 \).
3. So \( r = \sqrt{\frac23}R \), and \( \frac{d^2V}{dr^2} < 0 \) there, confirming a maximum.

Teacher's Note:
1. Use the right triangle formed by the radius, half the height and the sphere radius.
2. Maximise V as a function of r only.

Answer: (c) \( \frac16 \)

Explanation:
1. \( f'(x) = \frac{4 - x^2}{(4 + x + x^2)^2} \), which is positive on [-1, 1].
2. So f is increasing and its maximum is at the right end point: \( f(1) = \frac16 \).

Teacher's Note:
1. When there is no critical point inside the interval, check the sign of the derivative.
2. An increasing function attains its maximum at the right end point.

Answer: (a) 2

Explanation:
1. \( f'(x) = (1 - x)(1 - 3x) = 0 \Rightarrow x = \frac13 \) or 1, giving \( f\left(\frac13\right) = \frac{4}{27} \) and f(1) = 0.
2. End points: f(0) = 0 and f(2) = 2.
3. The greatest value on [0, 2] is 2.

Teacher's Note:
1. On a closed interval compare critical values with the end-point values.
2. A local maximum is not always the greatest value.

Answer: (d) 139

Explanation:
1. \( f'(x) = 6x^2 - 24 = 0 \Rightarrow x = \pm2 \).
2. f(-3) = 125, f(-2) = 139, f(2) = 75, f(3) = 89.
3. The maximum is 139.

Teacher's Note:
1. Evaluate at every critical point and both end points.
2. Choose the largest of these values.

Answer: (a) 120

Explanation:
1. \( f'(x) = 4x^3 - 124x + a = 0 \) at x = 1: \( 4 - 124 + a = 0 \Rightarrow a = 120 \).
2. Check: \( f''(1) = 12 - 124 < 0 \), so x = 1 is indeed a maximum.

Teacher's Note:
1. A maximum at an interior point needs a zero first derivative there.
2. Confirm with the second derivative test.

Answer: (a) -128

Explanation:
1. \( y' = 6x^2 - 42x + 36 = 6(x - 1)(x - 6) \).
2. \( y'' = 12x - 42 \): at x = 6, y'' = 30 > 0, a local minimum.
3. \( y(6) = 432 - 756 + 216 - 20 = -128 \).

Teacher's Note:
1. The cubic has one local maximum (x = 1) and one local minimum (x = 6).
2. Use the second derivative test to choose the minimum.

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

Explanation:
1. \( f'(x) = \frac{(1 + x\tan x) - x(\tan x + x\sec^2x)}{(1 + x\tan x)^2} = \frac{1 - x^2\sec^2x}{(1 + x\tan x)^2} \).
2. f' = 0 gives \( x\sec x = 1 \) for x > 0, i.e. \( x = \cos x \).
3. For x just below this value f' > 0 and just above it f' < 0, so it is a maximum.

Teacher's Note:
1. Differentiate using the quotient rule and simplify the numerator.
2. The sign change of the derivative confirms the maximum.

Answer: (c) \( -\frac98 \)

Explanation:
1. The vertex is at \( x = -\frac14 \).
2. Value \( = 2\cdot\frac{1}{16} - \frac14 - 1 = \frac18 - \frac14 - 1 = -\frac98 \).

Teacher's Note:
1. The minimum of ax² + bx + c is c minus b²/(4a).
2. Here that is -1 minus 1/8.

Answer: (b) 16

Explanation:
1. \( y = 8 - x \), so \( xy = 8x - x^2 \).
2. \( \frac{d}{dx} = 8 - 2x = 0 \Rightarrow x = 4 \), and the second derivative is negative.
3. Maximum \( xy = 4 \times 4 = 16 \).

Teacher's Note:
1. Eliminate y using the given sum.
2. Two numbers with a fixed sum have the greatest product when they are equal.

Answer: (a) \( \frac12 \)

Explanation:
1. \( B = \frac\pi2 - A \), so \( \cos A\cos B = \cos A\sin A = \frac12\sin 2A \).
2. The maximum of \( \sin 2A \) is 1, so the maximum is \( \frac12 \).

Teacher's Note:
1. Use the complementary-angle identity cos(pi/2 minus A) = sin A.
2. Then the double-angle formula reduces the product to one sine.

Answer: (b) \( 2c\sqrt{ab} \)

Explanation:
1. With \( y = \frac{c^2}{x} \): \( f(x) = ax + \frac{bc^2}{x} \).
2. \( f'(x) = a - \frac{bc^2}{x^2} = 0 \Rightarrow x = c\sqrt{\frac ba} \), and f'' > 0 there.
3. Minimum \( = ac\sqrt{\frac ba} + bc\sqrt{\frac ab} = 2c\sqrt{ab} \).

Teacher's Note:
1. This is the AM-GM inequality in disguise.
2. The two terms are equal at the minimum.

Answer: (c) \( \frac{c^3}{\sqrt{2ab}} \)

Explanation:
1. By AM-GM, \( c^6 = a^2x^4 + b^2y^4 \ge 2abx^2y^2 \).
2. So \( x^2y^2 \le \frac{c^6}{2ab} \), giving \( xy \le \frac{c^3}{\sqrt{2ab}} \).
3. Equality holds when \( ax^2 = by^2 \), so the maximum is attained.

Teacher's Note:
1. Use AM-GM on the two fourth-power terms.
2. Check that equality is possible so that the bound is the true maximum.