JEE Mathematics Maxima and Minima MCQs Set 07

Answer: (d) 0

Explanation:
1. \( \int_0^x te^{-t^2}dt = \frac12\left(1 - e^{-x^2}\right) \).
2. Since \( e^{-x^2} \le 1 \), the value is at least 0, with equality at x = 0.
3. The minimum value is 0.

Teacher's Note:
1. Integrate first, then bound the result.
2. The derivative of the integral is x e to the minus x squared, which changes sign at 0.

Answer: (a) \( x = \tan^{-1}\sqrt{\frac pq} \)

Explanation:
1. \( f'(x) = \sin^{p-1}x\cos^{q-1}x\,[p\cos^2x - q\sin^2x] = 0 \).
2. So \( \tan^2x = \frac pq \Rightarrow x = \tan^{-1}\sqrt{\frac pq} \).
3. f' changes from positive to negative there, so it is a maximum.

Teacher's Note:
1. Factorise the derivative.
2. The bracket changes sign from positive to negative at the point.

Answer: (c) 1

Explanation:
1. \( 2x^2 - 2x + 1 = 2\left(x - \frac12\right)^2 + \frac12 > 0 \) and \( \sin^2x \ge 0 \), so the exponent is at least 0.
2. The exponent is 0 at x = 0, so the minimum value is \( e^0 = 1 \).

Teacher's Note:
1. Find the minimum of the exponent first, because e to the power t increases with t.
2. A positive quadratic times sin squared is never negative.

Answer: (b) (5, 15)

Explanation:
1. Let the cubed part be y, so the other part is 20 - y and \( P = (20 - y)y^3 \).
2. \( P' = y^2[3(20 - y) - y] = y^2(60 - 4y) = 0 \Rightarrow y = 15 \).
3. P' changes from positive to negative at y = 15, so the parts are 5 and 15.

Teacher's Note:
1. The part with the larger power gets the larger share.
2. The sign change of P' confirms the maximum.

Answer: (d) 1

Explanation:
1. \( \sqrt3\cos x + \sin x = 2\cos\left(x - \frac\pi6\right) \), so its minimum is -2.
2. The minimum exponent is \( 2 - 2 = 0 \), so the minimum value is \( e^0 = 1 \).

Teacher's Note:
1. Combine the sine and cosine into one cosine.
2. The exponential is smallest when its exponent is smallest.

Answer: (d) Does not exist

Explanation:
1. \( f'(x) = \frac{1 - \log x}{x^2} \), which is zero at x = e. f increases up to x = e and decreases afterwards.
2. As x tends to infinity, f tends to 0 from above but never reaches it.
3. On \( [2, \infty) \) the values are all positive and get arbitrarily close to 0, so no minimum is attained.

Teacher's Note:
1. Study the behaviour of the function at infinity.
2. An infimum that is not attained is not a minimum.

Answer: (d) \( \sqrt{\frac ba} \)

Explanation:
1. \( f'(x) = a - \frac{b}{x^2} = 0 \Rightarrow x = \sqrt{\frac ba} \).
2. \( f''(x) = \frac{2b}{x^3} > 0 \), so it is a minimum.

Teacher's Note:
1. Equate the derivative to zero.
2. AM-GM gives the same point where the two terms are equal.

Answer: (b) \( -\frac18 \)

Explanation:
1. \( f'(x) = \frac{16 - x^2}{(x^2 + 16)^2} = 0 \Rightarrow x = \pm4 \).
2. f' is negative for x < -4 and positive for -4 < x < 4, so x = -4 is a minimum.
3. \( f(-4) = \frac{-4}{32} = -\frac18 \).

Teacher's Note:
1. Take both signs of x into account when asking for the least value.
2. The positive critical point gives the maximum, one eighth.

Answer: (a) A minimum

Explanation:
1. This is the second derivative test: f'(a) = 0 and f''(a) > 0 means the graph is concave upward at a.
2. So f has a local minimum at x = a.

Teacher's Note:
1. A positive second derivative means concave up.
2. A negative second derivative would give a maximum.

Answer: (b) 2

Explanation:
1. For x > 0, \( x + \frac1x - 2 = \left(\sqrt x - \frac{1}{\sqrt x}\right)^2 \ge 0 \).
2. So the sum is at least 2, with equality at x = 1.

Teacher's Note:
1. AM-GM gives the same result.
2. The value 2 is attained, so it is the least value.

Answer: (d) 8, 8

Explanation:
1. \( y = 16 - x \), so \( S = x^2 + (16 - x)^2 \).
2. \( S' = 2x - 2(16 - x) = 4x - 32 = 0 \Rightarrow x = 8 \), and S'' = 4 > 0.
3. So x = y = 8.

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

Answer: (c) \( \frac1e \)

Explanation:
1. \( f'(x) = \frac{1 - \log x}{x^2} = 0 \Rightarrow x = e \).
2. f' changes from positive to negative at e, so it is a maximum.
3. \( f(e) = \frac{1}{e} \).

Teacher's Note:
1. This function is a standard example worth remembering.
2. The maximum value is 1 over e at x = e.

Answer: (b) 0, -4

Explanation:
1. \( f'(x) = (x + 2)^2 + 2(x - 1)(x + 2) = (x + 2)(3x) \), zero at x = -2 and x = 0.
2. f' changes from + to - at x = -2 (a local maximum, f = 0) and from - to + at x = 0 (a local minimum, f = -4).

Teacher's Note:
1. Factorise the derivative and track its sign.
2. The local maximum value is 0 and the local minimum is -4.

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

Explanation:
1. \( f'(x) = 6(x - 1)(x - 6) \) and \( f''(x) = 12x - 42 \).
2. f''(1) = -30 < 0, so there is a maximum at x = 1. f''(6) = 30 > 0, so there is a minimum at x = 6.

Teacher's Note:
1. Use the second derivative test at both critical points.
2. Options (a) and (b) swap the maximum and the minimum.

Answer: (d) \( |c| > |b|\sqrt2 \)

Explanation:
1. The minimum of f is \( 2c^2 - b^2 \) (at x = -b). The maximum of g is \( b^2 + c^2 \) (at x = -c).
2. \( 2c^2 - b^2 > b^2 + c^2 \Rightarrow c^2 > 2b^2 \).
3. So \( |c| > |b|\sqrt2 \).

Teacher's Note:
1. Use the vertex form of each quadratic to find its extreme value.
2. Compare the extreme values to get the required inequality.

Answer: (b) One maximum and one minimum

Explanation:
1. \( f'(x) = 6(x - 2)(x + 1) \), which changes sign at both roots.
2. f' goes from + to - at x = -1 (maximum) and from - to + at x = 2 (minimum).
3. So there is one maximum and one minimum.

Teacher's Note:
1. A cubic with two distinct critical points has one maximum and one minimum.
2. Check the sign changes of f'.

Answer: (a) \( \frac43 \)

Explanation:
1. \( 4x^2 + 2x + 1 = 4\left(x + \frac14\right)^2 + \frac34 \ge \frac34 \).
2. So f is maximum when the denominator is minimum: \( f_{max} = \frac{1}{3/4} = \frac43 \).

Teacher's Note:
1. Minimise the denominator to maximise a reciprocal.
2. Complete the square in the denominator.

Answer: (d) \( \frac\pi2 \)

Explanation:
1. Area \( = \frac12(PQ)(PR)\sin\theta \) with PQ and PR fixed.
2. The maximum of \( \sin\theta \) is 1, at \( \theta = \frac\pi2 \).

Teacher's Note:
1. Area with two fixed sides depends only on the included angle.
2. A right angle gives the largest area.

Answer: (d) \( \frac p4 \)

Explanation:
1. With radius r and arc length l: \( p = 2r + l \), so \( l = p - 2r \).
2. Area \( = \frac12rl = \frac12r(p - 2r) = \frac{pr}{2} - r^2 \).
3. \( \frac{dA}{dr} = \frac p2 - 2r = 0 \Rightarrow r = \frac p4 \), and the second derivative is -2 < 0.

Teacher's Note:
1. Perimeter of a sector includes both radii and the arc.
2. The area is a downward parabola in r.

Answer: (d) All of these

Explanation:
1. On [-1, 2], \( f'(x) = 6x + 12 > 0 \), so f is increasing.
2. At x = 2 the left value is 12 + 24 - 1 = 35 and the right value is 37 - 2 = 35, so f is continuous at 2.
3. For x > 2, \( f(x) = 37 - x \) is decreasing from 35, so f(2) = 35 is the greatest value on [-1, 3]. All three statements are true.

Teacher's Note:
1. Check continuity at the joining point first.
2. Study each piece separately for monotonicity.

Answer: (c) x = a is not a point of maximum or minimum

Explanation:
1. The factor \( (x - a)^{2n} \) is non-negative on both sides of a, so f' does not change sign at x = a.
2. Hence x = a is neither a maximum nor a minimum.
3. (At x = b the sign of f' changes from negative to positive, so b is a minimum.)

Teacher's Note:
1. An even power of a factor means no sign change at that root.
2. An odd power means the sign changes.