JEE Mathematics Application of Derivatives MCQs Set 09

Answer: (c) \( \cot^{-1}\sqrt2 \)

Explanation:
1. Height \( = a\sin\alpha \) and half the base diagonal \( = a\cos\alpha \), so the square base has area \( 2a^2\cos^2\alpha \).
2. \( V = \frac13 \cdot 2a^2\cos^2\alpha \cdot a\sin\alpha = \frac23 a^3 \sin\alpha\cos^2\alpha \).
3. \( \frac{dV}{d\alpha} = 0 \Rightarrow \cos^2\alpha = 2\sin^2\alpha \Rightarrow \cot\alpha = \sqrt2 \).

Teacher's Note:
1. A regular rectangular pyramid has a square base.
2. The lateral edge is the hypotenuse of a triangle with the height and the half diagonal.

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

Explanation:
1. Let the altitude be h and the circumradius of the base be R, so \( R^2 + h^2 = l^2 \).
2. The base area is \( \frac{3\sqrt3}{4}R^2 \), so \( V = \frac{3\sqrt3}{4}(l^2 - h^2)h \).
3. \( \frac{dV}{dh} = 0 \Rightarrow l^2 = 3h^2 \Rightarrow h = \frac{l}{\sqrt3} \).

Teacher's Note:
1. The area of an equilateral triangle in terms of its circumradius is \( \frac{3\sqrt3}{4}R^2 \).
2. Maximising \( (l^2 - h^2)h \) is a simple cubic problem.

Answer: (c) 3

Explanation:
1. Both pieces give \( f(1) = 1 \), so f is continuous.
2. For \( x < 1 \), \( f'(x) = \frac35 x^{-2/5} \) does not exist at x = 0. At x = 1 the left slope is \( \frac35 \) and the right slope is \( -3 \), so f' does not exist there either.
3. For \( x > 1 \), \( f'(x) = -3(x - 2)^2 = 0 \) at x = 2. So the critical points are 0, 1 and 2.

Teacher's Note:
1. Critical points include places where the derivative does not exist.
2. Always compare the one-sided slopes at the joining point.

Answer: (b) 4

Explanation:
1. The function is even, so count positive roots. For \( u = |x| > 0 \), take logs: \( g(u) = 2\ln u + 2 - u = 0 \).
2. \( g'(u) = \frac2u - 1 = 0 \) at u = 2, with \( g(2) = 2\ln 2 > 0 \), while \( g(u) \to -\infty \) as \( u \to 0^+ \) and as \( u \to \infty \).
3. So there are 2 positive roots and, by symmetry, 2 negative roots: 4 in all.

Teacher's Note:
1. Taking logarithms simplifies the product of powers and exponentials.
2. The sign of g at its maximum and at the two ends tells the number of roots.

Answer: (b) \( (3, -15) \)

Explanation:
1. At \( (h, k) \), the slope is \( m = 3h^2 - 6h - 7 \). The intercepts are \( x_0 = h - \frac km \) and \( y_0 = k - mh \).
2. We need \( x_0 > 0 \) and \( y_0 < 0 \) with \( x_0 = \frac{|y_0|}{2} \).
3. At (3, -15): \( m = 2 \), \( x_0 = \frac{21}{2} \), \( y_0 = -21 \), and \( 10.5 = \frac{21}{2} \) holds. At (-1, 9) the x-intercept is negative and at (1, -3) the y-intercept is positive.

Teacher's Note:
1. Write both intercepts in terms of the point of tangency.
2. Test each option against the sign conditions as well as the ratio.

Answer: (c) \( x < -1 \)

Explanation:
1. Zero gradient at (0, 1) gives \( c = 0 \) and \( d = 1 \).
2. Touching the x-axis at (-1, 0) gives \( a - b + 1 = 0 \) and \( y'(-1) = -4a + 3b = 0 \), so \( a = 3 \), \( b = 4 \).
3. \( y' = 12x^3 + 12x^2 = 12x^2(x + 1) < 0 \) exactly when \( x < -1 \).

Teacher's Note:
1. Touching the axis means both y and y' vanish at that point.
2. Factorise the derivative to read its sign.

Answer: (c) \( \frac{2}{3\sqrt3} \)

Explanation:
1. \( f'(x) = 3x^2 - 6x + 2 = 0 \Rightarrow x = 1 \pm \frac{1}{\sqrt3} \). The local maximum value is \( f\left(1 - \frac{1}{\sqrt3}\right) = \frac{2}{3\sqrt3} \).
2. For \( k = \frac{2}{3\sqrt3} \) the line y = k touches the curve at the local maximum (a repeated positive root) and cuts the left branch once at a negative x.
3. That gives exactly one positive and one negative solution. For the other options the count is different.

Teacher's Note:
1. A repeated root occurs where the horizontal line touches a local extremum.
2. Sketch the cubic: it has a local maximum at a positive x and a local minimum further right.

Answer: (d) 21

Explanation:
1. By the Mean Value Theorem, \( f(5) - f(-2) = 7f'(c) \) for some c in (-2, 5).
2. Since \( f'(c) \le 3 \), \( f(5) - f(-2) \le 21 \).
3. The value 21 is reached by \( f(x) = 3x \).

Teacher's Note:
1. The Mean Value Theorem links the change in f to the derivative bounds.
2. Use the upper bound of f' to get the greatest possible value.

Answer: (c) (1, 4)

Explanation:
1. f is increasing, so the inequality is equivalent to \( g(\alpha^2 - 2\alpha) > g(3\alpha - 4) \).
2. g is decreasing, so this means \( \alpha^2 - 2\alpha < 3\alpha - 4 \), i.e. \( \alpha^2 - 5\alpha + 4 < 0 \).
3. \( (\alpha - 1)(\alpha - 4) < 0 \Rightarrow \alpha \in (1, 4) \).

Teacher's Note:
1. Remove f and g one at a time, reversing the inequality for the decreasing function.
2. The sign of the final quadratic gives the interval.

Answer: (d) (2e, 0)

Explanation:
1. Touching the x-axis needs \( y = 0 \) and \( \frac{dy}{dt} = 0 \): \( -1 + ae^{at} = 0 \Rightarrow e^{at} = \frac1a \).
2. \( y = 0 \Rightarrow -t + \frac1a = 0 \Rightarrow t = \frac1a \). Then \( e^{at} = e = \frac1a \), so \( a = \frac1e \) and \( t = e \).
3. \( x = t + e^{at} = e + e = 2e \), so A = (2e, 0).

Teacher's Note:
1. Horizontal tangent on the x-axis needs both y = 0 and dy/dt = 0.
2. Solve for the parameter a before finding the point.

Answer: (d) The function is one-one in every continuous interval [a, b] defined on one side of the origin.

Explanation:
1. \( f'(x) = 1 + \frac{1}{x^2} > 0 \), so f is increasing on each of \( (-\infty, 0) \) and \( (0, \infty) \) separately, hence one-one on any interval on one side of the origin.
2. (a) is false: f(-1) = f(1) = 0, so f is not one-one on \( R - \{0\} \). (b) is false: \( f(-0.1) > f(0.1) \).
3. (c) is false: x = 0 and y = x are both asymptotes.

Teacher's Note:
1. A positive derivative on each branch makes the function monotonic on each branch.
2. The two branches together are not monotonic.

Answer: (c) 4

Explanation:
1. \( f(x) = \frac{(x^2 - 1)^2 - (x - 1)^2}{2} \), so \( f'(x) = (x^2 - 1)2x - (x - 1) = (x - 1)(2x^2 + 2x - 1) \).
2. For \( 1 < x \le 2 \), \( 2x^2 + 2x - 1 > 0 \), so f' > 0 and f is increasing.
3. The maximum is at x = 2: \( f(2) = \frac{9 - 1}{2} = 4 \).

Teacher's Note:
1. Use the fundamental theorem of calculus with variable limits.
2. An increasing function reaches its maximum at the right end point.

Answer: (b) \( \frac{3\sqrt3}{32} \)

Explanation:
1. Take the diameter from \( (-\frac12, 0) \) to \( (\frac12, 0) \), one vertex at the end of the diameter, and the other at \( P = (\frac12\cos\theta, \frac12\sin\theta) \) on the arc, with the right angle at the foot of the perpendicular from P.
2. Area \( = \frac12\left(\frac12 + \frac12\cos\theta\right)\frac12\sin\theta = \frac18(1 + \cos\theta)\sin\theta \).
3. The derivative is zero when \( \cos\theta = \frac12 \), i.e. \( \theta = \frac{\pi}{3} \), giving \( \frac18 \cdot \frac32 \cdot \frac{\sqrt3}{2} = \frac{3\sqrt3}{32} \).

Teacher's Note:
1. A single angle parameter gives a one-variable maximisation.
2. The vertex at the end of the diameter gives the largest triangle.

Answer: (b) \( F''(c) = 0 \) for some \( c \in \left(0, \frac{\pi}{2}\right) \)

Explanation:
1. With \( \arcsin(\cos x) = \frac{\pi}{2} - x \) and \( \arcsin(\sin x) = x \), \( F'(x) = -\sin x\, e^{(1 + \pi/2 - x)^2} - \cos x\, e^{(1 + x)^2} \).
2. At the end points, \( F'(0) = -e^{1} = -e \) and \( F'(\frac{\pi}{2}) = -e^{1} = -e \), so \( F'(0) = F'(\frac{\pi}{2}) \).
3. By Rolle's theorem applied to F', \( F''(c) = 0 \) for some c in \( (0, \frac{\pi}{2}) \). (F' is always negative, so (c) is false.)

Teacher's Note:
1. Use the limits sin x and cos x with the chain rule, and simplify the inverse sine.
2. Equal values of F' at the two ends point to Rolle's theorem on F'.

Answer: (a) \( \pi S \)

Explanation:
1. The circumcircle of a right triangle has the hypotenuse c as its diameter, so its area is \( \frac{\pi c^2}{4} \).
2. With legs p and q, \( c^2 = p^2 + q^2 \ge 2pq = 4S \).
3. So the area \( \ge \frac{\pi \cdot 4S}{4} = \pi S \), with equality for the isosceles right triangle.

Teacher's Note:
1. Use AM-GM on the squares of the legs.
2. Equality holds when the two legs are equal.

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

Explanation:
1. \( \frac{d}{dx}f(2x) = 2f'(2x) = f'(x) \), so \( f'(2x) = \frac12 f'(x) \).
2. \( f'(2) = \frac12 \) and \( f'(4) = \frac12 f'(2) = \frac14 \).
3. By the Mean Value Theorem for f', \( f''(c) = \frac{f'(4) - f'(2)}{4 - 2} = \frac{\frac14 - \frac12}{2} = -\frac18 \).

Teacher's Note:
1. Apply the Mean Value Theorem to f' instead of f.
2. The functional relation gives f' at the end points.

Answer: (c) \( (-1, 1) \)

Explanation:
1. The abscissa changes slower than the ordinate when \( \frac{dx}{dt} < \frac{dy}{dt} \), i.e. \( \frac{dy}{dx} > 1 \).
2. \( y = 3x^{1/3} \) gives \( \frac{dy}{dx} = x^{-2/3} \).
3. \( x^{-2/3} > 1 \Rightarrow |x| < 1 \), so the interval is (-1, 1).

Teacher's Note:
1. Slower rate of x means the ratio dy/dx exceeds 1.
2. Fractional powers of negative x are handled through the cube root.

Answer: (d) \( f(x) > g(x) \) for all \( x > x_0 \)

Explanation:
1. Let \( h = f - g \). Then \( h(x_0) = 0 \) and \( h'(x) > 0 \) for \( x > x_0 \).
2. So h is strictly increasing on \( [x_0, \infty) \).
3. Hence \( h(x) > 0 \), i.e. \( f(x) > g(x) \), for all \( x > x_0 \).

Teacher's Note:
1. Study the difference of the two functions.
2. A positive derivative from a zero start keeps the function positive.

Answer: (b) \( \sin\theta = \frac{\sqrt5 - 1}{2} \)

Explanation:
1. Triangle CPQ has sides a, a and \( 2a\sin\theta \), area \( a^2\sin\theta\cos\theta \) and semi-perimeter \( a(1 + \sin\theta) \). The inradius is \( r = \frac{a\sin\theta\cos\theta}{1 + \sin\theta} \).
2. With \( s = \sin\theta \), \( r = a s\sqrt{\frac{1 - s}{1 + s}} \), and \( \frac{d}{ds}\ln r = \frac1s - \frac{1}{1 - s^2} = 0 \Rightarrow s^2 + s - 1 = 0 \).
3. \( s = \frac{\sqrt5 - 1}{2} \) (the positive root).

Teacher's Note:
1. Inradius equals area divided by semi-perimeter.
2. Logarithmic differentiation keeps the algebra neat.

Answer: (b) 4

Explanation:
1. The function is defined only for \( x \ge 0 \) because of \( \sqrt{x^3} \).
2. \( f'(x) = \sqrt x - \frac12 + \frac12 + \frac12\cos 2x - \sqrt x = \frac12\cos 2x \).
3. \( \cos 2x = 0 \) for \( x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \) in \( [0, 2\pi] \): 4 points.

Teacher's Note:
1. Check the domain of the function before counting critical points.
2. The terms in the derivative cancel neatly, leaving cos 2x.

Answer: (b) (0, 1)

Explanation:
1. Intersections: \( mx = \frac{x}{x^2 + 1} \Rightarrow x = 0 \) or \( x^2 = \frac1m - 1 \).
2. Besides the origin, two further points (giving enclosed regions) exist only if \( \frac1m - 1 > 0 \), i.e. \( 0 < m < 1 \).
3. For \( m \le 0 \) or \( m \ge 1 \) the line meets the curve only at the origin.

Teacher's Note:
1. A bounded region needs at least two intersection points.
2. Solve for x² and require it to be positive.

Answer: (a) always zero

Explanation:
1. g is decreasing and f is increasing, so \( h = f \circ g \) is decreasing.
2. Since h maps into \( [0, \infty) \), \( h(x) \ge 0 \), and \( h(x) \le h(0) = 0 \) for \( x \ge 0 \).
3. So \( h(x) = 0 \) for all x, and \( h(x) - h(1) = 0 \) always.

Teacher's Note:
1. Combine monotonicity with the range restriction.
2. A decreasing, non-negative function that starts at 0 is identically zero.

Answer: (b) increasing in (0, 1)

Explanation:
1. Here \( t^2 = \frac1x - 1 \), so x takes values in (0, 1).
2. \( y = \frac{x}{t} = \frac{x^{3/2}}{(1 - x)^{1/2}} \), and \( \frac{d}{dx}\ln y = \frac{3}{2x} + \frac{1}{2(1 - x)} > 0 \).
3. So f is increasing on its whole domain (0, 1).

Teacher's Note:
1. Eliminate the parameter to find the domain first.
2. The domain of f is only (0, 1), so options with larger intervals are not possible.

Answer: (a) \( (-\infty, 4] \)

Explanation:
1. \( f'(x) = 4x - k \ge 0 \) must hold on [1, 2].
2. The smallest value of 4x is 4, at x = 1, so \( k \le 4 \).
3. k = 4 is allowed because f' is then zero only at the single point x = 1. So the set is \( (-\infty, 4] \).

Teacher's Note:
1. Increasing on an interval means the derivative is non-negative on all of it.
2. Check the left end, where 4x is smallest.

Answer: (b) \( (0, 1) \cup (1, 4) \)

Explanation:
1. \( \cos^2\frac x4 - \sin^2\frac x4 = \cos\frac x2 \), so \( f'(x) = -\frac{(a - 1)(a - 2)}{2}\sin\frac x2 + (a - 1) = (a - 1)\left[1 - \frac{a - 2}{2}\sin\frac x2\right] \).
2. No critical points means \( f'(x) \ne 0 \) for all x: we need \( a \ne 1 \) and \( \left|\frac{a - 2}{2}\right| < 1 \).
3. \( |a - 2| < 2 \Rightarrow 0 < a < 4 \), so \( a \in (0, 1) \cup (1, 4) \).

Teacher's Note:
1. Simplify with the identity \( \cos^2 A - \sin^2 A = \cos 2A \).
2. Factor out a - 1 and require the bracket never to vanish.