JEE Mathematics Application of Derivatives MCQs Set 08

Answer: (d) \( \frac{1}{\sqrt{8e}} \)

Explanation:
1. Area \( A = \tfrac12 x e^{-x^2} \).
2. \( A' = \tfrac12 e^{-x^2}(1 - 2x^2) = 0 \Rightarrow x = \tfrac{1}{\sqrt2} \), and A' changes from positive to negative there.
3. \( A_{max} = \tfrac12 \cdot \tfrac{1}{\sqrt2} e^{-1/2} = \tfrac{1}{\sqrt{8e}} \).

Teacher's Note:
1. Do not forget the factor 1/2 in the area of a triangle.
2. Setting the derivative to zero and checking its sign change confirms a maximum.

Answer: (b) \( \cot^{-1}(K^2) \)

Explanation:
1. At x = 0, the slope is \( y' = K^2 e^{0} = K^2 \).
2. The tangent makes angle \( \tan^{-1}K^2 \) with the x-axis, so it makes \( \frac{\pi}{2} - \tan^{-1}K^2 = \cot^{-1}(K^2) \) with the y-axis.

Teacher's Note:
1. The angle with the y-axis is the complement of the angle with the x-axis.
2. Always ask which axis the angle is measured from.

Answer: (b) \( (0, \pi) \)

Explanation:
1. Let \( g(x) = x\sin x \). Then \( g'(x) = \sin x + x\cos x \).
2. \( g(0) = 0 \) and \( g(\pi) = 0 \), and g is continuous and differentiable.
3. By Rolle's theorem, \( g'(c) = 0 \) for some c in \( (0, \pi) \).

Teacher's Note:
1. When an equation looks like a derivative, integrate it to find the hidden function.
2. Rolle's theorem then guarantees the root.

Answer: (d) 3

Explanation:
1. Put \( t = \tan x \). Then \( \tan(x + \frac{\pi}{6}) = \frac{\sqrt3 t + 1}{\sqrt3 - t} \) and the expression is \( \frac{\sqrt3 t + 1}{t(\sqrt3 - t)} \).
2. Differentiating shows the minimum occurs at \( t = \frac{1}{\sqrt3} \), i.e. \( x = \frac{\pi}{6} \).
3. The value there is \( \frac{\tan(\pi/3)}{\tan(\pi/6)} = \frac{\sqrt3}{1/\sqrt3} = 3 \).

Teacher's Note:
1. Substituting t = tan x turns the problem into a rational function.
2. Check the end behaviour: the expression tends to infinity at both ends of the interval, so 3 is the minimum.

Answer: (b) \( c + d - b - a \)

Explanation:
1. A sum of distances is least for x between the two middle points, \( b \le x \le c \).
2. There \( |x-b| + |x-c| = c - b \) and \( |x-a| + |x-d| = d - a \).
3. Minimum value \( = (c - b) + (d - a) = c + d - b - a \).

Teacher's Note:
1. For an even number of points, any x between the two middle ones gives the minimum.
2. Pair the outer and inner points to read off the value.

Answer: (c) \( \frac{27}{4}c^3 \)

Explanation:
1. At \( (x_0, y_0) \), \( y' = -\frac{2y_0}{x_0} \). The tangent is \( y - y_0 = -\frac{2y_0}{x_0}(x - x_0) \).
2. The x-intercept is \( a = \frac{3x_0}{2} \) and the y-intercept is \( b = 3y_0 \).
3. \( a^2 b = \frac{9x_0^2}{4} \cdot 3y_0 = \frac{27}{4}x_0^2 y_0 = \frac{27}{4}c^3 \).

Teacher's Note:
1. Use the curve relation \( x_0^2 y_0 = c^3 \) at the end.
2. The result is independent of the point, which is why it is a constant.

Answer: (c) at least 2 real roots

Explanation:
1. Expanding, \( f(x) = -21x^{10} + 7x^9 + 5x^8 - 3x^6 + 18x^5 + 6x^4 - 12x^2 \).
2. \( f(0) = 0 \) and \( f(1) = -21 + 7 + 5 - 3 + 18 + 6 - 12 = 0 \).
3. So x = 0 and x = 1 are both roots, hence at least two real roots.

Teacher's Note:
1. Check simple values like 0 and 1 before expanding fully.
2. Two verified roots are enough to rule out options (a), (b) and (d).

Answer: (b) e

Explanation:
1. \( f'(x) = \ln x - 1 = 0 \Rightarrow x = e \).
2. \( f(1) = -2 \), \( f(e) = -e \), \( f(e^2) = 0 \).
3. Greatest = 0, least = \( -e \), so the difference is e.

Teacher's Note:
1. Compare the values at critical points and at both end points.
2. Remember e is about 2.718, so \( -e < -2 \).

Answer: (c) is equal to 8

Explanation:
1. Equating, \( x^3 - 6x + 3 = 0 \), so \( x^3 = 6x - 3 \).
2. At every common point, \( y = x^3 + 2x - 1 = 6x - 3 + 2x - 1 = 8x - 4 \).
3. So all three points lie on the line \( y = 8x - 4 \), and the slope of the line through any two of them is 8.

Teacher's Note:
1. Substitute the intersection condition back into one curve to find the line.
2. Three collinear points give a unique slope.

Answer: (d) \( f(x) = \begin{cases} \frac{x^3 - 2x^2 - 5x + 6}{x-1}, & x \ne 1 \\ -6, & x = 1 \end{cases} \) on \( [-2, 3] \)

Explanation:
1. (a) is discontinuous at x = 1. (b) \( \lim_{x \to 0^-} \frac{\sin x}{x} = 1 \ne f(0) \), so it is discontinuous at 0. (c) is not defined at x = 1.
2. (d) Since \( x^3 - 2x^2 - 5x + 6 = (x-1)(x^2 - x - 6) \), \( f(x) = x^2 - x - 6 \) for every x, including \( f(1) = -6 \): continuous and differentiable.
3. \( f(-2) = 0 = f(3) \), so all three conditions hold.

Teacher's Note:
1. Check continuity, differentiability and equal end values in that order.
2. Factorising the numerator shows the apparent break at x = 1 is removable.

Answer: (c) 1

Explanation:
1. Let \( B = (0, b) \). The right angle is at A, so \( \vec{OA} \cdot \vec{AB} = 0 \): \( (t, t^2)\cdot(-t, b - t^2) = 0 \).
2. \( -t^2 + t^2 b - t^4 = 0 \Rightarrow b = 1 + t^2 \).
3. \( h = OB = 1 + t^2 \to 1 \) as \( t \to 0 \).

Teacher's Note:
1. Use the dot product of perpendicular vectors to find B.
2. The limit is found by letting t go to 0 in h.

Answer: (b) 1

Explanation:
1. For \( a > 1 \), \( a^x = x \) has a solution only if \( a \le e^{1/e} \approx 1.444 \).
2. No integer greater than 1 satisfies this.
3. For \( a = 1 \), \( y = 1 \) meets \( y = x \) at (1, 1). So only a = 1 works.

Teacher's Note:
1. The critical case is when \( y = a^x \) just touches the line y = x.
2. At that touching point a equals \( e^{1/e} \).

Answer: (a) 4

Explanation:
1. Let \( A = (a, a^3) \). The tangent is \( y - a^3 = 3a^2(x - a) \).
2. Substituting \( y = x^3 \): \( x^3 - 3a^2 x + 2a^3 = 0 \), i.e. \( (x - a)^2 (x + 2a) = 0 \), so B has \( x = -2a \).
3. Gradient at B \( = 3(4a^2) = 12a^2 = 4 \times 3a^2 \), so K = 4.

Teacher's Note:
1. A tangent meets the repeated root at the point of tangency.
2. Divide the cubic by \( (x - a)^2 \) to find the other intersection.

Answer: (b) f is monotonic

Explanation:
1. \( f(x) = 2\cos^{-1}|x| \) for \( |x| \le 1 \).
2. It is even, not differentiable at 0 (corner), and has its maximum value \( \pi \) at x = 0.
3. It increases on \( [-1, 0] \) and decreases on \( [0, 1] \), so it is not monotonic.

Teacher's Note:
1. Rewrite using the double-angle identity to see the structure.
2. Evenness plus a corner at 0 rules out monotonic behaviour.

Answer: (a) 13

Explanation:
1. The circle is \( (x+6)^2 + (y-10)^2 = 16 \): centre (-6, 10), radius 4.
2. Reflect (2, 5) in the x-axis to get (2, -5). The shortest path touching the x-axis has the same length as the straight line from (2, -5) to the circle.
3. Distance to the centre \( = \sqrt{8^2 + 15^2} = 17 \), so the shortest path \( = 17 - 4 = 13 \).

Teacher's Note:
1. Reflection turns a bent path into a straight line.
2. The shortest distance to a circle is along the line through its centre.

Answer: (c) are perpendicular to each other

Explanation:
1. The slope of the tangent is \( \frac{dy}{dx} = -\frac{3x + 2y}{2x + 5y} \).
2. On \( y = -\frac32 x \), \( 3x + 2y = 0 \), so the tangent at P has slope 0 (horizontal).
3. On \( y = -\frac25 x \), \( 2x + 5y = 0 \), so the tangent at Q is vertical.
4. A horizontal and a vertical line are perpendicular.

Teacher's Note:
1. Differentiate implicitly and substitute the line equations.
2. The curve is symmetric about the origin, so both points on one line give parallel tangents.

Answer: (a) \( 4 + 3\sqrt3 \)

Explanation:
1. Place the triangle with base from (-3, 0) to (3, 0) and apex (0, 4).
2. The shortest Y-network meets at the Fermat point, where each angle is 120 degrees. It lies on the axis at \( (0, y) \), and the lines to (-3, 0) and (3, 0) make 120 degrees with each other, so each makes 60 degrees with the axis: \( \tan 60^\circ = \frac{3}{y} \Rightarrow y = \sqrt3 \).
3. Length \( = (4 - \sqrt3) + 2\sqrt{9 + 3} = 4 - \sqrt3 + 4\sqrt3 = 4 + 3\sqrt3 \).

Teacher's Note:
1. The point minimising total distance to three vertices is the Fermat point.
2. A quick check: \( 4 + 3\sqrt3 \approx 9.196 \), less than the perimeter-based options.

Answer: (d) \( \frac52 \)

Explanation:
1. \( f'(t) = 12t^2 - 2t - 2 = 2(3t + 1)(2t - 1) \), so on [0, 1] f decreases up to \( t = \frac12 \) and then increases.
2. \( g(\tfrac14) = f(\tfrac14) = \frac{1}{16} - \frac{1}{16} - \frac12 + 1 = \frac12 \), and \( g(\tfrac34) = f(\tfrac12) = \frac12 - \frac14 - 1 + 1 = \frac14 \).
3. \( g(\tfrac54) = 3 - \frac54 = \frac74 \), so the sum is \( \frac12 + \frac14 + \frac74 = \frac52 \).

Teacher's Note:
1. g(x) is the running minimum of f, so find where f turns.
2. Use the formula 3 - x only for x greater than 1.

Answer: (a) f, g, h

Explanation:
1. f is not differentiable at \( x = \frac12 \) (the cube-root type term has a vertical tangent there).
2. g(x) = 0 for \( 0 < x < 1 \) but g(0) = 1, so g is discontinuous at 0.
3. h(x) = {x} jumps from near 1 to 0 at x = 1, so it is discontinuous at the end point.
4. k is continuous and differentiable on [0, 1] because x + 3 > 0, so LMVT applies to k only.

Teacher's Note:
1. LMVT needs continuity on [a, b] and differentiability on (a, b).
2. A vertical tangent or a jump breaks the conditions.

Answer: (d) \( -6 \)

Explanation:
1. Horizontal tangent: \( f'(x) = 4x^3 + 2bx + 8 = 0 \). Inflexion: \( f''(x) = 12x^2 + 2b = 0 \Rightarrow b = -6x^2 \).
2. Substituting: \( 4x^3 - 12x^3 + 8 = 0 \Rightarrow x^3 = 1 \Rightarrow x = 1 \).
3. Then \( b = -6 \).

Teacher's Note:
1. Both conditions must hold at the same x, so solve them together.
2. Eliminate b using the second-derivative condition.

Answer: (c) increasing in \( (-\infty, \infty) \)

Explanation:
1. \( f'(x) = 2 - \frac{1}{1 + x^2} - \frac{1}{\sqrt{1 + x^2}} \).
2. For \( x \ne 0 \), both \( \frac{1}{1 + x^2} < 1 \) and \( \frac{1}{\sqrt{1 + x^2}} < 1 \), so \( f'(x) > 0 \); at x = 0, \( f'(0) = 0 \).
3. Hence \( f' \ge 0 \) with equality only at one point, so f is increasing everywhere.

Teacher's Note:
1. Bound each fraction by 1 to settle the sign of the derivative.
2. A derivative that is zero at a single point does not stop f from being increasing.

Answer: (d) \( \frac{16}{9\pi} \)

Explanation:
1. Volume in the pot is \( V = \pi r^2 h \) with \( r = 7.5 \).
2. \( \frac{dV}{dt} = \pi (7.5)^2 \frac{dh}{dt} = 100 \).
3. \( \frac{dh}{dt} = \frac{100}{56.25\pi} = \frac{16}{9\pi} \) cm per minute, whatever the current depth.

Teacher's Note:
1. In a cylinder the level rises at a constant rate for a constant inflow.
2. The depth of 10 cm is extra information that is not needed.

Answer: (d) \( (0, 1) \cup (1, \infty) \)

Explanation:
1. \( f'(x) = \ln x \), so f decreases on (0, 1) and increases on \( (1, \infty) \).
2. The minimum is \( f(1) = 0 \).
3. So \( f(x) > 0 \) for every x in the domain except x = 1.

Teacher's Note:
1. A function that has its minimum value 0 is positive everywhere else.
2. The domain of ln x is x > 0.

Answer: (c) FTTF

Explanation:
1. Statement-1 is false: without continuity at c, the value f(c) could be lower than nearby values, so c need not be a local maximum.
2. Statement-2 is true: at a local maximum the curve is concave down on both sides, so concavity does not change.
3. Statement-3 is true: \( f(x) = |x|^3 \), \( f'(x) = 3x|x| \), \( f''(x) = 6|x| \) exists at 0.
4. Statement-4 is false: \( f^{-1} \) is differentiable at f(c) only if \( f'(c) \ne 0 \). The order is FTTF.

Teacher's Note:
1. Look for missing conditions such as continuity or a non-zero derivative.
2. Test each statement with a simple counterexample.

Answer: (a) \( \left(0, \frac1e\right) \)

Explanation:
1. Roots are the intersections of \( y = |\ln x| \) with the line \( y = px \).
2. For \( x > 1 \) the line meets \( \ln x \) twice exactly when it is below the tangent from the origin, whose slope is \( \frac1e \) (touching at x = e).
3. For \( 0 < x < 1 \) there is always exactly one intersection with \( -\ln x \). So three roots need \( 0 < p < \frac1e \).

Teacher's Note:
1. Count intersections of the curve and the line graphically.
2. The tangent from the origin to \( y = \ln x \) touches at x = e with slope 1/e.