JEE Mathematics Application of Derivatives MCQs Set 10

Answer: (c) 1

Explanation:
1. At \( (x_1, y_1) \), \( \frac{dy}{dx} = -\frac{y_1}{n x_1} \). The tangent meets the axes at \( X = x_1(1 + n) \) and \( Y = y_1\left(1 + \frac1n\right) \).
2. The area is \( \frac12 |XY| = \frac12 \frac{(1 + n)^2}{n}x_1 y_1 \).
3. For this to stay constant as \( (x_1, y_1) \) moves, \( x_1 y_1 \) must be constant, which needs n = 1 (the curve \( xy = a^2 \)); then the area is \( 2a^2 \).

Teacher's Note:
1. For a hyperbola xy = constant, every tangent cuts off a triangle of the same area.
2. Use the curve equation to see which n makes the product constant.

Answer: (a) exactly one statement is correct

Explanation:
1. I is false: at a maximum \( f''(c) \le 0 \), and \( f''(c) = 0 \) is possible (for example \( f = -x^4 \)).
2. II is false: \( \int(1 + \sin x)\,dx \) is not periodic. IV is false: f' only changes from positive to negative and need not be monotonic.
3. III is true, since the integral of a periodic function over any full period is the same.

Teacher's Note:
1. Test each statement with a standard counterexample.
2. Only the period-integral property is always true.

Answer: (a) 4

Explanation:
1. f is continuous at x = -1 (both pieces give 1). For \( -1 < x < 1 \), \( f' = \frac{x^2 - 2x - 1}{(x^2 + 1)^2} = 0 \) at \( x = 1 - \sqrt2 \).
2. f' does not exist at x = 1 (corner) or at x = -1 (left slope -2, right slope \( \frac12 \)). For \( x > 1 \), \( f' = \frac{-x^2 + 2x + 1}{(x^2 + 1)^2} = 0 \) at \( x = 1 + \sqrt2 \).
3. The critical points are \( -1, 1 - \sqrt2, 1, 1 + \sqrt2 \): 4 in all.

Teacher's Note:
1. Include points where the derivative fails to exist as well as where it is zero.
2. Check the one-sided slopes at the joining point x = -1.

Answer: (c) 2

Explanation:
1. \( f'(x) = \frac13 x^{-2/3}(x - 4) + x^{1/3} = \frac{4(x - 1)}{3x^{2/3}} \).
2. f' is zero at x = 1 and does not exist at x = 0, where f is defined.
3. So there are 2 critical points.

Teacher's Note:
1. Both zeros and non-existence points of f' count as critical points.
2. Combine the terms over a common denominator to see both clearly.

Answer: (a) \( e^{-1} \)

Explanation:
1. Area \( A = x \cdot y = \frac{\ln x}{x} \).
2. \( A' = \frac{1 - \ln x}{x^2} = 0 \Rightarrow x = e \), and A' changes sign from positive to negative.
3. \( A_{max} = \frac1e \).

Teacher's Note:
1. One vertex at the origin means the area is simply x times y.
2. The sign change of the derivative confirms the maximum.

Answer: (b) a < 0

Explanation:
1. For three distinct real roots, \( f'(x) = 3x^2 + a \) must have two real zeros, so \( a < 0 \).
2. If \( a \ge 0 \), f is monotonic and has only one real root.
3. Hence a < 0 (and also \( 4|a|^3 > 27b^2 \)), so options (a), (c) and (d) are all impossible.

Teacher's Note:
1. A cubic with three distinct real roots needs a local maximum and a local minimum.
2. Look at the discriminant of the derivative.

Answer: (c) LMVT is applicable to f, and Rolle's theorem is applicable to g with b = 1

Explanation:
1. On [1, 2], \( f(x) = x - 1 \), which is continuous and differentiable, so LMVT applies to f. Also \( f(1) = 0 \ne f(2) = 1 \), so Rolle's theorem does not apply to f.
2. \( g(1) = 0 + b = b \) and \( g(2) = 1 + b\sin\pi = 1 \).
3. Rolle's theorem needs \( g(1) = g(2) \), so b = 1.

Teacher's Note:
1. Rolle's theorem needs equal values at the end points.
2. On this interval |1 - x| is just x - 1.

Answer: (d) \( \left(\sqrt{\frac32}, \frac{5}{\sqrt6}\right) \)

Explanation:
1. \( g(x) = f'(x) = x + \frac1x \) and \( g'(x) = 1 - \frac{1}{x^2} \).
2. \( g(3) = \frac{10}{3} \) and \( g(\frac12) = \frac52 \), so the chord slope is \( \frac{10/3 - 5/2}{3 - 1/2} = \frac13 \).
3. \( 1 - \frac{1}{x^2} = \frac13 \Rightarrow x = \sqrt{\frac32} \), and \( g = \sqrt{\frac32} + \sqrt{\frac23} = \frac{5}{\sqrt6} \).

Teacher's Note:
1. This is the Mean Value Theorem applied to g.
2. The fundamental theorem of calculus gives f' directly.

Answer: (a) \( \frac14(\pi + 2t) \)

Explanation:
1. \( \frac{dx}{dt} = a(1 + \cos 2t) = 2a\cos^2 t \) and \( \frac{dy}{dt} = 2a(1 + \sin t)\cos t \).
2. \( \tan\psi = \frac{1 + \sin t}{\cos t} = \tan\left(\frac{\pi}{4} + \frac t2\right) \).
3. So \( \psi = \frac{\pi}{4} + \frac t2 = \frac14(\pi + 2t) \).

Teacher's Note:
1. Recognise \( \frac{1 + \sin t}{\cos t} \) as a tangent of a half-sum angle.
2. The question asks for the angle, not the slope.

Answer: (a) 3

Explanation:
1. \( f'(2) = 4 \cdot 2 + 3 = 11 \).
2. \( \frac{f(a) - f(1)}{a - 1} = \frac{2a^2 + 3a - 5}{a - 1} = 2a + 5 \).
3. \( 2a + 5 = 11 \Rightarrow a = 3 \).

Teacher's Note:
1. Factorise \( 2a^2 + 3a - 5 = (a - 1)(2a + 5) \).
2. The slope of the chord must equal f' at the given point.

Answer: (d) only (i), (ii) and (iv) are true

Explanation:
1. (i) holds because a continuous function on a closed interval is bounded. (ii) holds by the intermediate value theorem.
2. (iv) holds by the Mean Value Theorem: the average slope \( \frac{f(b) - f(a)}{b - a} > 0 \).
3. (iii) can fail since extrema may occur at end points, and (v) can fail (for example f(x) = x). So only (i), (ii) and (iv) are always true.

Teacher's Note:
1. Closed interval, continuity and MVT give the first, second and fourth statements.
2. A single counterexample such as f(x) = x disproves statements (iii) and (v).

Answer: (b) H = 1 and V = 2

Explanation:
1. \( \frac{dy}{dt} = 4t + 3 = 0 \) gives one value \( t = -\frac34 \), where \( \frac{dx}{dt} \ne 0 \), so H = 1.
2. \( \frac{dx}{dt} = 3t^2 - 8t - 3 = (3t + 1)(t - 3) = 0 \) gives \( t = -\frac13 \) and \( t = 3 \), where \( \frac{dy}{dt} \ne 0 \), so V = 2.
3. Hence H = 1 and V = 2.

Teacher's Note:
1. Horizontal tangent means dy/dt = 0 and vertical tangent means dx/dt = 0.
2. Check that the other derivative is non-zero at those parameter values.

Answer: (c) 2

Explanation:
1. The slope of the normal is \( -\frac{1}{n a^{n-1}} \), so \( b = a^n + \frac{a}{n a^{n-1}} = a^n + \frac{a^{2-n}}{n} \).
2. As \( a \to 0 \), \( a^n \to 0 \), and \( \frac{a^{2-n}}{n} \) has a finite non-zero limit only when \( n = 2 \), giving \( \frac12 \).
3. For n = 1 the limit is 0 and for n > 2 it is infinite, so n = 2.

Teacher's Note:
1. Write the equation of the normal and set x = 0.
2. Only one value of n gives a finite non-zero limit.

Answer: (b) 4

Explanation:
1. The tangent at \( (t, 3 - t^2) \) is \( y - (3 - t^2) = -2t(x - t) \), with y-intercept \( 3 + t^2 \) and x-intercept \( \frac{3 + t^2}{2t} \) (for t > 0).
2. Area \( = \frac{(3 + t^2)^2}{4t} \), and \( \frac{dA}{dt} = 0 \Rightarrow 3t^2 - 3 = 0 \Rightarrow t = 1 \).
3. \( A_{min} = \frac{16}{4} = 4 \).

Teacher's Note:
1. Both intercepts must be positive, so take the tangent at a point with t > 0.
2. Minimise the area as a function of the point of tangency.

Answer: (c) \( \left(-\infty, \frac2e\right] \)

Explanation:
1. \( f'(x) = \frac{2 - \ln x}{2x^{3/2}} = 0 \Rightarrow x = e^2 \), a maximum with \( f(e^2) = \frac{2}{e} \).
2. As \( x \to 0^+ \), \( f \to -\infty \), and as \( x \to \infty \), \( f \to 0^+ \).
3. So the range is \( \left(-\infty, \frac2e\right] \).

Teacher's Note:
1. Find the maximum and examine the behaviour at both ends of the domain.
2. Since f is continuous, every value between is attained.

Answer: (d) \( \frac{32}{27} \)

Explanation:
1. \( y' = 3x^2 + 2x - 1 = (3x - 1)(x + 1) = 0 \Rightarrow x = \frac13, -1 \).
2. \( y\left(\frac13\right) = \frac1{27} + \frac19 - \frac13 = -\frac{5}{27} \) and \( y(-1) = -1 + 1 + 1 = 1 \).
3. The distance \( = 1 + \frac{5}{27} = \frac{32}{27} \).

Teacher's Note:
1. Horizontal tangents occur where the derivative is zero.
2. The distance between two horizontal lines is the difference of their y-values.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. \( f'(x) = 60(x^4 - x^3 + x^2 - x + 1) > 0 \) for all x, so f is strictly increasing. Being an odd-degree polynomial, f takes all real values, so the Assertion is true.
2. The Reason is false: \( e^x \) is continuous and strictly increasing on R but its range is only \( (0, \infty) \).

Teacher's Note:
1. A monotonic continuous function has range R only if it is unbounded on both sides.
2. Test a general statement with a counterexample such as the exponential function.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. \( f'(x) = \frac{1}{x + 2} - \frac{4}{(x + 3)^2} = \frac{(x + 1)^2}{(x + 2)(x + 3)^2} \ge 0 \) on the domain \( x > -2 \), so f is strictly increasing and \( f(-1) = 0 \) gives exactly one root. The Assertion is true.
2. The Reason is false: a strictly monotonic continuous function need not change sign on (a, b), so it may have no root there.

Teacher's Note:
1. Strict monotonicity gives at most one root, not necessarily one.
2. The sign of the end values decides the existence of a root.

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. \( f'(x) = x^6 - x^5 + x^4 - x^3 + x^2 - x + 1 > 0 \) for all x (it equals \( \frac{x^7 + 1}{x + 1} \) for \( x \ne -1 \)), so f is strictly increasing and has at most one root.
2. Because f is strictly increasing, \( f(a) \ne f(b) \) for \( a \ne b \), so Rolle's theorem can never be applied. The Reason explains the Assertion.

Teacher's Note:
1. A strictly increasing function cannot take the same value twice.
2. That is exactly why Rolle's theorem fails and there cannot be two roots.

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. \( f'(x) = 1 - \cos x \ge 0 \), so f is increasing everywhere; \( g'(x) = 4x^3 \) is positive for \( x > 0 \) and negative for \( x < 0 \). The Assertion is true.
2. The Reason is a true general result about odd and even functions, and it explains both parts of the Assertion since f is odd and g is even.

Teacher's Note:
1. Odd functions keep their monotonic direction on both sides, even functions reverse it.
2. Use the symmetry of the graph to decide.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. \( |x\cos\frac1x| \le |x| \to 0 \), so f is continuous at 0 and hence on [-1, 1].
2. The difference quotient \( \frac{f(h) - f(0)}{h} = \cos\frac1h \) has no limit as \( h \to 0 \), so f is not differentiable at 0. LMVT therefore fails (Assertion true), but the Reason is false because f is continuous.

Teacher's Note:
1. Continuity and differentiability are separate conditions.
2. Squeeze theorem shows the continuity at 0.

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. At x = 1, the left limit is 2 but the right limit is 1, so f is not continuous. It is not differentiable at x = 1, and \( f(0) = 0 \ne f(4) = 1 \): all three conditions fail.
2. Yet \( f'(2) = 2(2 - 2) = 0 \) with 2 in (0, 4). The Reason explains why this can happen.

Teacher's Note:
1. Rolle's theorem is an if-then statement, and the converse need not be true.
2. Always check each of the three conditions separately.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. \( f'(x) = \frac1x > 0 \) and \( g'(x) = -\frac{1}{x^2} < 0 \) on (0, 10), so the Assertion is true.
2. The Reason is false: \( f(x) = x^3 \) is increasing, but \( f'(x) = 3x^2 \) is not decreasing everywhere.

Teacher's Note:
1. Increase of a function says nothing about the monotonicity of its derivative.
2. Use a simple cubic as a counterexample.

Answer: (a) is equal to 0

Explanation:
1. \( g(x) = x\ln(1 + x) - x\ln x \).
2. As \( x \to 0^+ \), \( x\ln(1 + x) \to 0 \) and \( x\ln x \to 0 \).
3. So the limit is 0.

Teacher's Note:
1. The limit of x ln x as x tends to 0 from the right is zero.
2. Split the logarithm of a quotient into a difference.