JEE Mathematics Application of Derivatives MCQs Set 11

Answer: (d) is monotonic

Explanation:
1. \( \ln f = x\ln\left(1 + \frac1x\right) \), and \( \frac{f'}{f} = \ln\left(1 + \frac1x\right) - \frac{1}{x + 1} \).
2. This is positive for x > 0, because \( \ln(1 + t) > \frac{t}{1 + t} \) with \( t = \frac1x \).
3. Hence f is increasing for all x > 0, so f is monotonic.

Teacher's Note:
1. Take logarithms first to differentiate a variable exponent.
2. The inequality \( \ln(1 + t) > \frac{t}{1 + t} \) is a standard one.

Answer: (d) \( \sqrt e \)

Explanation:
1. Taking logarithms, the limit is \( \exp\left(\lim \frac1n\sum \ln f\left(\frac kn\right)\right) = \exp\int_0^1 x\ln\left(1 + \frac1x\right)dx \).
2. \( \int_0^1 x\ln(1 + x)\,dx = \frac14 \) and \( \int_0^1 x\ln x\,dx = -\frac14 \), so the integral is \( \frac14 + \frac14 = \frac12 \).
3. The limit is \( e^{1/2} = \sqrt e \).

Teacher's Note:
1. A product of equally spaced values under an nth root becomes an integral of the logarithm.
2. Evaluate each integral by parts.

Answer: (b) \( (-\infty, -1) \cup (-1, 0) \)

Explanation:
1. \( f(x) = 1 + \frac{1}{x^2 - 1} \), so \( f'(x) = -\frac{2x}{(x^2 - 1)^2} \).
2. f' > 0 exactly when x < 0, excluding the points x = -1 where f is undefined.
3. So f is increasing in \( (-\infty, -1) \cup (-1, 0) \).

Teacher's Note:
1. Divide to write the function as 1 plus a simple fraction.
2. Remove the points where the function is not defined.

Answer: (a) local maxima but no local minima

Explanation:
1. \( f'(x) = -\frac{2x}{(x^2 - 1)^2} \) changes sign from positive to negative at x = 0.
2. So x = 0 is a local maximum with f(0) = 0.
3. f' has no other zero, so there is no local minimum.

Teacher's Note:
1. A sign change of the derivative from + to - signals a local maximum.
2. Count all the sign changes of f' in the domain.

Answer: (b) 5049

Explanation:
1. \( f'(x) = 24x^2 + 8ax + 2 \) must change sign, so its discriminant is positive: \( 64a^2 - 192 > 0 \Rightarrow a^2 > 3 \).
2. For a in [1, 100] this excludes only a = 1.
3. The sum of the integers from 2 to 100 is \( 5050 - 1 = 5049 \).

Teacher's Note:
1. Non-monotonic means f' changes sign, so f' has two distinct real roots.
2. Subtract the excluded values from the full sum.

Answer: (d) -256

Explanation:
1. The sum of the base 2 logarithms of the roots is the base 2 logarithm of their product, so the product is \( 2^5 = 32 \).
2. For \( 8x^3 + 4ax^2 + 2bx + a = 0 \), the product of the roots is \( -\frac{a}{8} \).
3. \( -\frac a8 = 32 \Rightarrow a = -256 \).

Teacher's Note:
1. A sum of logarithms is the logarithm of the product.
2. The product of roots comes from the constant term and the leading coefficient.

Answer: (c) the curve is concave down on \( (-\infty, 2) \) and on \( (2, \infty) \).

Explanation:
1. From the sign of f', f increases on \( (-\infty, -5) \) and \( (2, 4) \) and decreases on \( (-5, 2) \) and \( (4, \infty) \); (b) states the reverse, so it is false.
2. f'' is negative everywhere except at x = 2, so the curve is concave down on each side of 2 and never concave up on an interval; (d) is false.
3. There is no sign change of f'', so there is no point of inflexion and (a) is false; (c) is true.

Teacher's Note:
1. The sign of f'' decides concavity.
2. Compare the intervals of increase and decrease with the given sign pattern of f'.

Answer: (a) 1, 2, 3 and 4

Explanation:
1. \( f(x) = x^{4/3} - x^{1/3} \), \( f'(x) = \frac{4x - 1}{3x^{2/3}} \), so f decreases for \( x < \frac14 \) and increases for \( x > \frac14 \): statement 2 is correct.
2. \( f''(x) = \frac{2(2x + 1)}{9x^{5/3}} \) is zero at \( x = -\frac12 \), undefined at 0, and changes sign at both. It is negative on \( (-\frac12, 0) \) and there are 2 inflexion points: statements 1 and 3 are correct.
3. The area for \( 0 < x < 1 \) is \( \int_0^1 (x^{1/3} - x^{4/3})dx = \frac34 - \frac37 = \frac{9}{28} \): statement 4 is correct.

Teacher's Note:
1. Use f'' to find both the inflexion points and the concavity.
2. A vertical tangent at x = 0 still allows an inflexion point.

Answer: (d) 1 and 4 only

Explanation:
1. The curve is the hyperbola \( xy = K^2 \). At \( (x_0, y_0) \), \( \frac{dy}{dx} = -\frac{y_0}{x_0} \), and the tangent has intercepts \( a = 2x_0 \) and \( b = 2y_0 \).
2. Since \( x_0 y_0 = K^2 > 0 \), \( x_0 \) and \( y_0 \) have the same sign.
3. So a and b have the same sign: both positive or both negative.

Teacher's Note:
1. The tangent to xy = constant has intercepts that are double the coordinates of the point.
2. Equal signs follow from the positive product.

Answer: (c) 1, 2, 3 and 4

Explanation:
1. The integrand is real only for \( |t| \le 1 \), so f is defined on [-1, 1].
2. \( f'(x) = \sqrt{1 - x^4} \ge 0 \), so f is increasing. The integrand is even, so f is odd.
3. \( f''(x) = \frac{-2x^3}{\sqrt{1 - x^4}} \) changes sign at 0, so (0, 0) is a point of inflexion.

Teacher's Note:
1. Differentiate under the integral sign using the fundamental theorem.
2. The integral of an even function from 0 to x is odd.

Answer: (d) 1 and 2 only

Explanation:
1. Intercepts equal in magnitude and opposite in sign mean the tangent is \( \frac xa - \frac ya = 1 \), i.e. its slope is 1.
2. \( f'(x) = x^2 - 5x + 7 = 1 \Rightarrow x^2 - 5x + 6 = 0 \Rightarrow x = 2 \) or 3.
3. \( f(2) = \frac83 \) and \( f(3) = \frac72 \), and the tangents there do not pass through the origin. At x = 1 the slope is 3, so (1, 5/6) fails.

Teacher's Note:
1. Intercepts a and -a correspond to the line \( y = x - a \).
2. Solve f'(x) = 1 and then find the points.

Answer: (c) 2, 3 and 4 only

Explanation:
1. \( f'(x) = 100x^{99} + \cos x \).
2. On (0, 1) and \( (0, \frac\pi2) \) both terms are positive. On \( (\frac\pi2, \pi) \), \( 100x^{99} \) is huge compared with \( |\cos x| \le 1 \), so f' is positive.
3. On (-1, 1), near x = -1 we get \( f'(-1) = -100 + \cos 1 < 0 \), so it is not increasing there.

Teacher's Note:
1. Compare the size of the two terms of the derivative.
2. A very large power dominates the cosine term for x greater than 1.

Answer: (b) 1, 2, 3 and 4

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; the left slope is 24 and the right slope is -1, so f'(2) does not exist.
3. f rises to 35 at x = 2 and then falls, so the maximum is at x = 2. All four statements are correct.

Teacher's Note:
1. Check the joining point for continuity and for equal one-sided slopes.
2. The maximum occurs where the slope changes from positive to negative.

Answer: (d) 1 and 3 only

Explanation:
1. \( f'(x) = x(2 - \cos x) \) has the sign of x, so f decreases for x < 0 and increases for x > 0, with minimum \( f(0) = -1 \). Statement 2 is false.
2. So f(x) = k has no solution for k < -1 (statement 1 true), and \( f \to \infty \) at both ends (statement 3 true).
3. f is even, so its zeros come in pairs on both sides of the origin: statement 4 is false.

Teacher's Note:
1. Factor the derivative to see its sign.
2. An even function has a symmetrical set of zeros.

Answer: (b) 1, 3 and 4 only

Explanation:
1. An inverse exists only if f is one-one, so statement 3 holds.
2. The inverse of an increasing function is increasing and the inverse of a decreasing function is decreasing, so statement 1 holds and statement 2 does not.
3. g maps the range of f back onto the domain of f, so g is onto: statement 4 holds.

Teacher's Note:
1. Reflecting in the line y = x preserves the direction of monotonicity.
2. An inverse function is always onto its codomain, the domain of f.

Answer: (c) 1, 2 and 3 only

Explanation:
1. The domain is \( 0 < x < e \), and \( f'(x) = -\frac{1}{x(1 - \ln x)} < 0 \) throughout.
2. So f decreases on the whole of (0, e): statements 1 and 2 fail. f'(1) = -1, not zero, and f' exists at 1, so x = 1 is not critical: statement 3 fails.
3. \( f \to \infty \) as \( x \to 0^+ \) and \( f \to -\infty \) as \( x \to e^- \), so there are two vertical asymptotes: statement 4 holds. The ones that do not hold are 1, 2 and 3.

Teacher's Note:
1. Read the question carefully: it asks which statements do not hold.
2. Find the domain before looking at monotonicity.

Answer: (d) 1 and 4 only

Explanation:
1. At x = -1 both sides give 1, and at x = 1 both give 1, so f is continuous everywhere: statement 1 is true.
2. f is differentiable at points such as x = 0, so statement 2 is false.
3. The sign of f' changes at x = -1 (plus to minus), 0 (minus to plus), 1 (plus to minus) and 2 (minus to plus): four times, so statement 3 is false and statement 4 is true (maxima at -1 and 1, minima at 0 and 2).

Teacher's Note:
1. List the sign of f' on each piece to count the changes.
2. The corners at -1 and 1 are local maxima because f' changes from + to -.

Answer: (b) 2, 3 and 4 only

Explanation:
1. \( \sin(\cot^{-1}x) = \frac{1}{\sqrt{1 + x^2}} = u \), and \( \cos^2(\tan^{-1}u) = \frac{1}{1 + u^2} = \frac{1 + x^2}{2 + x^2} \).
2. This is even, has range \( [\frac12, 1) \) (so statement 1 is false), has f'(0) = 0, and tends to 1 as \( x \to \pm\infty \).
3. Hence statements 2, 3 and 4 are correct.

Teacher's Note:
1. Simplify the composition step by step from the inside.
2. The minimum value at x = 0 is 1/2, not 0.

Answer: (d) 1 and 3 only

Explanation:
1. Let the line be \( y = mx + c \). Tangency to \( y = x^2 + 1 \) needs \( m^2 = 4(1 - c) \), and tangency to \( y = -x^2 \) needs \( m^2 = 4c \).
2. Hence \( 4c = 4 - 4c \Rightarrow c = \frac12 \) and \( m^2 = 2 \).
3. The lines are \( y = \pm\sqrt2 x + \frac12 \): statements 1 and 3.

Teacher's Note:
1. Use the discriminant condition for a line to touch a parabola.
2. The same line must satisfy both conditions.

Answer: (d) 3 and 4 only

Explanation:
1. \( \cos t\cos(x - t) = \cos x\cos^2 t + \sin x\sin t\cos t \). Integrating over \( [0, \pi] \): \( \int\cos^2 t = \frac\pi2 \) and \( \int\sin t\cos t = 0 \), so \( f(x) = \frac\pi2\cos x \).
2. f is smooth, so statement 1 is false. The maximum is \( \frac\pi2 \), not \( \pi \), so statement 2 is false.
3. \( f'(x) = -\frac\pi2\sin x = 0 \) at \( x = \pi \) (statement 3 true), and the minimum is \( -\frac\pi2 \) (statement 4 true).

Teacher's Note:
1. Expand the cosine of a difference before integrating.
2. After integration the function is just a multiple of cos x.

Answer: (c) 1, 3 and 4 only

Explanation:
1. 1: \( \frac{d}{dx}(x - \tan^{-1}x) = \frac{x^2}{1 + x^2} > 0 \) and the value is 0 at 0, so it is true.
2. 2: \( \cos x > 1 - \frac{x^2}{2} \) for x > 0, so the stated inequality is false. 3: let \( h = 1 + x\ln(x + \sqrt{1 + x^2}) - \sqrt{1 + x^2} \); \( h' = \ln(x + \sqrt{1 + x^2}) > 0 \) and \( h(0) = 0 \): true.
3. 4: \( \frac{d}{dx}\left(\ln(1 + x) - x + \frac{x^2}{2}\right) = \frac{x^2}{1 + x} > 0 \) and the value is 0 at 0: true.

Teacher's Note:
1. Compare each pair of functions by differentiating their difference.
2. Start from the equality at x = 0.

Answer: (d) 2, 3 and 4 only

Explanation:
1. \( f'(x) = \frac{2 - x}{x^3} \), which is positive on (0, 2) and negative on \( (-\infty, 0) \) and \( (2, \infty) \): statement 2 is true. At x = 2 there is a maximum (value 1/4), so statement 1 is false.
2. \( f''(x) = \frac{2x - 6}{x^4} \) is negative for x < 3 (x not zero): statement 3 is true.
3. f'' changes sign at x = 3, so x = 3 is a point of inflexion: statement 4 is true.

Teacher's Note:
1. Use f' for monotonicity and f'' for concavity.
2. Check the sign change of f'' before calling a point an inflexion point.

Answer: (a) 2, 3 and 4 only

Explanation:
1. f is continuous on a closed interval, so it has a minimum there: statement 2 is true. Positive f'' means concave upwards: statement 3 is true.
2. If f(3) = f(4), Rolle's theorem gives a point with f' = 0: statement 4 is true.
3. Statement 1 is not always true: for \( f(x) = e^x \) there is no relative minimum inside (-3, 4).

Teacher's Note:
1. A convex function can be monotonic, so it need not have an interior minimum.
2. Rolle's theorem needs two equal values.

Answer: (a) 2, 3 and 4 only

Explanation:
1. 1 is false: \( f(x) = x\sin x \) has no horizontal asymptote and no infinite limit. 2 is true by the intermediate value theorem since \( 3 < \pi < 4 \).
2. 3 is true: for large positive x, \( \frac{x + 1}{x} > 1 \) is outside the domain of \( \sin^{-1} \), so the limit does not exist.
3. 4 is true: the circle is \( (x - 1)^2 + (y - 1)^2 = 1 \) with centre (1, 1), and every line \( y - 1 = m(x - 1) \) passes through the centre, hence cuts the circle at right angles.

Teacher's Note:
1. Use counterexamples for general claims.
2. A line through the centre meets a circle orthogonally.