JEE Mathematics Application of Derivatives MCQs Set 16

Answer: (d) \( -\left(\frac{d^2y}{dx^2}\right)\left(\frac{dy}{dx}\right)^{-3} \)

Explanation:
1. \( \frac{dx}{dy}=\left(\frac{dy}{dx}\right)^{-1} \).
2. Differentiating with respect to \( y \): \( \frac{d^2x}{dy^2}=-\left(\frac{dy}{dx}\right)^{-2}\frac{d^2y}{dx^2}\cdot\frac{dx}{dy}=-\frac{d^2y}{dx^2}\left(\frac{dy}{dx}\right)^{-3} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( t_4>t_3>t_1>t_2 \)

Explanation:
1. Here \( \tan\theta<1<\cot\theta \). For base \( \cot\theta>1 \) a larger exponent gives a larger value: \( t_4>t_3>1 \).
2. For base \( \tan\theta<1 \): \( t_1<1 \) and the larger exponent gives the smaller value: \( t_2<t_1 \).
3. So \( t_4>t_3>t_1>t_2 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( \left(\frac98,\frac92\right) \)

Explanation:
1. \( 2y\frac{dy}{dt}=18\frac{dx}{dt} \) and \( \frac{dy}{dt}=2\frac{dx}{dt} \) give \( 4y=18 \), \( y=\frac92 \).
2. Then \( x=\frac{y^2}{18}=\frac98 \).

Teacher's Note:
1. Differentiate the curve with respect to time.
2. Use the given ratio of rates.

Answer: (b) \( \left(\pm\frac4{\sqrt3},2\right) \)

Explanation:
1. Differentiating: \( 3y^2y'+6x=12y' \), so \( y'=\frac{6x}{12-3y^2} \). The tangent is vertical when \( 12-3y^2=0 \), i.e. \( y=\pm2 \).
2. For \( y=2 \): \( 8+3x^2=24 \), \( x=\pm\frac4{\sqrt3} \). For \( y=-2 \) we would need \( 3x^2=-16 \) (impossible).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( (2+a)^2f''(1)+(2-a)^2f''(-1)=0 \)

Explanation:
1. \( f'(x)=\frac{2a(x^2-1)}{(x^2+ax+1)^2} \). Differentiating again and evaluating: \( f''(1)=\frac{4a}{(2+a)^2} \) and \( f''(-1)=-\frac{4a}{(2-a)^2} \).
2. Hence \( (2+a)^2f''(1)+(2-a)^2f''(-1)=4a-4a=0 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) 2

Explanation:
1. Let \( t=x+2 \): \( \frac{x^2+2x+4}{x+2}=\frac{t^2-2t+4}{t}=t+\frac4t-2 \).
2. For \( t>0 \) AM-GM gives \( t+\frac4t\ge4 \), so the value is at least \( 2 \) (at \( t=2 \)), and the negative values for \( t<0 \) are excluded.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) decreasing in \( \left(-t,-\frac14\right) \) and increasing in \( \left(-\frac14,t\right) \)

Explanation:
1. \( f'(x)=2+6x+12x^2 \) is an upward parabola with vertex at \( x=-\frac14 \).
2. Hence it decreases for \( x<-\frac14 \) and increases for \( x>-\frac14 \), and \( -t<-\frac14<t \) because \( t>\frac12 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( \left[0,\frac\pi2\right] \)

Explanation:
1. \( \phi'(x)=\cos^3x \), which is non-negative exactly when \( \cos x\ge0 \).
2. On \( \left[0,\frac\pi2\right] \) this holds; on the other intervals \( \cos x\le0 \) (decreasing).

Teacher's Note:
1. Reduce the expression to a standard form before reading off the answer.
2. Verify with a simple check of the final result.

Answer: (a) 3

Explanation:
1. By the triangle inequality \( \left||\tfrac x2+1|-|\tfrac x2-1|\right|\le\left|(\tfrac x2+1)-(\tfrac x2-1)\right|=2 \).
2. So \( f\le2+1=3 \), reached for \( x\ge2 \).

Teacher's Note:
1. Reduce the problem to a standard result and then substitute.
2. Verify the final value with a simple case.

Answer: (a) 1

Explanation:
1. At \( x=1 \) both curves pass through \( (1,0) \).
2. Slopes: \( \frac{d}{dx}\left(3^{x-1}\ln x\right)=1 \) and \( \frac{d}{dx}(x^x-1)=x^x(1+\ln x)=1 \) at \( x=1 \).
3. The slopes are equal, so the angle is 0 and its cosine is 1.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) 6

Explanation:
1. Secant is convex on \( (0,\pi/2) \), so by Jensen \( \sum\sec A\ge3\sec\frac{A+B+C}{3}=3\sec60^\circ=6 \).
2. Equality holds for the equilateral triangle.

Teacher's Note:
1. Reduce the expression to a standard form before reading off the answer.
2. Verify with a simple check of the final result.

Answer: (d) \( (0,1] \)

Explanation:
1. The minimum of a quadratic \( Ax^2+Bx+C \) is \( C-\frac{B^2}{4A} \), so \( m(b)=1-\frac{b^2}{1+b^2}=\frac1{1+b^2} \).
2. As \( b \) ranges over the reals, \( 1+b^2\in[1,\infty) \), so \( m(b)\in(0,1] \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) all four statements are correct

Explanation:
1. From \( f(x)=f(1-x) \): \( f'(x)=-f'(1-x) \), so \( f'(\frac12)=0 \) (ii) and \( f'(\frac34)=-f'(\frac14)=0 \). With \( f'(\frac14)=f'(\frac12)=f'(\frac34)=0 \), \( f'' \) vanishes at least twice on \( [0,1] \) by Rolle (i).
2. (iii): \( f(x+\frac12) \) is even in \( x \) and \( \sin x \) is odd, so the integral over a symmetric interval is 0.
3. (iv): substitute \( t\to1-t \) in the right integral; \( \sin\pi(1-t)=\sin\pi t \) and the limits map onto \( [0,\frac12] \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( 6a,\ 4b,\ 9c \) are in GP

Explanation:
1. \( f'(x)=3ax^2+2bx+c=3a(x-a)(x-2a) \).
2. Comparing: \( -\frac{2b}{3a}=3a \) gives \( b=-\frac{9a^2}{2} \), and \( \frac{c}{3a}=2a^2 \) gives \( c=6a^3 \).
3. Then \( 6a,\ 4b=-18a^2,\ 9c=54a^3 \) have common ratio \( -3a \): a GP.

Teacher's Note:
1. The extrema are the roots of f'(x), so use sum and product of roots.
2. Test the ratios of consecutive terms.

Answer: (a) \( \pm1 \)

Explanation:
1. \( f'(x)=\sqrt{2-x^2} \), so \( x^2=\sqrt{2-x^2} \).
2. Squaring: \( x^4+x^2-2=0 \), i.e. \( (x^2+2)(x^2-1)=0 \), so \( x=\pm1 \) (both satisfy the original equation).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( yy''+y'^2+1=0 \)

Explanation:
1. Differentiate: \( 2x+2yy'=0 \), i.e. \( x+yy'=0 \).
2. Differentiate again: \( 1+y'^2+yy''=0 \).

Teacher's Note:
1. Differentiate implicitly twice, using the product rule for y y'.
2. The constant 1 on the left comes from d/dx of x.

Answer: (a) (i), (ii) and (iii) only

Explanation:
1. \( f'(x)=\cos\frac1x+\frac1x\sin\frac1x\to1 \) (i).
2. \( f''(x)=-\frac{\cos(1/x)}{x^3}<0 \) for \( x\ge1 \) (iii), so \( f'(x)>1 \) for all \( x \) (it decreases towards 1).
3. Hence \( f(x+2)-f(x)=\int_x^{x+2}f'>2 \) (ii) and (iv) is false.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) 2

Explanation:
1. Denominator \( =3+2\cos2\theta+\frac32\sin2\theta \), whose minimum is \( 3-\sqrt{4+\tfrac94}=3-\tfrac52=\tfrac12 \).
2. So the maximum of the expression is \( \frac1{1/2}=2 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( f\left(\frac12\right)<\frac12 \) and \( f\left(\frac13\right)<\frac13 \)

Explanation:
1. Differentiating: \( \sqrt{1-f'^2}=f \), so \( f'^2=1-f^2 \) and with \( f(0)=0,\ f\ge0 \): \( f(x)=\sin x \).
2. Since \( \sin x<x \) for \( x>0 \): \( f(\frac12)<\frac12 \) and \( f(\frac13)<\frac13 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( \frac\pi3 \)

Explanation:
1. \( 3\sin x-4\sin^3x=\sin3x \).
2. It increases when \( \cos3x\ge0 \), i.e. \( 3x\in\left[-\frac\pi2,\frac\pi2\right] \) (plus periods), an interval of length \( \frac\pi3 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) Statement 1 is true and Statement 2 is false

Explanation:
1. Let \( g(x)=\frac{\tan x}{x} \). Since \( \tan \) is convex on \( (0,\frac\pi2) \) with \( \tan0=0 \), \( g \) is increasing, so \( \frac{\tan\beta}{\beta}>\frac{\tan\alpha}{\alpha} \): Statement 1 is true.
2. Statement 2 is false: convexity does not force an increasing function (e.g. \( f(x)=x^2 \) with \( x_1=-1>x_2=-2 \) gives \( f(x_1)=1<4=f(x_2) \)).

Teacher's Note:
1. tan x / x increases because tan x is convex and passes through the origin.
2. A convex function may decrease on part of its domain.

Answer: (d) \( \frac{1}{2\sqrt3} \)

Explanation:
1. At the point, \( x=-2 \) and \( y=\frac{2\pi}{3} \) (principal value of \( \sec^{-1} \) lies in \( [0,\pi] \)).
2. \( \frac{dx}{dy}=\sec y\tan y=(-2)(-\sqrt3)=2\sqrt3 \).
3. So \( \frac{dy}{dx}=\frac{1}{2\sqrt3} \).

Teacher's Note:
1. Differentiate x with respect to y first and then invert.
2. tan(2 pi / 3) is negative.

Answer: (a) \( \text{increasing on }\left[-\tfrac12,1\right] \)

Explanation:
1. \( f'(x)=e^{x-x^2}(1+x-2x^2)=e^{x-x^2}(1-x)(1+2x) \).
2. This is non-negative for \( -\frac12\le x\le1 \), so \( f \) is increasing there (and decreasing outside).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) (a, 0)

Explanation:
1. The curve is \( (x-a)^2+y^2=a^2 \), a circle with centre \( (a,0) \).
2. Every normal of a circle passes through its centre.

Teacher's Note:
1. Eliminate theta to identify the curve.
2. Normals to a circle are its radii.

Answer: (c) 24

Explanation:
1. \( F'=F\left(\frac{f'}f+\frac{g'}g+\frac{h'}h\right) \) (logarithmic differentiation).
2. So \( 21=4-7+k \) and \( k=24 \).

Teacher's Note:
1. Logarithmic differentiation of a product adds the ratios f'/f.
2. Solve the linear equation for k.

Answer: (b) 4

Explanation:
1. \( \left|\frac{f(x)-f(y)}{x-y}\right|<|x-y|^2 \); letting \( y\to x \) gives \( f'(x)=0 \) for all x.
2. So \( f \) is constant and \( f(4)=f(2)=4 \).

Teacher's Note:
1. Reduce the problem to a standard result and then substitute.
2. Verify the final value with a simple case.

Answer: (c) on the left of \( x=c \)

Explanation:
1. Tangent: \( y=e^c\{1+(x-c)\} \). The line through the two points: \( y=e^c\{\cosh1+\sinh1\,(x-c)\} \).
2. Setting equal: \( x-c=\frac{\cosh1-1}{1-\sinh1} \), and \( \cosh1-1\approx0.543>0 \), \( 1-\sinh1\approx-0.175<0 \), so \( x-c<0 \) (about \( -3.1 \)).
3. The intersection lies to the left of \( x=c \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) 1

Explanation:
1. The slope of the normal is \( \tan\frac{3\pi}4=-1 \).
2. The tangent is perpendicular, so \( f'(3)=-\frac{1}{-1}=1 \).

Teacher's Note:
1. Product of the slopes of tangent and normal is -1.
2. tan(3 pi / 4) = -1.

Answer: (a) 2

Explanation:
1. On \( (-3,-1] \) the function \( (2+x)^3 \) is increasing up to \( f(-1)=1 \). On \( (-1,2) \), \( x^{2/3} \) decreases from 1 to 0 on \( (-1,0) \) and then increases.
2. So \( x=-1 \) is a local maximum and \( x=0 \) (a cusp) is a local minimum: 2 in all.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) a local maximum at \( x=1+\log2 \) and a local minimum at \( x=e \)

Explanation:
1. \( g'(x)=f(x) \). On \( (1,2] \): \( f=2-e^{x-1} \) is positive for \( x<1+\ln2 \) and negative after, so \( g \) has a local maximum at \( 1+\ln2 \).
2. On \( (2,3] \): \( f=x-e \) is negative before \( e \) and positive after, giving a local minimum at \( e \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.