JEE Mathematics Application of Derivatives MCQs Set 29

Answer: (d) \(\dfrac\pi6\)

Explanation:
1. The tangent is \(\dfrac{x\cos\theta}{3\sqrt3}+y\sin\theta=1\), with intercepts \(\dfrac{3\sqrt3}{\cos\theta}\) and \(\dfrac1{\sin\theta}\).
2. Setting the derivative of the sum to zero: \(\dfrac{3\sqrt3\sin\theta}{\cos^2\theta}=\dfrac{\cos\theta}{\sin^2\theta}\), so \(\tan^3\theta=\dfrac1{3\sqrt3}\) and \(\theta=\dfrac\pi6\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (a) \(0<x<\dfrac\pi4\)

Explanation:
1. \(f'(x)\) has the sign of \(\cos x-\sin x\).
2. This is positive for \(0<x<\dfrac\pi4\), but negative for \(\dfrac\pi4<x<\dfrac\pi2\) and for many \(x<0\) or \(x>0\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (d) \(x+2y=\dfrac\pi2\)

Explanation:
1. With \(s=\sin(x+y)\), \(y'=-\dfrac s{1+s}\). Setting \(y'=-\dfrac12\) gives \(s=1\), so \(x+y=\dfrac\pi2\) and \(y=\cos\dfrac\pi2=0\).
2. The point is \(\left(\dfrac\pi2,0\right)\) and the tangent is \(y=-\dfrac12\left(x-\dfrac\pi2\right)\), i.e. \(x+2y=\dfrac\pi2\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (a) \(\dfrac\pi3\)

Explanation:
1. The hypotenuse is \(b\), so the legs are \(a\) and \(c=\sqrt{b^2-a^2}=\sqrt{16-8a}\).
2. Area \(=\dfrac12a\sqrt{16-8a}\); maximise \(a^2(16-8a)\): \(32a-24a^2=0\), so \(a=\dfrac43\), \(b=\dfrac83\).
3. \(\sin A=\dfrac ab=\dfrac12\), so \(A=\dfrac\pi6\) and \(C=\dfrac\pi3\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (b) \(\left(4,\dfrac83\right)\)

Explanation:
1. Parametrise \(x=t^2\), \(y=\dfrac{t^3}3\) (so \(x\ge0\)). Then \(\dfrac{dy}{dx}=\dfrac t2\) and the normal slope is \(-\dfrac2t\).
2. Equal intercepts means normal slope \(-1\): \(t=2\), the point \(\left(4,\dfrac83\right)\) (for \(t=-2\) the intercepts are equal in size but opposite in sign).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (a) I and III only

Explanation:
1. The two expressions are equal at \(x=\pm2\). For \(-3\le x\le0\), \(f=\sqrt{1+x^2}\) for \(x\le-2\) and \(\sqrt{9-x^2}\) for \(-2\le x\le0\); \(f(0^-)=3\).
2. For \(0\le x\le3\), \(f=\sqrt{1+x^2}\) up to \(2\), then \(\sqrt{9-x^2}\); \(f(0^+)=1\), so \(f\) is discontinuous at \(0\).
3. \(f\) decreases then increases around \(-2\) (minimum) and increases then decreases around \(2\) (maximum). So I and III.

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (a) II and III only

Explanation:
1. \(f'(x)=\cos x-\sin x=\sqrt2\cos\left(x+\dfrac\pi4\right)\), positive on \(\left[0,\dfrac\pi4\right)\cup\left(\dfrac{5\pi}4,2\pi\right]\) and negative on \(\left(\dfrac\pi4,\dfrac{5\pi}4\right)\).
2. So II is true, I is false, III is true (it is a subset of the increasing intervals) and IV is false.

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (b) \(5\)

Explanation:
1. \(f'(x)=(x-5)^{54}(x-6)^{65}(121x-660)\); the factor \((x-5)^{54}\) is an even power, so \(f'\) does not change sign at \(5\).
2. Near \(5\), \(f\approx C(x-5)^{55}\), an odd power, so \(f\) changes concavity there: \(x=5\) is the point of inflexion. At \(x=6\) and \(x=\dfrac{60}{11}\) there are extrema.

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (c) \(4\)

Explanation:
1. \(1\le|\sin x|+|\cos x|\le\sqrt2<2\), so \(y=1\) for every \(x\).
2. The curves meet at \((\pm2,1)\); the radius at \((2,1)\) has slope \(\dfrac12\), so the circle's tangent has slope \(-2\).
3. The line \(y=1\) has slope \(0\), so \(\tan\theta=2\) and \(\tan^2\theta=4\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (d) \(2\)

Explanation:
1. \(f'(x)=\dfrac13\left[(x+1)^{-2/3}-(x-1)^{-2/3}\right]\) and for \(0<x<1\), \((x+1)^{-2/3}<(1-x)^{-2/3}\).
2. So \(f\) is decreasing and its greatest value is \(f(0)=1-(-1)=2\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (a) \(\dfrac{\sqrt3\pi}{24}\)

Explanation:
1. For \(\alpha=\pi\): \(r=\dfrac12\) and \(h=\sqrt{1-\dfrac14}=\dfrac{\sqrt3}2\).
2. \(V=\dfrac13\pi r^2h=\dfrac13\pi\cdot\dfrac14\cdot\dfrac{\sqrt3}2=\dfrac{\sqrt3\pi}{24}\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (b) once

Explanation:
1. \(f'<0\) on \((a,b)\), so \(f\) is strictly decreasing on \([a,b]\).
2. A strictly decreasing function takes the value \(0\) only once, at \(x=c\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (d) \(\dfrac ab\)

Explanation:
1. At \(x=0\), \(y=b\) and \(y'=-\dfrac ba\).
2. The normal slope is \(-\dfrac1{y'}=\dfrac ab\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (a) \(a^2b^2\)

Explanation:
1. \(\dfrac{dy}{dx}=-\dfrac yx\), so the subnormal is \(\left|y\dfrac{dy}{dx}\right|=\dfrac{y^2}{|x|}=p\).
2. \(\dfrac{|y|^3}p=|x||y|=a^2b^2\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (b) \(\dfrac{a^2}{b^2}\)

Explanation:
1. \(\dfrac{dy}{dx}=-\dfrac{b^2x}{a^2y}\), subtangent \(p=\left|\dfrac y{dy/dx}\right|=\dfrac{a^2y^2}{b^2|x|}\).
2. So \(\dfrac{p|x|}{y^2}=\dfrac{a^2}{b^2}\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (d) \(0.72\pi\) cm\(^2\)

Explanation:
1. \(dA=2\pi r\,dr=2\pi(6)(0.06)=0.72\pi\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (a) \(6\%\)

Explanation:
1. \(\dfrac{dV}V=3\dfrac{dx}x=3\times2\%=6\%\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (d) \(5\)

Explanation:
1. \(\dfrac d{dt}\left(\dfrac{x^2}2-2x+5\right)=(x-2)\dfrac{dx}{dt}\).
2. Setting \((x-2)\dfrac{dx}{dt}=3\dfrac{dx}{dt}\) with \(\dfrac{dx}{dt}\ne0\) gives \(x=5\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.

Answer: (c) \(\dfrac{3\sqrt3}2\) cm\(^2\)/s

Explanation:
1. \(A=\dfrac{\sqrt3}4s^2\), so \(\dfrac{dA}{dt}=\dfrac{\sqrt3}2s\dfrac{ds}{dt}=\dfrac{\sqrt3}2(30)(0.1)=\dfrac{3\sqrt3}2\).

Teacher's Note:
1. Translate the word problem into a function of one variable, find its critical points and confirm with the second derivative or the end values.
2. For monotonicity examine the sign of the derivative; for tangents and normals use the slope dy/dx at the point.