JEE Mathematics Application of Derivatives MCQs Set 32

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

Explanation:
1. Differentiating, \(2y\frac{dy}{dt}=18\frac{dx}{dt}\). With \(\frac{dy}{dt}=2\frac{dx}{dt}\): \(4y=18\), so \(y=\frac92\).
2. Then \(x=\frac{y^2}{18}=\frac{81/4}{18}=\frac98\).

Teacher's Note:
1. Relate the rates using the equation of the curve, then solve for the point.

Answer: (b) (x-1)^3

Explanation:
1. \(f'(x)=3(x-1)^2+c_1\). The tangent slope at \(x=2\) is 3: \(3+c_1=3\), so \(c_1=0\).
2. \(f(x)=(x-1)^3+c_2\), and \(f(2)=1\) gives \(1+c_2=1\), so \(c_2=0\).
3. So \(f(x)=(x-1)^3\).

Teacher's Note:
1. Integrate twice and use the slope and the point to fix both constants.

Answer: (d) \(80\pi\) cm\(^2\)/s

Explanation:
1. \(A=\pi r^2\), so \(\frac{dA}{dt}=2\pi r\frac{dr}{dt}\).
2. With \(r=8\) and \(\frac{dr}{dt}=5\): \(\frac{dA}{dt}=2\pi\cdot8\cdot5=80\pi\) cm\(^2\)/s.

Teacher's Note:
1. The speed of the wave is the rate of change of the radius.

Answer: (c) Both (a) and (b)

Explanation:
1. \(6\frac{dy}{dt}=3x^2\frac{dx}{dt}\), so \(\frac{dy}{dt}=\frac{x^2}{2}\frac{dx}{dt}\).
2. Setting \(\frac{x^2}{2}=8\) gives \(x=\pm4\).
3. Then \(y=\frac{64+2}{6}=11\) or \(y=\frac{-64+2}{6}=-\frac{31}{3}\). Both statements are true.

Teacher's Note:
1. Differentiate the relation with respect to time to link the rates.

Answer: (b) -72 units/s

Explanation:
1. Slope \(m=5-6x^2\), so \(\frac{dm}{dt}=-12x\frac{dx}{dt}\).
2. At \(x=3\): \(\frac{dm}{dt}=-12\cdot3\cdot2=-72\) units/s.

Teacher's Note:
1. The slope itself is a function of x; differentiate it with respect to time.

Answer: (d) None of these

Explanation:
1. \(V=\pi r^2h\), so \(\frac{dV}{dt}=\pi\left(2rh\frac{dr}{dt}+r^2\frac{dh}{dt}\right)\).
2. \(=\pi(2\cdot4\cdot6\cdot3+16\cdot(-4))=\pi(144-64)=80\pi\) m\(^3\)/s.
3. This is not among the first three options.

Teacher's Note:
1. The altitude is decreasing, so its rate is negative.

Answer: (a) \(\left(-\frac\pi3,\frac\pi3\right)\)

Explanation:
1. \(f'(x)=\sec^2x-4<0\) means \(\sec^2x<4\), i.e. \(|\cos x|>\frac12\).
2. Within \((-\frac\pi2,\frac\pi2)\) this gives \(-\frac\pi3<x<\frac\pi3\).

Teacher's Note:
1. Convert the inequality for the derivative into a condition on cosine.

Answer: (b) \([-2,-1]\)

Explanation:
1. \(f'(x)=6x^2+18x+12=6(x+1)(x+2)\).
2. \(f'(x)\le0\) for \(-2\le x\le-1\).

Teacher's Note:
1. Factorise the derivative and read the sign between the roots.

Answer: (c) Both (a) and (b) are true

Explanation:
1. The slope of the tangent is \(\frac{dy}{dx}\) by definition.
2. The slope is also \(\tan\theta\), where \(\theta\) is the angle with the positive X-axis.
3. So both statements are true.

Teacher's Note:
1. Slope means tangent of the inclination.

Answer: (c) \(\cos x\)

Explanation:
1. \(\frac{d}{dx}\cos x=-\sin x<0\) on \((0,\frac\pi2)\), so \(\cos x\) is decreasing.
2. \(\sin2x\) increases on \((0,\frac\pi4)\), \(\tan x\) increases, and \(\cos3x\) rises on \((\frac\pi3,\frac\pi2)\).

Teacher's Note:
1. Test each option on the whole interval, not just a part of it.

Answer: (d) None of these

Explanation:
1. \(\frac{dy}{dx}=-\frac{1}{(x-3)^2}\), which is always negative.
2. It can never equal 2, so no tangent has slope 2.
3. Hence none of the given lines is possible.

Teacher's Note:
1. A negative-valued derivative can never equal a positive slope.

Answer: (b) (1, 0)

Explanation:
1. Differentiating: \(1+y'=e^{xy}(y+xy')\), so \(y'(1-xe^{xy})=ye^{xy}-1\).
2. A tangent parallel to the Y-axis needs \(1-xe^{xy}=0\).
3. At \((1,0)\): \(1\cdot e^0=1\), and the point is on the curve since \(1+0=e^0\). At \((0,1)\), \(xe^{xy}=0\ne1\).

Teacher's Note:
1. A vertical tangent means the denominator of \(rac{dy}{dx}\) vanishes.

Answer: (b) 2%

Explanation:
1. \(S=4\pi r^2\), so \(\frac{dS}{S}=2\frac{dr}{r}\).
2. The percentage error in S is \(2\times1\%=2\%\).

Teacher's Note:
1. Relative errors multiply by the power of the radius.

Answer: (d) all of the above are true

Explanation:
1. By definition, f has a maximum value in I, the value \(f(c)\) is called the maximum value, and c is the point of maximum.
2. So all three statements are true.

Teacher's Note:
1. Distinguish the maximum value from the point where it occurs.

Answer: (b) f has no point of maximum value in R

Explanation:
1. \(|x|\) has its minimum value 0 at \(x=0\), not at \(x=1\), so (a) is false.
2. \(|x|\) grows without bound, so it has no maximum value; (b) is true.

Teacher's Note:
1. Check each statement separately.

Answer: (d) None of these

Explanation:
1. \(f'(x)=4x^3-124x+a\), and \(f'(1)=0\) gives \(4-124+a=0\), so \(a=120\).
2. 120 is not among the options 110, 10, 55.

Teacher's Note:
1. Substitute the point into the derivative equation.

Answer: (d) \(\frac1e\)

Explanation:
1. \(f'(x)=\frac{1-\ln x}{x^2}=0\) gives \(x=e\), and \(f'\) changes from + to -.
2. \(f(e)=\frac1e\).

Teacher's Note:
1. The point \(x=epprox2.718\) lies inside \((2,\infty)\).

Answer: (b) \(\pi\)

Explanation:
1. \(f'(x)=2\cos2x-1=0\) gives \(x=\pm\frac\pi6\), where \(f=\pm\left(\frac{\sqrt3}{2}-\frac\pi6\right)\).
2. End points: \(f(\frac\pi2)=-\frac\pi2\) and \(f(-\frac\pi2)=\frac\pi2\), which are larger in magnitude.
3. The greatest is \(\frac\pi2\), the least is \(-\frac\pi2\), and the difference is \(\pi\).

Teacher's Note:
1. Compare end-point values with those at the critical points.

Answer: (c) not an extreme point

Explanation:
1. With \(f'=f''=0\) and \(f'''\ne0\), the first non-vanishing derivative is of odd order (3).
2. Odd order means a point of inflexion, not an extremum.

Teacher's Note:
1. Extrema need the first non-zero derivative to be of even order.

Answer: (c) \(\frac{dx}{dy}=0\)

Explanation:
1. A normal parallel to the x-axis means the tangent is vertical.
2. A vertical tangent has \(\frac{dx}{dy}=0\) (the slope is infinite).

Teacher's Note:
1. The normal is perpendicular to the tangent.

Answer: (c) \(p=2,q=-7\)

Explanation:
1. The point lies on the curve: \(9=8p+q\).
2. \(2y\frac{dy}{dx}=3px^2\) gives \(6\cdot4=12p\), so \(p=2\).
3. Then \(q=9-16=-7\).

Teacher's Note:
1. Use both that the point is on the curve and that the slopes match.

Answer: (c) \(500\pi\) cm\(^3\)/s

Explanation:
1. After 1 s the radius is 5 cm. \(\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\).
2. \(=4\pi\cdot25\cdot5=500\pi\) cm\(^3\)/s.

Teacher's Note:
1. Find the radius at the given instant first.

Answer: (d) (1, 2)

Explanation:
1. \(f'(x)=2x-2>0\) when \(x>1\).
2. Among the options only \((1,2)\) lies in \(x>1\).

Teacher's Note:
1. Find where the derivative is positive.

Answer: (b) 0

Explanation:
1. At \(x=1\): \(3t^2=0\), so \(t=0\).
2. \(\frac{dy}{dx}=\frac{3t^2}{6t}=\frac t2\), which is 0 at \(t=0\).

Teacher's Note:
1. Find the parameter value for the given x, then evaluate the slope.

Answer: (d) any value of n

Explanation:
1. The curve gives \(\frac{n}{a}\left(\frac xa\right)^{n-1}+\frac nb\left(\frac yb\right)^{n-1}y'=0\), so at \((a,b)\): \(y'=-\frac ba\).
2. The line also has slope \(-\frac ba\), and both pass through \((a,b)\) for every n.
3. So it touches for any n.

Teacher's Note:
1. Compare the slopes at the common point.

Answer: (b) \(\frac14\)

Explanation:
1. \(f'(x)=x^{24}(1-x)^{74}(25-100x)=0\) gives \(x=\frac14\).
2. \(f'\) changes from + to -, so it is a maximum.

Teacher's Note:
1. For \(x^p(1-x)^q\) the maximum is at \(rac{p}{p+q}\).

Answer: (a) \(\frac\pi2\)

Explanation:
1. For \(y=x^2\), slope \(m_1=2x=2\). For \(6y=7-x^3\), slope \(m_2=-\frac{x^2}{2}=-\frac12\).
2. \(m_1m_2=-1\), so the curves cut at right angles.

Teacher's Note:
1. Two curves are orthogonal when the product of slopes at the point is -1.