JEE Mathematics Application of Derivatives MCQs Set 18

Answer: (c) 2

Explanation:
1. \( \frac{dx}{dt}=2t\cos t \) and \( \frac{dy}{dt}=2t\sin t \), so the tangent slope is \( \tan t \).
2. The normal has slope \(-\cot t\): \( (X-x)\cos t+(Y-y)\sin t=0 \), i.e. \( X\cos t+Y\sin t=x\cos t+y\sin t \).
3. The right side is \( 2\cos^2t+2\sin^2t=2 \), so the normal is \( X\cos t+Y\sin t=2 \), whose distance from the origin is 2.

Teacher's Note:
1. Write the normal in the form \( X\cos t+Y\sin t=p \); then p is the distance from the origin.

Answer: (b) \(-1\)

Explanation:
1. Rolle's theorem needs \( f(-1)=f(1) \): \( -2+b-c=2+b+c \), so \( c=-2 \).
2. \( f'(x)=6x^2+2bx+c \) and \( f'\left(\frac12\right)=0 \) gives \( \frac32+b+c=0 \), so \( b=\frac12 \).
3. Then \( 2b+c=1-2=-1 \).

Teacher's Note:
1. Use \( f(-1)=f(1) \) first and then \( f'(c)=0 \).

Answer: (d) \( 4\sqrt2 \)

Explanation:
1. A tangent from P is \( y=mx+h \) with \( \frac{h}{\sqrt{1+m^2}}=4 \), so \( m^2=\frac{h^2}{16}-1 \). It meets the x-axis at \( x=-\frac hm \).
2. \( AB=\frac{2h}{|m|} \), so the area is \( \frac12\cdot\frac{2h}{|m|}\cdot h=\frac{h^2}{|m|}=\frac{4h^2}{\sqrt{h^2-16}} \).
3. Put \( t=h^2 \) and minimise \( \frac t{\sqrt{t-16}} \): the derivative vanishes when \( 2(t-16)=t \), i.e. \( t=32 \). So \( h=4\sqrt2 \).

Teacher's Note:
1. Express the area in terms of h alone and set the derivative to zero.

Answer: (d) \( \left(2^{-0.4},1\right) \)

Explanation:
1. The sign of \( f' \) is that of \( (1+x)^{-0.4}\left(1+x^{0.6}\right)-(1+x)^{0.6}x^{-0.4} \), which has the sign of \( 1+x^{0.6}-\frac{1+x}{x^{0.4}}=1-x^{-0.4} \).
2. For \( 0\lt x\lt 1 \), \( x^{-0.4}\gt 1 \), so \( f'\lt 0 \) and f is decreasing.
3. Hence the maximum is \( f(0)=1 \) and the minimum is \( f(1)=\frac{2^{0.6}}2=2^{-0.4} \).

Teacher's Note:
1. Find the sign of the derivative on the whole interval, then compare end values.

Answer: (b) \(-1\)

Explanation:
1. \( f(-1)=f(1) \) gives \( -2+a-b=2+a+b \), so \( b=-2 \).
2. \( f'\left(\frac12\right)=\frac32+a+b=0 \) gives \( a=\frac12 \).
3. So \( 2a+b=1-2=-1 \).

Teacher's Note:
1. Rolle's theorem requires equal end values and a zero of the derivative at c.

Answer: (c) \( \pi/4 \)

Explanation:
1. \( \frac{dy}{d\theta}=3\cos2\theta \) and \( \frac{dx}{d\theta}=e^\theta(\sin\theta+\cos\theta) \).
2. The tangent is horizontal when \( \frac{dy}{d\theta}=0 \) and \( \frac{dx}{d\theta}\ne0 \). \( \cos2\theta=0 \) gives \( \theta=\frac\pi4,\frac{3\pi}4 \).
3. At \( \theta=\frac{3\pi}4 \), \( \sin\theta+\cos\theta=0 \) so \( \frac{dx}{d\theta}=0 \) too, and the slope is not zero there. Hence \( \theta=\frac\pi4 \).

Teacher's Note:
1. A horizontal tangent needs the numerator of the slope to be zero while the denominator is not.

Answer: (a) \( 260/\sqrt{37} \)

Explanation:
1. By the cosine rule, \( AB^2=a^2+b^2+ab \) where \( a=OA,\ b=OB \) and \( \cos120^\circ=-\frac12 \). At the instant, \( AB^2=64+36+48=148 \), so \( AB=2\sqrt{37} \).
2. Differentiating: \( 2AB\cdot AB'=2aa'+2bb'+a'b+ab'=320+360+120+240=1040 \).
3. So \( AB'=\frac{1040}{4\sqrt{37}}=\frac{260}{\sqrt{37}} \).

Teacher's Note:
1. Write the distance using the cosine rule and differentiate with respect to time.

Answer: (b) (ii) and (iii) only

Explanation:
1. f is increasing and g is decreasing.
2. (i): \( g(x-1)\gt g(x) \), so \( f(g(x-1))\gt f(g(x)) \); (i) is false. (ii): \( g(x+1)\lt g(x) \), so \( f(g(x+1))\lt f(g(x)) \); (ii) is true.
3. (iii): \( f(x+1)\gt f(x) \), so \( g(f(x+1))\lt g(f(x)) \); (iii) is true. (iv): \( f(x-1)\lt f(x) \) gives \( g(f(x-1))\gt g(f(x)) \); (iv) is false.

Teacher's Note:
1. An increasing function preserves inequalities while a decreasing function reverses them.

Answer: (c) \( \frac1{36} \)

Explanation:
1. \( V=\frac43\pi r^3=288\pi \) gives \( r^3=216 \), so \( r=6 \).
2. \( \frac{dV}{dt}=4\pi r^2\frac{dr}{dt} \), so \( 4\pi=4\pi\cdot36\cdot\frac{dr}{dt} \).
3. Hence \( \frac{dr}{dt}=\frac1{36} \) cm/sec.

Teacher's Note:
1. Differentiate the volume formula with respect to time.

Answer: (a) Statement I is true; Statement II is true; Statement II is a correct explanation for Statement I.

Explanation:
1. Let \( h(x)=x\log x+x-2 \). Then \( h(1)=-1\lt 0 \) and \( h(2)=2\log2\gt 0 \).
2. f is increasing and g is decreasing on \([1,2]\), so \( h \) is increasing, continuous and changes sign, giving exactly one root in \( (1,2) \).
3. So both statements are true, and Statement II (the graphs intersect in [1,2] because one rises and the other falls from \( f\lt g \) to \( f\gt g \)) explains Statement I.

Teacher's Note:
1. Use the intermediate value theorem on the difference of the two functions.

Answer: (c) Statement I is true; Statement II is false.

Explanation:
1. For \( f(x)=x^2\left(e^x+e^{-x}\right) \), \( f'(x)=2x\left(e^x+e^{-x}\right)+x^2\left(e^x-e^{-x}\right)\gt 0 \) for \( x\gt 0 \), so Statement I is true.
2. For \( x^2e^{-x} \), the derivative is \( (2x-x^2)e^{-x} \), which is negative for \( x\gt 2 \). So \( x^2e^{-x} \) is not increasing for all \( x\gt 0 \), and Statement II is false.

Teacher's Note:
1. Check the derivative of each part separately; the sum can increase even when one part does not.

Answer: (b) 20 cm/min

Explanation:
1. At depth y the radius is \( \frac y2 \), so \( V=\frac13\pi\left(\frac y2\right)^2y=\frac{\pi y^3}{12} \).
2. \( \frac{dV}{dt}=\frac{\pi y^2}4\frac{dy}{dt} \). With \( \frac{dV}{dt}=77\ \text{litres/min}=77000\ \text{cm}^3/\text{min} \) and \( y=70 \) cm: \( 77000=\frac{22}{7}\cdot\frac{4900}4\cdot\frac{dy}{dt}=3850\frac{dy}{dt} \).
3. So \( \frac{dy}{dt}=20 \) cm/min.

Teacher's Note:
1. Keep all lengths in centimetres and use similar triangles to express the radius in terms of the depth.

Answer: (a) 4

Explanation:
1. Let the horizontal distance be x and the cord length s, with \( s^2=x^2+300^2 \). When \( s=500 \), \( x=400 \).
2. \( s\frac{ds}{dt}=x\frac{dx}{dt} \), so \( \frac{ds}{dt}=\frac{400}{500}\cdot5 \).
3. \( \frac{ds}{dt}=4 \) km/hr.

Teacher's Note:
1. Relate the cord length to the horizontal distance with Pythagoras and differentiate.

Answer: (b) 1

Explanation:
1. On \( n\le x\lt n+1 \) the curve is \( y=e^{x-n} \). The tangent at \( x=t \) passes through \( \left(-\frac12,0\right) \) when \( 0-e^{t-n}=e^{t-n}\left(-\frac12-t\right) \).
2. This gives \( t=\frac12 \), which does not depend on n.
3. The point of tangency must lie in \( [n,n+1) \), so only \( n=0 \) works. Hence exactly 1 tangent.

Teacher's Note:
1. The curve is a shifted copy of \( e^x \) on each unit interval; find the tangency point and see in which interval it lies.

Answer: (c) \( (2,2) \) and \( (-2,2) \)

Explanation:
1. For \( \left(x,\frac{x^2}2\right) \), \( d^2=x^2+\left(\frac{x^2}2-3\right)^2 \). With \( t=x^2 \): \( \frac{d(d^2)}{dt}=1+\left(\frac t2-3\right)=0 \) gives \( t=4 \).
2. So \( x=\pm2 \) and \( y=2 \), giving \( d^2=4+1=5 \), which is smaller than the value 6 at \( t=2 \).
3. The nearest points are \( (2,2) \) and \( (-2,2) \).

Teacher's Note:
1. The parabola is symmetric about the y-axis, so there are two nearest points.

Answer: (b) (ii) and (iii) only

Explanation:
1. \( g'(x)=f'\left(\frac x2\right)-f'(1-x) \). Since \( f''\lt 0 \), \( f' \) is decreasing.
2. So \( g'(x)\gt 0 \) when \( \frac x2\lt 1-x \), i.e. \( x\lt \frac23 \), and \( g'(x)\lt 0 \) when \( x\gt \frac23 \).
3. Hence g increases in \( \left[0,\frac23\right] \) and decreases in \( \left[\frac23,1\right] \).

Teacher's Note:
1. \( f''\lt 0 \) means \( f' \) is decreasing, so compare the arguments of \( f' \).

Answer: (b) (i) and (iii) only

Explanation:
1. On \( [0,1] \), f increases from \(-1\) to \( 0 \); for \( x\gt 1 \), \( f=-x\lt -1 \). So the maximum value 0 is attained at \( x=1 \), and (i) is true, (ii) is false.
2. f is one-one with range \( (-\infty,0] \). The inverse is \( \sqrt{1-y^2} \) for \( -1\le y\le0 \) and \(-y\) for \( y\lt -1 \).
3. The smallest value of \( f^{-1} \) is 0, attained at \( y=-1 \). So (iii) is true and (iv) is false.

Teacher's Note:
1. Find the range of f and then read off the inverse, whose minimum is the smallest x-value of f.

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

Explanation:
1. \( f'(x)=\frac1{1+x^2}-\frac1{2x}=-\frac{(x-1)^2}{2x(1+x^2)}\le0 \) for \( x\gt 0 \), so f decreases on \( (0,\infty) \) and (iii) is true.
2. Hence on \( \left[\frac1{\sqrt3},\sqrt3\right] \) the greatest value is \( f\left(\frac1{\sqrt3}\right)=\frac\pi6+\frac14\ln3 \) and the least is \( f(\sqrt3)=\frac\pi3-\frac14\ln3 \), so (i) and (ii) are true.
3. f is defined only for \( x\gt 0 \), so (iv) is false.

Teacher's Note:
1. Factor the derivative as a perfect square to see that it never changes sign.

Answer: (d) \( \frac12 \) of the radius of the cone

Explanation:
1. Let the cone have radius R and height H, and the cylinder radius r and height h. By similar triangles \( h=H\left(1-\frac rR\right) \).
2. Curved surface \( S=2\pi rh=2\pi H\left(r-\frac{r^2}R\right) \).
3. \( \frac{dS}{dr}=2\pi H\left(1-\frac{2r}R\right)=0 \) gives \( r=\frac R2 \).

Teacher's Note:
1. Use similar triangles to eliminate the height of the cylinder.

Answer: (d) \( -e^{-x} \)

Explanation:
1. For \( f=-e^{-x} \): \( f'=e^{-x}\gt 0 \) and \( f''=-e^{-x}\lt 0 \) for all x.
2. \( e^x \) has \( f''\gt 0 \), \( e^{-x} \) has \( f'\lt 0 \), and \( \sin x \) changes the signs of both derivatives.
3. So only \( -e^{-x} \) is increasing and concave downward everywhere.

Teacher's Note:
1. Compute the first and second derivatives of each option and check their signs.

Answer: (d) \( a=-\frac32,\ b=\frac92 \)

Explanation:
1. The point lies on the curve: \( a+b=3 \).
2. At a point of inflection \( y''=6ax+2b=0 \), so \( 6a+2b=0 \) and \( b=-3a \).
3. Then \( a-3a=3 \) gives \( a=-\frac32 \) and \( b=\frac92 \).

Teacher's Note:
1. Use the point on the curve and the condition \( y''=0 \).