JEE Mathematics Application of Derivatives MCQs Set 20

Answer: (c) \(-1\)

Explanation:
1. \( f(-1)=k+2 \). For \( x\lt -1 \), \( f=k-2x \) is decreasing, so values to the left are larger than \( k+2 \).
2. For \( x\gt -1 \), \( f(x)=2x+3\gt 1 \), with limit 1 at \(-1\).
3. A local minimum at \(-1\) needs \( k+2\le1 \), i.e. \( k\le-1 \). Among the options only \( k=-1 \) works.

Teacher's Note:
1. Compare the value at \(-1\) with the values on both sides.

Answer: (c) \( \frac29 \)

Explanation:
1. After 49 minutes, \( V=4500\pi-49\cdot72\pi=972\pi \). Then \( \frac43\pi r^3=972\pi \) gives \( r^3=729 \), \( r=9 \).
2. \( \frac{dV}{dt}=4\pi r^2\frac{dr}{dt} \), so \( -72\pi=4\pi\cdot81\frac{dr}{dt} \).
3. \( \frac{dr}{dt}=-\frac29 \), so the radius decreases at \( \frac29 \) m/min.

Teacher's Note:
1. First find the volume and hence the radius at the given instant.

Answer: (d) Does not exist

Explanation:
1. Let \( f(x)=2x^3+3x+k \). \( f'(x)=6x^2+3\gt 0 \), so f is strictly increasing.
2. A strictly increasing function can take the value 0 at most once.
3. So it can never have two distinct roots in \( [0,1] \), and no such k exists.

Teacher's Note:
1. Show the function is monotonic to limit the number of roots.

Answer: (c) \( \frac1{18\pi} \) cm/min

Explanation:
1. The outer radius is \( r=10+\text{thickness}=15 \) cm. The volume of the ice-coated ball changes at the rate \( \frac{dV}{dt}=-50 \).
2. \( \frac{dV}{dt}=4\pi r^2\frac{dr}{dt} \), so \( -50=4\pi\cdot225\frac{dr}{dt} \).
3. \( \frac{dr}{dt}=-\frac1{18\pi} \), so the thickness decreases at \( \frac1{18\pi} \) cm/min.

Teacher's Note:
1. The thickness of the ice changes at the same rate as the outer radius.

Answer: (a) Statement 1 is true, Statement 2 is true and Statement 2 is a correct explanation for Statement 1

Explanation:
1. By AM-GM, \( e^x+2e^{-x}\ge2\sqrt{2} \), so \( 0\lt f(x)\le\frac1{2\sqrt2} \). Statement 2 is true.
2. f is continuous with range \( \left(0,\frac1{2\sqrt2}\right] \), and \( 0\lt \frac13\lt \frac1{2\sqrt2}\approx0.354 \). So f takes the value \( \frac13 \), and Statement 1 is true.
3. Statement 1 follows from the range given in Statement 2, so Statement 2 explains it.

Teacher's Note:
1. Find the range of the function using AM-GM and check whether the given value lies in it.

Answer: (d) \( f'(c)=2g'(c) \)

Explanation:
1. Apply Cauchy's mean value theorem: \( \frac{f'(c)}{g'(c)}=\frac{f(1)-f(0)}{g(1)-g(0)} \).
2. \( f(1)-f(0)=6-2=4 \) and \( g(1)-g(0)=2-0=2 \).
3. So \( \frac{f'(c)}{g'(c)}=2 \), i.e. \( f'(c)=2g'(c) \).

Teacher's Note:
1. Use Cauchy's mean value theorem for two functions.

Answer: (c) \( \alpha=2,\ \beta=-\frac12 \)

Explanation:
1. \( f'(x)=\frac\alpha x+2\beta x+1 \). At \( x=-1 \): \( -\alpha-2\beta+1=0 \). At \( x=2 \): \( \frac\alpha2+4\beta+1=0 \).
2. From the first, \( \alpha=1-2\beta \). Substituting: \( \frac12-\beta+4\beta+1=0 \), so \( \beta=-\frac12 \).
3. Then \( \alpha=1+1=2 \).

Teacher's Note:
1. Set \( f'=0 \) at both points to get two linear equations.

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

Explanation:
1. \( f'(x)=\frac1x+2bx+a \). At \( x=-1 \): \( -1-2b+a=0 \). At \( x=2 \): \( \frac12+4b+a=0 \). Subtracting gives \( -6b-\frac32=0 \), so \( b=-\frac14 \) and \( a=\frac12 \). Statement 2 gives \( b=-\frac12 \), so it is false.
2. \( f''(x)=-\frac1{x^2}+2b=-\frac1{x^2}-\frac12\lt 0 \), so both extreme points are local maxima. Statement 1 is true.

Teacher's Note:
1. Solve for a and b first, then check the sign of the second derivative.

Answer: (a) 0

Explanation:
1. The limit gives \( \frac{f(x)}{x^2}\to2 \), so \( f(0)=f'(0)=0 \) and the coefficient of \( x^2 \) is 2: \( f(x)=ax^4+bx^3+2x^2 \).
2. \( f'(x)=x\left(4ax^2+3bx+4\right) \) has roots 1 and 2, so \( 4ax^2+3bx+4=4a(x-1)(x-2) \), giving \( 8a=4 \), \( a=\frac12 \) and \( 3b=-12a \), \( b=-2 \).
3. \( f(2)=8-16+8=0 \).

Teacher's Note:
1. Use the limit to find the lowest-order terms of the polynomial.

Answer: (b) 1

Explanation:
1. The maximum possible value is 2, attained only if \( \cos x=1 \) and \( \cos\sqrt2x=1 \).
2. That needs \( x=2\pi m \) and \( \sqrt2x=2\pi n \), so \( \sqrt2=\frac nm \) unless \( x=0 \).
3. Since \( \sqrt2 \) is irrational this only happens at \( x=0 \), so there is exactly 1 point.

Teacher's Note:
1. Both cosines must equal 1 at the same time; irrationality of \( \sqrt2 \) rules out all other points.

Answer: (c) 5

Explanation:
1. For \( |x|\lt 1 \), \( f(x)=|x|+1-x^2 \), which has V-shaped local minima at \( x=0 \) (value 1). On \( 0\lt x\lt 1 \), \( f'=1-2x=0 \) at \( x=\frac12 \), a local maximum; similarly at \( x=-\frac12 \).
2. At \( x=\pm1 \), f decreases to the left and increases to the right (for \( x\gt 1 \), \( f=x^2+x-1 \)), so \( x=\pm1 \) are local minima.
3. Total: \( 0,\ \pm\frac12,\ \pm1 \), which is 5 points.

Teacher's Note:
1. Treat the intervals \( |x|\lt 1 \) and \( |x|\gt 1 \) separately and check the corners.

Answer: (c) 9

Explanation:
1. The least degree is 3 with \( p'(x)=k(x-1)(x-3) \), where \( k\gt 0 \) for a maximum at 1. Then \( p(x)=k\left(\frac{x^3}3-2x^2+3x\right)+C \).
2. \( p(3)=C=2 \) and \( p(1)=\frac{4k}3+2=6 \) give \( k=3 \).
3. \( p'(0)=3k=9 \).

Teacher's Note:
1. A cubic is the lowest degree with one local maximum and one local minimum.

Answer: (b) 2

Explanation:
1. Let \( f(x)=x^4-4x^3+12x^2+x-1 \). \( f'(x)=4x^3-12x^2+24x+1 \) and \( f''(x)=12\left(x^2-2x+2\right)\gt 0 \), so \( f' \) is increasing and has exactly one real root, near \( x\approx-0.04 \).
2. So f has a single minimum there, with value about \(-1.02\lt 0\), and \( f\to\infty \) at both ends.
3. Hence f has exactly 2 real roots.

Teacher's Note:
1. Show that the derivative has only one real root, so the graph has a single dip.

Answer: (b) 1

Explanation:
1. \( g=e^f \) has local maxima exactly where f has local maxima.
2. The sign of \( f' \) changes only at \( x=2009 \) (odd power) and \( x=2011 \) (odd power); the even powers at 2010 and 2012 do not change the sign.
3. \( f'\gt 0 \) for \( x\lt 2009 \), \( f'\lt 0 \) on \( (2009,2011) \), \( f'\gt 0 \) for \( x\gt 2011 \). So there is a local maximum at 2009 and a local minimum at 2011: 1 local maximum.

Teacher's Note:
1. Only factors with odd powers change the sign of the derivative.

Answer: (b) 2

Explanation:
1. The denominator is \( \frac{1-\cos2\theta}2+\frac32\sin2\theta+\frac{5(1+\cos2\theta)}2=3+2\cos2\theta+\frac32\sin2\theta \).
2. Its minimum is \( 3-\sqrt{4+\frac94}=3-\frac52=\frac12 \).
3. So the maximum value of the expression is 2.

Teacher's Note:
1. Reduce the denominator to the form \( c+A\cos2\theta+B\sin2\theta \).

Answer: (b) 3

Explanation:
1. Let \( g(x)=x^3-3x \), with local maximum 2 at \( x=-1 \) and local minimum \(-2\) at \( x=1 \).
2. The equation \( g(x)=-c \) has three distinct real roots when \( -2\lt -c\lt 2 \), i.e. \(-2\lt c\lt 2\).
3. The integers are \(-1,0,1\), so there are 3.

Teacher's Note:
1. A cubic has three distinct real roots when the line lies strictly between its local maximum and minimum values.

Answer: (b) 1

Explanation:
1. For \( x\lt 2 \): \( f=4x^3+x^2-2x \), \( f'=12x^2+2x-2=0 \) gives \( x=\frac13 \) (or \(-\frac12\), outside the interval), and \( f''\gt 0 \) so \( x=\frac13 \) is a minimum.
2. At \( x=0 \), \( f'(0)=-2\lt 0 \), so \( x=0 \) is an end-point maximum, not a minimum.
3. For \( x\ge2 \): \( f'=12x^2-2x+2\gt 0 \), and at \( x=2 \) the derivative is positive on both sides, so there is no extremum. Hence 1 point of minimum.

Teacher's Note:
1. Split the modulus at \( x=2 \), and check the end points separately.

Answer: (b) 1

Explanation:
1. Let \( g(x)=3\tan x+x^3-2 \). \( g'(x)=3\sec^2x+3x^2\gt 0 \), so g is increasing.
2. \( g(0)=-2\lt 0 \) and \( g\left(\frac\pi4\right)=3+\frac{\pi^3}{64}-2\gt 0 \).
3. So g has exactly one zero in the interval.

Teacher's Note:
1. A continuous increasing function with a sign change has exactly one zero.

Answer: (b) 2

Explanation:
1. The normal at P has slope \( -\frac1{na^{n-1}} \): \( y-a^n=-\frac{x-a}{na^{n-1}} \). Putting \( x=0 \): \( b=a^n+\frac{a^{2-n}}n \).
2. For \( b\to\frac12 \) as \( a\to0 \), the term \( a^{2-n} \) must tend to a constant, so \( n=2 \), giving \( b=a^2+\frac12\to\frac12 \).
3. For \( n=1 \), \( b\to0 \); for \( n\ge3 \), \( b\to\infty \).

Teacher's Note:
1. Find the y-intercept of the normal as a function of a, then examine the limit for each n.

Answer: (b) 4

Explanation:
1. The tangent at \( \left(t,3-t^2\right) \), \( t\gt 0 \), is \( y=3+t^2-2tx \), with y-intercept \( 3+t^2 \) and x-intercept \( \frac{3+t^2}{2t} \).
2. The area is \( A=\frac{\left(3+t^2\right)^2}{4t} \), and \( \frac{dA}{dt}\propto\left(3+t^2\right)\left(3t^2-3\right)=0 \) gives \( t=1 \).
3. \( A=\frac{16}4=4 \).

Teacher's Note:
1. Write the tangent line in terms of the point of contact and minimise the area.