JEE Mathematics Application of Derivatives MCQs Set 15

Answer: (a) \( \frac32 \)

Explanation:
1. \( f(1)=-1 \), \( f(2)=0 \), so \( \dfrac{f(2)-f(1)}{2-1}=1 \).
2. \( f'(c)=2c-2=1 \) gives \( c=\frac32 \).
3. \( \frac32\in(1,2) \).

Teacher's Note:
1. Set \( f'(c) \) equal to the slope of the chord.
2. Check that c lies inside the open interval.

Answer: (b) \( \frac12 \)

Explanation:
1. \( f(0)=0 \), \( f(1)=1 \), so the chord slope is 1.
2. \( f'(c)=2-2c=1 \) gives \( c=\frac12 \).
3. \( \frac12\in(0,1) \).

Teacher's Note:
1. Compute the chord slope from the end values.
2. Solve the linear equation for c.

Answer: (c) \( \frac13 \)

Explanation:
1. \( f(0)=3 \), \( f(1)=1 \), so the chord slope is \( -2 \).
2. \( f'(c)=3c^2-4c-1=-2 \) gives \( 3c^2-4c+1=0 \), so \( c=\frac13 \) or \( c=1 \).
3. Only \( c=\frac13 \) lies in the open interval \( (0,1) \).

Teacher's Note:
1. The theorem gives c in the open interval.
2. Reject the root c = 1.

Answer: (d) 3

Explanation:
1. \( f(1)=0 \), \( f(4)=6 \), so the chord slope is \( \frac63=2 \).
2. \( f'(x)=3x^2-12x+11=2 \) gives \( 3x^2-12x+9=0 \), so \( x=1 \) or \( x=3 \).
3. Only \( x=3 \) is in the open interval \( (1,4) \), so \( c=3 \).

Teacher's Note:
1. Expand f to differentiate it.
2. The point x = 1 is an end point, not allowed.

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

Explanation:
1. \( f'(x)=\dfrac{2a(x^2-1)}{(x^2+ax+1)^2} \), so \( f'(1)=f'(-1)=0 \). This rules out the last two options (the product is 0).
2. \( f''(1)=\dfrac{4a}{(2+a)^2} \) and \( f''(-1)=-\dfrac{4a}{(2-a)^2} \) (the other term vanishes because \( x^2-1=0 \)).
3. Then \( (2+a)^2f''(1)=4a \) and \( (2-a)^2f''(-1)=-4a \), whose sum is 0. The second option gives \( 4a+4a\ne0 \).

Teacher's Note:
1. First find where the first derivative vanishes.
2. At a root of the numerator only the derivative of the numerator survives in the second derivative.

Answer: (b) The graph of f must have a sharp corner at \( x=2 \)

Explanation:
1. At \( x=-5 \) and \( x=4 \) the derivative changes from positive to negative, so both are relative maxima.
2. At \( x=2 \) it changes from negative to positive, so \( x=2 \) is a relative minimum, and \( f'(2) \) does not exist.
3. A relative minimum at a point where f' does not exist, with f concave down on both sides, must be a sharp corner (cusp-like point).

Teacher's Note:
1. Read the sign change of f' at each critical point.
2. Concave down on both sides of a minimum forces a corner.

Answer: (c) the curve is always concave down

Explanation:
1. \( f''(x)<0 \) for all \( x\ne2 \), so f is concave down on \( (-\infty,2) \) and on \( (2,\infty) \).
2. There is no change of concavity, hence no point of inflection, and the curve is never concave up.
3. The last option reverses the actual monotonicity: f increases for \( x<-5 \) and on \( (2,4) \).

Teacher's Note:
1. Concavity is read from the second derivative, monotonicity from the first.
2. A point of inflection needs a sign change of f''.

Answer: (d) 4

Explanation:
1. \( f''(x)=12x^2+6ax+3 \). Concave upward everywhere needs \( f''(x)\ge0 \) for all x (without changing sign).
2. Discriminant \( 36a^2-144\le0 \) gives \( |a|\le2 \).
3. Non-zero integers: \( -2,-1,1,2 \), that is 4 values.

Teacher's Note:
1. A quadratic is non-negative for all x iff its discriminant is at most zero.
2. Exclude a = 0 as the question says non-zero.

Answer: (a) 140

Explanation:
1. Height \( h=500\tan\theta \), so \( \dfrac{dh}{dt}=500\sec^2\theta\,\dfrac{d\theta}{dt} \).
2. At \( \theta=\frac\pi4 \), \( \sec^2\theta=2 \).
3. \( \dfrac{dh}{dt}=500\times2\times0.14=140 \) ft/min.

Teacher's Note:
1. Relate h to the angle using the fixed horizontal distance.
2. Differentiate with respect to time.

Answer: (b) continuous but not differentiable at \( x=1 \)

Explanation:
1. \( f'(t)=3t^2-2t+1>0 \) for all t (discriminant negative), so f is increasing and \( g(x)=f(x) \) on \( [0,1] \).
2. \( g(1)=f(1)=2 \) and \( \lim_{x\to1^+}(3-x)=2 \), so g is continuous at 1.
3. Left derivative \( f'(1)=2 \); right derivative \( -1 \). They differ, so g is not differentiable at 1.

Teacher's Note:
1. First show that f is increasing so that the maximum equals f(x).
2. Compare the one-sided derivatives.

Answer: (c) 9

Explanation:
1. Horizontal distance \( x=15\times\frac2{60}=\frac12 \) km. Distance from P: \( l=\sqrt{x^2+h^2}=\sqrt{\frac14+\frac49}=\frac56 \) km.
2. \( l^2=x^2+h^2 \) with h constant gives \( l\dfrac{dl}{dt}=x\dfrac{dx}{dt} \).
3. \( \dfrac{dl}{dt}=\dfrac xl\cdot15=\dfrac{1/2}{5/6}\cdot15=9 \) km/h.

Teacher's Note:
1. The height is constant, only x changes.
2. Use \( l^2=x^2+h^2 \) and differentiate.

Answer: (d) \( f(x)\le g(x) \) for all \( x\ge0 \)

Explanation:
1. Let \( F(x)=f(x)-g(x) \). Then \( F'(x)\le0 \) for \( x\ge0 \), so F is non-increasing there.
2. Hence \( F(x)\le F(0)=f(0)-g(0)=0 \) for \( x\ge0 \).
3. So \( f(x)\le g(x) \) for all \( x\ge0 \).

Teacher's Note:
1. Study the difference of the two functions.
2. A non-increasing function starting at 0 stays at or below 0.

Answer: (a) \( e \)

Explanation:
1. Perimeter: \( 2r+r\theta=4e \), so \( r\theta=4e-2r \).
2. Area \( =\frac12r^2\theta=\frac12r(4e-2r)=r(2e-r) \).
3. This quadratic in r is maximum at \( r=e \), where the area is \( e^2 \).

Teacher's Note:
1. Eliminate the angle using the perimeter condition.
2. A downward parabola has its maximum at the midpoint of its roots.

Answer: (b) 2 and 0

Explanation:
1. \( y'=(x-1)(3x-1) \), so the critical points are \( x=\frac13 \) and \( x=1 \).
2. Values: \( y(0)=0 \), \( y(\frac13)=\frac4{27} \), \( y(1)=0 \), \( y(2)=2 \).
3. Largest is 2 (at \( x=2 \)); smallest is 0 (at \( x=0 \) and \( x=1 \)).

Teacher's Note:
1. On a closed interval compare critical values with the end values.
2. The local maximum \( \frac4{27} \) is not the absolute maximum.

Answer: (c) \( [-8,72] \)

Explanation:
1. \( f'(x)=12x^2-12=0 \) gives \( x=\pm1 \); both lie in \( [-1,3] \).
2. \( f(-1)=8 \), \( f(1)=-8 \), \( f(3)=108-36=72 \).
3. Least value \( -8 \), greatest value 72; f is continuous so the image is \( [-8,72] \).

Teacher's Note:
1. Compare critical values with the end values.
2. Continuity gives every value in between.

Answer: (d) \( \left[\frac13,3\right] \)

Explanation:
1. \( f'(x)=\dfrac{2(x^2-1)}{(x^2+x+1)^2} \), zero at \( x=\pm1 \).
2. f increases on \( (-\infty,-1) \), decreases on \( (-1,1) \) and increases on \( (1,\infty) \). So \( f(-1)=3 \) is a local maximum and \( f(1)=\frac13 \) a local minimum, and \( f\to1 \) as \( x\to\pm\infty \).
3. Hence the range is \( [\frac13,3] \).

Teacher's Note:
1. Find the greatest and least values from the critical points.
2. Check the behaviour for large x.

Answer: (a) \( (-\infty,1)\cup[9,\infty) \)

Explanation:
1. Put \( y=\dfrac{x^2+x-2}{x^2+x} \): \( (y-1)(x^2+x)+2=0 \).
2. For \( y=1 \) this reads \( 2=0 \), impossible, so \( y\ne1 \). For \( y\ne1 \) the equation \( x^2+x+\dfrac2{y-1}=0 \) needs a real root: \( 1-\dfrac8{y-1}\ge0 \).
3. This gives \( y<1 \) or \( y\ge9 \); the value 9 is attained at \( x=-\frac12 \). Range \( (-\infty,1)\cup[9,\infty) \).

Teacher's Note:
1. Treat y as a parameter and demand a real root x.
2. Check separately whether the boundary values are attained.

Answer: (b) \( x=\cos x \)

Explanation:
1. \( f=\dfrac1{g} \) where \( g(x)=\dfrac1x+\tan x>0 \). f is maximum where g is minimum.
2. \( g'(x)=-\dfrac1{x^2}+\sec^2x=0 \) gives \( x=\cos x \) (for \( x>0 \)); g decreases before and increases after this point.
3. The equation \( x=\cos x \) has a root near 0.739, which is not \( \frac\pi4 \); the other equations have no root in \( (0,\frac\pi2) \) apart from the trivial one.

Teacher's Note:
1. Reciprocal turns a maximum into a minimum.
2. The equation \( x=\cos x \) cannot be solved in closed form.

Answer: (c) the x-axis and the y-axis

Explanation:
1. Near \( x=0 \), \( 1-\cos x\approx\frac{x^2}2 \), so \( y\approx\frac12x^{7/3} \).
2. \( \dfrac{y}{x}\approx\frac12x^{4/3}\to0 \), so \( \dfrac{dy}{dx}\Big|_{x=0}=0 \).
3. The tangent at the origin is the horizontal line \( y=0 \) and the normal is the vertical line \( x=0 \).

Teacher's Note:
1. Use the definition of the derivative at 0 since \( x^{1/3} \) is not differentiable there.
2. A horizontal tangent has a vertical normal.

Answer: (d) \( \dfrac{y^2}{c} \)

Explanation:
1. \( \dfrac{dy}{dx}=\sinh\dfrac xc \), so the normal length is \( y\sqrt{1+\left(\frac{dy}{dx}\right)^2} \).
2. \( 1+\sinh^2\frac xc=\cosh^2\frac xc=\dfrac{y^2}{c^2} \).
3. Length \( =y\cdot\dfrac yc=\dfrac{y^2}c \).

Teacher's Note:
1. Normal length is \( y\sqrt{1+y'^2} \).
2. Use the identity \( \cosh^2-\sinh^2=1 \).