JEE Mathematics Application of Derivatives MCQs Set 13

Answer: (d) \( \frac{16}{9\pi} \)

Explanation:
1. Volume of coffee in the pot \( V=\pi r^2h \) with \( r=\frac{15}{2} \) constant.
2. \( \dfrac{dV}{dt}=\pi r^2\dfrac{dh}{dt} \), so \( 100=\pi\cdot\dfrac{225}{4}\dfrac{dh}{dt} \).
3. \( \dfrac{dh}{dt}=\dfrac{400}{225\pi}=\dfrac{16}{9\pi} \) cm/min, independent of the depth.

Teacher's Note:
1. The pot is a cylinder, so the rate of rise depends only on its radius.
2. The depth of 10 cm is extra information.

Answer: (b) 40

Explanation:
1. Let \( \theta \) be the angle turned at the centre and x the position of the shadow on the fence: \( x=r\tan\theta \).
2. \( \dfrac{dx}{dt}=r\sec^2\theta\,\dfrac{d\theta}{dt}=v\sec^2\theta \), where \( v=r\omega=20 \).
3. One eighth of the circle means \( \theta=\frac{\pi}{4} \), so \( \sec^2\theta=2 \).
4. Speed of the shadow \( =20\times 2=40 \) km/hr.

Teacher's Note:
1. The shadow of a point moving on a circle is where the line from the centre through the horse meets the fence.
2. Note \( r\,d\theta/dt \) is the horse's own speed.

Answer: (a) \( -2\le f(2)-f(-1)\le 1 \)

Explanation:
1. Integrating over [-1, 0]: \( 0\le f(0)-f(-1)\le 1 \).
2. Integrating over [0, 2]: \( -2\le f(2)-f(0)\le 0 \).
3. Adding the two: \( -2\le f(2)-f(-1)\le 1 \).

Teacher's Note:
1. Split the interval at 0 where the bounds on f' change.
2. Add the resulting inequalities.

Answer: (c) LMVT is applicable to f and Rolle's theorem is applicable to g with b = 1

Explanation:
1. On [1, 2], \( f(x)=x-1 \) is continuous and differentiable, so LMVT applies; but \( f(1)=0\ne f(2)=1 \), so Rolle's theorem does not apply to f.
2. \( g(1)=0+b\sin\frac{\pi}{2}=b \) and \( g(2)=1+0=1 \). Rolle's theorem needs \( g(1)=g(2) \), so b = 1.
3. With b = 1, \( g'(x)=1+\frac{\pi}{2}\cos\frac{\pi x}{2}=0 \) has the solution \( \cos\frac{\pi x}{2}=-\frac{2}{\pi} \) in (1, 2). So Rolle applies to g for b = 1.

Teacher's Note:
1. Rolle's theorem needs equal end values; LMVT does not.
2. Equate g(1) and g(2) to find b.

Answer: (d) Only (i), (ii) and (iv) are true

Explanation:
1. (i) True: a continuous function on a closed interval is bounded.
2. (ii) True: by the intermediate value theorem, since f changes sign.
3. (iii) False: the extremes may occur at the end points.
4. (iv) True: by the MVT, \( f'(c)=\frac{f(b)-f(a)}{b-a}>0 \) for some c.
5. (v) False: an increasing function such as \( f(x)=x \) on \( [-1,1] \) has no point with negative derivative.
6. So (i), (ii), (iv) only.

Teacher's Note:
1. Use the intermediate value theorem for existence of zeros and the MVT for the sign of f'.
2. A strictly increasing example kills (iii) and (v).

Answer: (d) 6

Explanation:
1. \( g=(ff')' \). f has zeros at a, e and one each in (b, c) and (c, d): 4 zeros.
2. By Rolle, f' vanishes between consecutive zeros of f: at least 3 further zeros. So \( h=ff' \) has at least 7 zeros.
3. By Rolle again, \( h'=g \) has at least 6 zeros.

Teacher's Note:
1. Notice that g is the derivative of \( ff' \).
2. Apply Rolle's theorem twice.

Answer: (a) \( \frac{240}{7} \)

Explanation:
1. After t seconds the car is \( 800-20t \) from O and the runner is 5t from O, with \( 60^\circ \) between the lines.
2. \( d^2=(800-20t)^2+(5t)^2-2(800-20t)(5t)\cos60^\circ=525t^2-36000t+640000 \).
3. \( \frac{d(d^2)}{dt}=1050t-36000=0 \Rightarrow t=\frac{240}{7} \) s, which is less than 40 s, so the car has not yet reached O.

Teacher's Note:
1. Use the cosine rule to find the distance between the two.
2. Minimise the square of the distance.

Answer: (a) at least one root

Explanation:
1. \( |a+b|<|a|+|b| \) means a and b have opposite signs: f(1) and 3 differ in sign, so \( f(1)<0 \).
2. \( |a+b|=|a|+|b| \) means same sign (or zero): since \( f(2)\ne0 \), \( f(2)>0 \).
3. f is continuous and changes sign, so it has at least one root in (1, 2) (the number of roots is not fixed).

Teacher's Note:
1. Equality in the triangle inequality means the two numbers have the same sign.
2. Then use the intermediate value theorem.

Answer: (a) \( f(x)=\begin{cases}\frac12-x,&x<\frac12\\ \left(\frac12-x\right)^2,&x\ge\frac12\end{cases} \)

Explanation:
1. The MVT needs continuity on [0, 1] and differentiability on (0, 1).
2. In option A the left derivative at \( x=\frac12 \) is -1 and the right derivative is 0, so f is not differentiable at \( \frac12\in(0,1) \).
3. The other three are differentiable on (0, 1): \( \frac{\sin x}{x} \) is continuous at 0 with value 1, \( x|x| \) is smooth, and \( |x| \) has its corner at 0, which is not in the open interval.

Teacher's Note:
1. Check the points where the formula changes inside the open interval.
2. A corner at an end point does not affect the MVT.

Answer: (a) 0

Explanation:
1. A degree 4 polynomial with only the roots 1, 2, 3 must have exactly one of them repeated.
2. At a repeated root the derivative also vanishes.
3. So one of \( f'(1),f'(2),f'(3) \) is 0 and the product is 0.

Teacher's Note:
1. A repeated root of f is a root of f' as well.
2. Count the roots: 4 roots with only 3 distinct values.

Answer: (b) 5 real roots

Explanation:
1. Roots of \( f(|x|)=0 \) come in pairs \( \pm x_0 \) (and x = 0 would be a single root).
2. 8 real roots mean f has 4 positive roots and \( f(0)\ne0 \).
3. f has degree 5, so its fifth root is also real (non-real roots occur in pairs). Hence f(x) = 0 has 5 real roots.

Teacher's Note:
1. Because \( f(|x|) \) is even, its roots are symmetric.
2. Complex roots of a real polynomial come in conjugate pairs.

Answer: (c) \( f(\frac12)<\frac12 \) and \( f(\frac13)<\frac13 \)

Explanation:
1. Differentiating both sides: \( \sqrt{1-f'^2}=f \), so \( f^2+f'^2=1 \) with \( f(0)=0 \), \( f\ge0 \).
2. Then \( f(x)=\sin x \) (the constant solution f = 0 does not satisfy the original equation).
3. Since \( \sin x<x \) for \( x>0 \): \( f(\frac12)<\frac12 \) and \( f(\frac13)<\frac13 \).

Teacher's Note:
1. Differentiate under the integral sign (fundamental theorem of calculus).
2. Use \( \sin x<x \) for positive x.

Answer: (d) \( \frac{2\pi}{5} \)

Explanation:
1. \( V=\pi r^2h \), so \( \dfrac{dV}{dt}=\pi\left(2rh\dfrac{dr}{dt}+r^2\dfrac{dh}{dt}\right) \).
2. \( =\pi\left(2\cdot2\cdot3\cdot0.1+4\cdot(-0.2)\right)=\pi(1.2-0.8)=0.4\pi \).
3. \( 0.4\pi=\dfrac{2\pi}{5} \) cm³/min.

Teacher's Note:
1. Differentiate V with respect to time using the product rule.
2. The height is decreasing, so its rate is negative.

Answer: (a) 9

Explanation:
1. The tangent at (x, y) has y-intercept \( y-x\frac{dy}{dx} \), so \( y-xy'=x^3 \).
2. Hence \( \frac{d}{dx}\left(\frac yx\right)=\frac{xy'-y}{x^2}=-x \), so \( \frac yx=-\frac{x^2}{2}+C \) and \( y=-\frac{x^3}{2}+Cx \).
3. f(1) = 1 gives \( C=\frac32 \), so \( f(x)=\frac{3x-x^3}{2} \).
4. \( f(-3)=\frac{-9+27}{2}=9 \).

Teacher's Note:
1. Write the tangent equation and read off its intercept.
2. The resulting equation is a first-order linear differential equation.

Answer: (d) \( f(x)=\begin{cases}x^2+\sin x,&x<0\\ x^3-5x,&x\ge0\end{cases} \), \( x\in[-2,3] \)

Explanation:
1. A: \( \frac{\sin3x}{x}\to3=f(0) \), so f is continuous on [0, 5] and differentiable inside. B: tan is smooth on the interval. C: \( x^3\sin\frac1x \) is continuous and differentiable on (0, 1), and at 0 it is continuous.
2. D: \( f'(0^-)=\lim(2x+\cos x)=1 \) but \( f'(0^+)=\lim(3x^2-5)=-5 \). f is not differentiable at \( 0\in(-2,3) \), so LMVT fails.

Teacher's Note:
1. Check continuity on the closed interval and differentiability on the open interval.
2. Look for a corner inside the interval.

Answer: (b) \( b=-5,\ c=8 \)

Explanation:
1. Rolle's theorem needs \( f(1)=f(2) \): \( 1+b+c=8+4b+2c\Rightarrow3b+c=-7 \).
2. \( f'\left(\frac43\right)=0 \): \( 3\cdot\frac{16}{9}+\frac{8b}{3}+c=0\Rightarrow16+8b+3c=0 \).
3. Substituting \( c=-7-3b \): \( 16+8b-21-9b=0\Rightarrow b=-5 \), and \( c=8 \).

Teacher's Note:
1. Two conditions give two equations for the two unknowns.
2. Verify by checking f(1) = 4 and f(2) = 4.

Answer: (b) at least one root in [0, 2]

Explanation:
1. The condition gives \( 6a+4b+3c+3d=0 \).
2. Let \( F(x)=\frac{ax^4}{4}+\frac{bx^3}{3}+\frac{cx^2}{2}+dx \). \( F(2)=4a+\frac{8b}{3}+2c+2d=\frac23(6a+4b+3c+3d)=0=F(0) \).
3. By Rolle's theorem \( F'(x)=ax^3+bx^2+cx+d=0 \) has a root in (0, 2).

Teacher's Note:
1. Spot the integral of the cubic whose values at 0 and 2 are equal.
2. Rolle's theorem is applied to the antiderivative.

Answer: (b) f(6) ≥ 8

Explanation:
1. By the MVT, \( f(6)-f(1)=5f'(c) \) for some \( c\in(1,6) \).
2. Since \( f'(c)\ge2 \), \( f(6)\ge f(1)+10=-2+10=8 \).

Teacher's Note:
1. Apply the mean value theorem on [1, 6].
2. The lower bound on f' gives a lower bound on the increase of f.

Answer: (a) at least (k - 1) roots of \( P'(x)=0 \)

Explanation:
1. Between any two consecutive roots of P, Rolle's theorem gives a root of \( P' \).
2. k roots form \( k-1 \) consecutive pairs, so \( P' \) has at least \( k-1 \) roots in (a, b).

Teacher's Note:
1. Rolle's theorem is applied to every pair of consecutive roots.
2. The statement is a lower bound, not an exact count.

Answer: (a) 7

Explanation:
1. By the MVT, \( f(2)-f(0)=2f'(c) \le 10 \).
2. So \( f(2)\le-3+10=7 \), attained by \( f(x)=5x-3 \).

Teacher's Note:
1. Use the mean value theorem on [0, 2].
2. The bound is achieved by a linear function.

Answer: (a) \( x\in(-1,1) \)

Explanation:
1. The ordinate changes faster than the abscissa when \( \left|\frac{dy}{dx}\right|>1 \).
2. \( 3y^2\frac{dy}{dx}=27\Rightarrow\frac{dy}{dx}=\frac{9}{y^2}>1\Rightarrow y^2<9\Rightarrow|y|<3 \).
3. Since \( x=\frac{y^3}{27} \), this is \( |x|<1 \). At x = 0 the tangent is vertical and y changes at an infinite rate, so x = 0 is included: \( (-1,1) \).

Teacher's Note:
1. Compare the rates of change dy/dt and dx/dt through dy/dx.
2. At x = 0 the abscissa momentarily does not change.

Answer: (b) 900 cm³/sec

Explanation:
1. \( V=x^3 \), so \( \frac{dV}{dt}=3x^2\frac{dx}{dt} \).
2. \( =3\cdot100\cdot3=900 \) cm³/sec.

Teacher's Note:
1. Differentiate the volume formula with respect to time.
2. Substitute the edge length and its rate.