JEE Mathematics Limits Continuity and Differentiability MCQs Set 12

Answer: (d) meets the curve again in the fourth quadrant

Explanation:
1. The curve factorises as \( (x-y)(x+3y)=0 \), a pair of lines. The point \( (1,1) \) lies on \( y=x \), so the slope of the tangent there is 1 and the normal is \( x+y=2 \).
2. It meets the other line \( x=-3y \) where \( -3y+y=2 \), i.e. \( y=-1 \), \( x=3 \). The point \( (3,-1) \) lies in the fourth quadrant. Answer: (d)

Teacher's Note:
1. Factorise the equation of the curve.
2. Intersect the normal with the other line.

Answer: (d) \( \pi \)

Explanation:
1. \( \sin(\pi\cos^2x)=\sin(\pi-\pi\sin^2x)=\sin(\pi\sin^2x) \).
2. \( \frac{\sin(\pi\sin^2x)}{x^2}=\frac{\sin(\pi\sin^2x)}{\pi\sin^2x}\cdot\pi\cdot\frac{\sin^2x}{x^2}\to1\cdot\pi\cdot1=\pi \). Answer: (d)

Teacher's Note:
1. Use \( \\cos^2x=1-\\sin^2x \) and \( \\sin(\\pi-\\theta)=\\sin\\theta \).
2. Apply \( \lim\frac{\sin t}t=1 \).

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

Explanation:
1. Let \( h(x)=f(x)-2g(x) \). Then \( h(0)=2-0=2 \) and \( h(1)=6-2\cdot2=2 \).
2. By Rolle's theorem there is \( c\in(0,1) \) with \( h'(c)=0 \), i.e. \( f'(c)-2g'(c)=0 \), so \( f'(c)=2g'(c) \). Answer: (d)

Teacher's Note:
1. Look for a combination \( f-kg \) taking equal values at 0 and 1.
2. Apply Rolle's theorem.

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

Explanation:
1. \( f'(x)=\frac\alpha x+2\beta x+1 \). At the extreme points \( f'(-1)=0 \) and \( f'(2)=0 \).
2. \( -\alpha-2\beta+1=0 \) and \( \frac\alpha2+4\beta+1=0 \). Solving: \( \alpha=2,\ \beta=-\frac12 \). Answer: (c)

Teacher's Note:
1. Differentiate and set \( f'=0 \) at both points.
2. Solve the two linear equations.

Answer: (a) \( \pm1 \)

Explanation:
1. \( \frac{dy}{dx}=|x|=2 \), so \( x=\pm2 \). The points are \( (2,2) \) and \( (-2,-2) \) because \( \int_0^{\pm2}|t|\,dt=\pm2 \).
2. The tangents are \( y-2=2(x-2) \) and \( y+2=2(x+2) \). Putting \( y=0 \) gives \( x=1 \) and \( x=-1 \). Answer: (a)

Teacher's Note:
1. Differentiate the integral.
2. Find the points of tangency and the tangent lines.

Answer: (a) \( \frac1{\sqrt2} \)

Explanation:
1. Let \( \tan^{-1}x=\theta \), so \( x=\tan\theta \) and \( y=\sec\theta=\sqrt{1+x^2} \).
2. \( \frac{dy}{dx}=\frac x{\sqrt{1+x^2}} \), which at \( x=1 \) equals \( \frac1{\sqrt2} \). Answer: (a)

Teacher's Note:
1. Simplify \( \\sec(\\tan^{-1}x) \) using a right triangle.
2. Differentiate.

Answer: (c) \( \frac29 \)

Explanation:
1. After 49 minutes the volume is \( 4500\pi-72\pi\cdot49=972\pi \), so \( \frac43\pi r^3=972\pi\Rightarrow r^3=729\Rightarrow r=9 \).
2. \( \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}=-72\pi \), so \( \left|\frac{dr}{dt}\right|=\frac{72\pi}{4\pi\cdot81}=\frac29 \) m/min. Answer: (c)

Teacher's Note:
1. Find the radius after 49 minutes.
2. Differentiate \( V=\frac43\pi r^3 \) with respect to time.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. \( f'(x)=\frac1x+2bx+a \). From \( f'(-1)=0 \): \( a-2b=1 \); from \( f'(2)=0 \): \( a+4b=-\frac12 \). So \( a=\frac12,\ b=-\frac14 \), and R is true.
2. \( f''(x)=-\frac1{x^2}+2b=-\frac1{x^2}-\frac12\lt0 \) for all \( x\ne0 \), so both extreme points are local maxima and A is true.
3. R only gives the values of \( a,b \); it does not by itself explain why the points are maxima. Both are true but II is not the correct explanation of I. Answer: (b)

Teacher's Note:
1. Use \( f'(-1)=f'(2)=0 \) to find \( a,b \).
2. Check the sign of \( f'' \).

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. For \( 2\le x\le5 \): \( f(x)=(x-2)+(5-x)=3 \), a constant. So \( f'(4)=0 \) and A is true.
2. \( f \) is continuous on \( [2,5] \), differentiable on \( (2,5) \) and \( f(2)=3=f(5) \), so R is true.
3. R gives the conditions of Rolle's theorem, which only guarantees some point with zero derivative; it does not explain why \( f'(4)=0 \) (that is because \( f \) is constant on the interval). So both are true but II is not the correct explanation. Answer: (b)

Teacher's Note:
1. Write \( f \) as a piecewise function.
2. Compare what R actually implies.

Answer: (c) \( -\left(\frac{d^2y}{dx^2}\right)\left(\frac{dy}{dx}\right)^{-3} \)

Explanation:
1. \( \frac{dx}{dy}=\left(\frac{dy}{dx}\right)^{-1} \).
2. \( \frac{d^2x}{dy^2}=\frac d{dy}\left(\frac{dy}{dx}\right)^{-1}=-\left(\frac{dy}{dx}\right)^{-2}\frac d{dx}\left(\frac{dy}{dx}\right)\frac{dx}{dy}=-\left(\frac{dy}{dx}\right)^{-3}\frac{d^2y}{dx^2} \). Answer: (c)

Teacher's Note:
1. Start from \( dx/dy=(dy/dx)^{-1} \).
2. Differentiate again with respect to \( y \) using the chain rule.

Answer: (a) 3

Explanation:
1. The denominator tends to 0 and the limit is finite, so the numerator must tend to 0: \( \left(\lim f(x)\right)^2-9=0 \).
2. Hence \( \lim f(x)=\pm3 \). Since \( f(x)\ge0 \), the limit is \( 3 \). Answer: (a)

Teacher's Note:
1. The numerator must vanish if the denominator does.
2. Use the range of \( f \) to choose the sign.

Answer: (d) \( 2af(a)-a^2f'(a) \)

Explanation:
1. The form is \( \frac00 \). Apply L'Hopital's rule: the derivative of the numerator is \( 2xf(a)-a^2f'(x) \) and of the denominator is 1.
2. At \( x=a \) the limit is \( 2af(a)-a^2f'(a) \). Answer: (d)

Teacher's Note:
1. Check the 0/0 form.
2. Differentiate numerator and denominator.

Answer: (c) A is true, but R is false.

Explanation:
1. \( f(x)=x\sin\frac1x \) for \( x\ne0 \) and \( f(0)=0 \). Since \( \left|x\sin\frac1x\right|\le|x|\to0 \), \( f \) is continuous at 0, so A is true.
2. \( f_2(x)=\sin\frac1x \) has no limit as \( x\to0 \) (it oscillates between \( -1 \) and \( 1 \)), so \( f_2 \) is not continuous at 0. R is false.
3. A is true but R is false. Answer: (c)

Teacher's Note:
1. Use the squeeze theorem for \( x\sin(1/x) \).
2. Check whether \( \sin(1/x) \) has a limit at 0.

Answer: (b) -4

Explanation:
1. \( g'(x)=2f(2f(x)+2)\cdot f'(2f(x)+2)\cdot2f'(x) \).
2. At \( x=0 \): \( 2f(0)+2=0 \), so \( g'(0)=2f(0)\,f'(0)\cdot2f'(0)=2(-1)(1)(2)(1)=-4 \). Answer: (b)

Teacher's Note:
1. Apply the chain rule three times.
2. Note that \( 2f(0)+2=0 \).

Answer: (a) 1

Explanation:
1. Since \( f \) is positive and increasing, \( 0\lt f(x)\lt f(2x)\lt f(3x) \) for \( x\gt0 \).
2. Dividing by \( f(x) \): \( 1\lt\frac{f(2x)}{f(x)}\lt\frac{f(3x)}{f(x)} \). The right side tends to 1, so by the squeeze theorem the limit is 1. Answer: (a)

Teacher's Note:
1. Use monotonicity to bound the ratio.
2. Apply the squeeze theorem.

Answer: (d) y = 3

Explanation:
1. \( \frac{dy}{dx}=1-\frac8{x^3}=0\Rightarrow x^3=8\Rightarrow x=2 \).
2. Then \( y=2+\frac44=3 \), so the tangent is \( y=3 \). Answer: (d)

Teacher's Note:
1. A horizontal tangent has zero slope.
2. Find the \( y \)-coordinate at that point.

Answer: (d) -1

Explanation:
1. \( f(-1)=k+2 \). For \( x\lt-1 \), \( f(x)=k-2x\gt k+2 \) automatically. For \( x\gt-1 \) near \( -1 \), \( f(x)=2x+3 \) takes values just above \( 1 \).
2. For a local minimum at \( -1 \) we need \( f(-1)\le1 \), i.e. \( k+2\le1 \), so \( k\le-1 \).
3. Among the options only \( k=-1 \) satisfies this. Answer: (d)

Teacher's Note:
1. Compare \( f(-1) \) with values on both sides.
2. Use the right-hand limit \( 2(-1)+3=1 \).

Answer: (a) -1

Explanation:
1. At \( x=1 \): \( 1-2\cot y-1=0\Rightarrow\cot y=0\Rightarrow y=\frac\pi2 \).
2. Differentiate: \( 2x^{2x}(1+\log x)-2\left[x^x(1+\log x)\cot y-x^x\operatorname{cosec}^2y\,y'\right]=0 \).
3. At \( \left(1,\frac\pi2\right) \): \( 2+2y'=0 \), so \( y'=-1 \). Answer: (a)

Teacher's Note:
1. Find \( y \) at \( x=1 \).
2. Differentiate implicitly using logarithmic differentiation for \( x^x \).

Answer: (c) A is true, but R is false.

Explanation:
1. \( (g\circ f)(x)=\sin(x|x|)=-\sin x^2 \) for \( x\lt0 \) and \( \sin x^2 \) for \( x\ge0 \). So \( (g\circ f)'(x)=-2x\cos x^2 \) for \( x\lt0 \) and \( 2x\cos x^2 \) for \( x\ge0 \).
2. Both sides give 0 at \( x=0 \), so the derivative exists and is continuous: A is true.
3. Second derivative: left value at 0 is \( -2 \), right value is \( 2 \). They differ, so \( g\circ f \) is not twice differentiable at 0: R is false. A is true but R is false. Answer: (c)

Teacher's Note:
1. Write \( g\circ f \) piecewise.
2. Compare the one-sided second derivatives at 0.

Answer: (b) The cubic has minima at \( \sqrt{\frac p3} \) and maxima at \( -\sqrt{\frac p3} \)

Explanation:
1. \( f'(x)=3x^2-p=0\Rightarrow x=\pm\sqrt{\frac p3} \), and \( f''(x)=6x \).
2. At \( x=\sqrt{p/3} \), \( f''\gt0 \) (minimum); at \( x=-\sqrt{p/3} \), \( f''\lt0 \) (maximum). Answer: (b)

Teacher's Note:
1. Find the critical points.
2. Use the second derivative test.

Answer: (d) hyperbola

Explanation:
1. The normal is \( Y-y=-\frac{dx}{dy}(X-x) \); putting \( Y=0 \) gives \( G=\left(x+y\frac{dy}{dx},0\right) \).
2. \( OG=2x\Rightarrow x+y\frac{dy}{dx}=2x\Rightarrow y\,dy=x\,dx \).
3. Integrating: \( \frac{y^2}2=\frac{x^2}2+C \), i.e. \( x^2-y^2=-2C \), which is a hyperbola. Answer: (d)

Teacher's Note:
1. Write the equation of the normal and find its \( X \)-intercept.
2. Form and solve the differential equation.

Answer: (b) \( \left(-\frac\pi2,\frac\pi4\right) \)

Explanation:
1. \( f'(x)=\frac{\cos x-\sin x}{1+(\sin x+\cos x)^2}=\frac{\sqrt2\cos\left(x+\frac\pi4\right)}{1+(\sin x+\cos x)^2} \).
2. \( f'\gt0 \) when \( -\frac\pi2\lt x+\frac\pi4\lt\frac\pi2 \), i.e. \( -\frac{3\pi}4\lt x\lt\frac\pi4 \). Only the interval \( \left(-\frac\pi2,\frac\pi4\right) \) lies inside this range. Answer: (b)

Teacher's Note:
1. Differentiate and simplify the numerator.
2. Compare the increasing range with each option.

Answer: (c) \( f(x) \) is differentiable everywhere

Explanation:
1. Since \( x\le|x| \) for every real \( x \), \( \min\{x+1,|x|+1\}=x+1 \).
2. So \( f(x)=x+1 \), a straight line, which is differentiable everywhere (and is not always \( \ge1 \)). Answer: (c)

Teacher's Note:
1. Compare \( x \) and \( |x| \).
2. Simplify the minimum.

Answer: (d) 1

Explanation:
1. \( \frac1x-\frac2{e^{2x}-1}=\frac{e^{2x}-1-2x}{x(e^{2x}-1)} \), a \( \frac00 \) form.
2. L'Hopital: \( \frac{2e^{2x}-2}{(e^{2x}-1)+2xe^{2x}} \), again \( \frac00 \); again: \( \frac{4e^{2x}}{4e^{2x}+4xe^{2x}}\to1 \).
3. For continuity \( f(0)=1 \). Answer: (d)

Teacher's Note:
1. Combine the fractions.
2. Apply L'Hopital's rule twice.

Answer: (b) \( (-\infty,\infty) \)

Explanation:
1. For \( x\ne0 \), \( f \) is a quotient of differentiable functions with non-zero denominator, so it is differentiable.
2. At \( x=0 \): \( \frac{f(h)-f(0)}h=\frac1{1+|h|}\to1 \) from both sides, so \( f'(0)=1 \) exists.
3. So \( f \) is differentiable on \( (-\infty,\infty) \). Answer: (b)

Teacher's Note:
1. Treat \( x\ne0 \) and \( x=0 \) separately.
2. Use the definition of derivative at 0.

Answer: (d) \( x=2 \)

Explanation:
1. \( f'(x)=\frac12-\frac2{x^2}=0\Rightarrow x^2=4\Rightarrow x=\pm2 \).
2. \( f''(x)=\frac4{x^3} \): \( f''(2)=\frac12\gt0 \) (minimum) and \( f''(-2)=-\frac12\lt0 \) (maximum). Answer: (d)

Teacher's Note:
1. Find the critical points.
2. Apply the second derivative test.