JEE Mathematics Limits Continuity and Differentiability MCQs Set 10

Answer: (a) \( 4\sqrt2 \)

Explanation:
1. Use \( 1+\cos x=2\cos^2\frac x2 \), so the denominator is \( \sqrt2\left(1-\cos\frac x2\right)=\sqrt2\cdot2\sin^2\frac x4 \).
2. The limit becomes \( \lim\frac{\sin^2x}{2\sqrt2\sin^2(x/4)}=\frac{x^2}{2\sqrt2\,(x/4)^2}=\frac{16}{2\sqrt2}=4\sqrt2 \). Answer: (a)

Teacher's Note:
1. Rewrite the denominator using half-angle identities.
2. Use \( \sin t\approx t \) as \( t\to0 \).

Answer: (b) \( x-\frac\pi6 \)

Explanation:
1. Divide numerator and denominator by \( \cos x \): \( \frac{\sqrt3\cos x+\sin x}{\cos x-\sqrt3\sin x}=\frac{\sqrt3+\tan x}{1-\sqrt3\tan x}=\tan\left(\frac\pi3+x\right) \).
2. So \( \cot^{-1}\left(\tan\left(\frac\pi3+x\right)\right)=\frac\pi2-\frac\pi3-x=\frac\pi6-x \), valid because \( \frac\pi6-x\in(0,\frac\pi6) \) lies in \( (0,\pi) \).
3. Hence \( 2y=\left(\frac\pi6-x\right)^2 \) and \( 2\frac{dy}{dx}=-2\left(\frac\pi6-x\right) \), giving \( \frac{dy}{dx}=x-\frac\pi6 \). Answer: (b)

Teacher's Note:
1. Convert the fraction to \( \tan(\pi/3+x) \).
2. Check the principal-value range of \( \cot^{-1} \).

Answer: (d) 33

Explanation:
1. Derivative \( =f'(f(f(x)))\,f'(f(x))\,f'(x)+2f(x)f'(x) \).
2. At \( x=1 \): \( f(1)=1 \), so \( f(f(1))=1 \) and \( f'(f(f(1)))=f'(1)=3 \), \( f'(f(1))=3 \). The value is \( 3\cdot3\cdot3+2\cdot1\cdot3=27+6=33 \). Answer: (d)

Teacher's Note:
1. Apply the chain rule twice.
2. Substitute \( f(1)=1 \) repeatedly.

Answer: (b) \( 2\sqrt3 \)

Explanation:
1. Let the half-height be \( h/2=3\cos\theta \) and radius \( r=3\sin\theta \), so \( V=\pi r^2h=54\pi\sin^2\theta\cos\theta \).
2. \( \frac{dV}{d\theta}=54\pi\sin\theta\,(2\cos^2\theta-\sin^2\theta)=0\Rightarrow\tan^2\theta=2 \), so \( \cos\theta=\frac1{\sqrt3} \).
3. \( h=6\cos\theta=\frac6{\sqrt3}=2\sqrt3 \). Answer: (b)

Teacher's Note:
1. Express radius and height through the angle at the centre.
2. Maximise the volume.

Answer: (d) two irrational and one rational number

Explanation:
1. Since the extreme points are \( -1,0,1 \), \( f'(x)=ax(x^2-1) \), so \( f(x)=\frac a4x^4-\frac a2x^2+C \).
2. \( f(x)=f(0)\Rightarrow\frac{x^4}4-\frac{x^2}2=0\Rightarrow x^2(x^2-2)=0\Rightarrow x=0,\pm\sqrt2 \).
3. So \( S=\{0,\sqrt2,-\sqrt2\} \): one rational and two irrational numbers. Answer: (d)

Teacher's Note:
1. Write \( f'(x) \) from its roots and integrate.
2. Solve \( f(x)=f(0) \).

Answer: (a) \( \frac12 \)

Explanation:
1. Continuity at \( \frac\pi4 \) requires \( k=\lim_{x\to\pi/4}\frac{\sqrt2\cos x-1}{\cot x-1} \), which is a \( \frac00 \) form.
2. By L'Hopital's rule, \( k=\lim\frac{-\sqrt2\sin x}{-\operatorname{cosec}^2x}=\sqrt2\sin^3\frac\pi4=\sqrt2\cdot\frac1{2\sqrt2}=\frac12 \). Answer: (a)

Teacher's Note:
1. Continuity means the value equals the limit.
2. Differentiate numerator and denominator.

Answer: (b) \( (2,-2) \)

Explanation:
1. The curve passes through \( (1,-5) \): \( -5=1+a-b \), so \( b-a=6 \).
2. The slope of the tangent at \( x=1 \) is \( 3+a \). The line has slope 1, so \( (3+a)(1)=-1 \) and \( a=-4 \), giving \( b=2 \).
3. The curve is \( y=x^3-4x-2 \). At \( x=2 \), \( y=8-8-2=-2 \), so \( (2,-2) \) lies on it. Answer: (b)

Teacher's Note:
1. Use the point to get one equation in \( a,b \).
2. Use the perpendicularity condition for the slope.

Answer: (b) \( \{5,10,15\} \)

Explanation:
1. \( g(x)=15-\big|15-|x-10|-10\big|=15-\big|5-|x-10|\big| \).
2. Kinks occur where \( |x-10|=5 \), that is \( x=5,15 \), and where the inner \( |x-10| \) has its kink, \( x=10 \) (there \( 5-|x-10|=5\ne0 \), so the outer modulus does not remove it).
3. So \( g \) is not differentiable at \( 5,10,15 \). Answer: (b)

Teacher's Note:
1. Simplify \( f(f(x)) \).
2. List points where a modulus argument is zero or has a corner.

Answer: (c) \( \left\{-\frac13,1\right\} \)

Explanation:
1. \( f(1)=-2 \) and \( f(-1)=0 \), so the slope of the segment is \( \frac{-2-0}{2}=-1 \).
2. \( f'(x)=3x^2-2x-2=-1\Rightarrow3x^2-2x-1=0\Rightarrow(x-1)(3x+1)=0 \).
3. So \( x=1 \) or \( x=-\frac13 \). Answer: (c)

Teacher's Note:
1. Find the slope of the chord.
2. Equate it to \( f'(x) \).

Answer: (b) \( \frac1{5\pi} \)

Explanation:
1. With \( \tan\alpha=\frac12 \), the radius of the water surface is \( r=\frac h2 \), so \( V=\frac13\pi r^2h=\frac{\pi h^3}{12} \).
2. \( \frac{dV}{dt}=\frac{\pi h^2}4\frac{dh}{dt}\Rightarrow\frac{dh}{dt}=\frac{4}{\pi h^2}\cdot5 \).
3. At \( h=10 \): \( \frac{dh}{dt}=\frac{20}{100\pi}=\frac1{5\pi} \) m/min. Answer: (b)

Teacher's Note:
1. Express the volume in terms of the depth only.
2. Differentiate with respect to time.

Answer: (b) \( -\frac{\pi^2}{2\sqrt3} \)

Explanation:
1. The left side is \( \frac{d}{dx}(y\cos x) \), so \( \frac{d}{dx}(y\cos x)=6x \) and \( y\cos x=3x^2+C \).
2. \( y(\pi/3)=0\Rightarrow0=3\cdot\frac{\pi^2}9+C\Rightarrow C=-\frac{\pi^2}3 \).
3. At \( x=\frac\pi6 \): \( y\cdot\frac{\sqrt3}2=\frac{\pi^2}{12}-\frac{\pi^2}3=-\frac{\pi^2}4 \), so \( y=-\frac{\pi^2}{2\sqrt3} \). Answer: (b)

Teacher's Note:
1. Recognise the left side as an exact derivative.
2. Use the initial condition to find the constant.

Answer: (d) \( \frac2{5-\pi} \)

Explanation:
1. \( f(5)=a(5-\pi)+1 \) and the right limit is \( b(5-\pi)+3 \) (since \( 5\gt\pi \)).
2. Continuity gives \( a(5-\pi)+1=b(5-\pi)+3 \), so \( (a-b)(5-\pi)=2 \).
3. Hence \( a-b=\frac2{5-\pi} \). Answer: (d)

Teacher's Note:
1. Evaluate both pieces at \( x=5 \).
2. Remember \( 5\gt\pi \), so the modulus opens with a positive sign.

Answer: (d) \( \frac83 \)

Explanation:
1. Left side: \( \frac{x^4-1}{x-1}\to4\cdot1^3=4 \).
2. Right side: \( \frac{x^3-k^3}{x^2-k^2}=\frac{x^2+xk+k^2}{x+k}\to\frac{3k^2}{2k}=\frac{3k}2 \).
3. \( \frac{3k}2=4\Rightarrow k=\frac83 \). Answer: (d)

Teacher's Note:
1. Use \( \lim\frac{x^n-a^n}{x-a}=na^{n-1} \).
2. Cancel the common factor on the right.

Answer: (d) \( \left(-\frac32,\frac12\right) \)

Explanation:
1. Left limit: \( \lim_{x\to0^-}\frac{\sin(p+1)x+\sin x}{x}=(p+1)+1=p+2 \).
2. Right limit: \( \frac{\sqrt{x+x^2}-\sqrt x}{x^{3/2}}=\frac{\sqrt{1+x}-1}{x}\to\frac12 \).
3. Continuity: \( p+2=\frac12=q \), so \( p=-\frac32 \) and \( q=\frac12 \). Answer: (d)

Teacher's Note:
1. Use \( \lim\frac{\sin ax}x=a \).
2. Take out \( \sqrt x \) from the right-hand expression.

Answer: (c) -7

Explanation:
1. The limit is finite and the denominator vanishes at \( x=1 \), so \( 1-a+b=0 \), i.e. \( a=1+b \).
2. Then \( x^2-(1+b)x+b=(x-1)(x-b) \) and the limit is \( 1-b=5\Rightarrow b=-4 \), \( a=-3 \).
3. \( a+b=-7 \). Answer: (c)

Teacher's Note:
1. The numerator must vanish at \( x=1 \).
2. Factorise and cancel.

Answer: (b) \( a\alpha^2-b\alpha-a=1 \)

Explanation:
1. \( (f\circ g)(x)=\ln\left(\sin(\sin^{-1}e^{-x})\right)=\ln e^{-x}=-x \), valid since \( e^{-x}\in(0,1] \).
2. So \( a=(f\circ g)'(\alpha)=-1 \) and \( b=(f\circ g)(\alpha)=-\alpha \).
3. Then \( a\alpha^2-b\alpha-a=-\alpha^2+\alpha^2+1=1 \); the other options do not hold identically. Answer: (b)

Teacher's Note:
1. Simplify \( f(g(x)) \) using \( \sin(\sin^{-1}t)=t \).
2. Substitute \( a=-1,\ b=-\alpha \) in each option.

Answer: (b) \( \frac1{18\pi} \)

Explanation:
1. Let the thickness be \( x \) cm. The volume of ice is \( V=\frac43\pi[(10+x)^3-10^3] \).
2. \( \frac{dV}{dt}=4\pi(10+x)^2\frac{dx}{dt}=-50 \).
3. At \( x=5 \): \( 4\pi\cdot225\cdot\frac{dx}{dt}=-50\Rightarrow\frac{dx}{dt}=-\frac1{18\pi} \). The thickness decreases at \( \frac1{18\pi} \) cm/min. Answer: (b)

Teacher's Note:
1. Write the volume of the ice shell.
2. Differentiate with respect to time.

Answer: (b) \( \left(-\frac1e,\frac1{e^2}\right) \)

Explanation:
1. At \( x=0 \): \( e^y=e \), so \( y=1 \). Differentiating once: \( e^yy'+xy'+y=0 \), so at \( x=0 \), \( ey'+1=0 \) and \( y'=-\frac1e \).
2. Differentiating again: \( e^yy''+e^yy'^2+xy''+2y'=0 \). At \( x=0 \): \( ey''+e\cdot\frac1{e^2}-\frac2e=0 \), so \( ey''=\frac1e \).
3. Hence \( y''=\frac1{e^2} \) and the pair is \( \left(-\frac1e,\frac1{e^2}\right) \). Answer: (b)

Teacher's Note:
1. Find \( y \) at \( x=0 \) first.
2. Differentiate the relation twice implicitly.

Answer: (a) 18

Explanation:
1. \( g(x)=\frac{\int_6^{f(x)}4t^3dt}{x-2} \), which is a \( \frac00 \) form at \( x=2 \) because \( f(2)=6 \).
2. By L'Hopital's rule and Leibniz's rule, \( \lim g(x)=4(f(2))^3f'(2)=4\cdot216\cdot\frac1{48}=18 \). Answer: (a)

Teacher's Note:
1. Check that the limit is of the form 0/0.
2. Differentiate the integral with respect to its upper limit.

Answer: (c) \( \frac1{12} \)

Explanation:
1. Both roots have modulus less than 1, so the infinite GPs sum to \( \frac\alpha{1-\alpha}+\frac\beta{1-\beta} \).
2. This equals \( \frac{(\alpha+\beta)-2\alpha\beta}{1-(\alpha+\beta)+\alpha\beta} \) with \( \alpha+\beta=\frac1{15} \) and \( \alpha\beta=-\frac2{375} \).
3. Value \( =\frac{\frac1{15}+\frac4{375}}{1-\frac1{15}-\frac2{375}}=\frac{29}{348}=\frac1{12} \). Answer: (c)

Teacher's Note:
1. Sum each infinite geometric series.
2. Use the sum and product of the roots.

Answer: (b) \( (4,3\sqrt3) \)

Explanation:
1. \( f \) is defined for \( x\in[0,k] \). \( f'(x)=\frac{3kx-4x^2}{2\sqrt{kx-x^2}} \), which is non-negative when \( x\le\frac{3k}4 \).
2. For \( f \) to increase on \( [0,3] \) we need \( 3\le\frac{3k}4 \), so \( k\ge4 \) and \( m=4 \).
3. The maximum on \( [0,3] \) is \( f(3)=3\sqrt{12-9}=3\sqrt3 \). So \( (m,M)=(4,3\sqrt3) \). Answer: (b)

Teacher's Note:
1. Find \( f'(x) \) and its sign.
2. The maximum of an increasing function is at the right end.

Answer: (b) 2

Explanation:
1. Rationalise: the denominator becomes \( \frac{(x^2+2\sin x+1)-(\sin^2x-x+1)}{\sqrt{\cdot}+\sqrt{\cdot}}=\frac{x^2-\sin^2x+2\sin x+x}{\sqrt{\cdot}+\sqrt{\cdot}} \).
2. As \( x\to0 \) the sum of square roots tends to \( 1+1=2 \), so the limit is \( 2\lim\frac{x+2\sin x}{x^2-\sin^2x+2\sin x+x} \).
3. Dividing by \( x \): \( \frac{1+2\frac{\sin x}x}{x-\frac{\sin^2x}x+2\frac{\sin x}x+1}\to\frac{3}{0-0+2+1}=1 \). Hence the limit is \( 2\times1=2 \). Answer: (b)

Teacher's Note:
1. Multiply by the conjugate.
2. Divide numerator and denominator by \( x \).

Answer: (a) \( \frac12 \)

Explanation:
1. \( f \) is maximum at \( x=2 \), so \( \alpha=2 \); \( g \) is minimum at \( x=-1 \), so \( \beta=-1 \). Then \( -\alpha\beta=2 \).
2. \( \frac{(x-1)(x-2)(x-3)}{(x-2)(x-4)}=\frac{(x-1)(x-3)}{x-4} \), which at \( x=2 \) is \( \frac{1\cdot(-1)}{-2}=\frac12 \). Answer: (a)

Teacher's Note:
1. Locate the extreme points of the modulus functions.
2. Factorise and cancel \( (x-2) \).

Answer: (b) \( 2\sqrt3\pi \)

Explanation:
1. With height \( h \) and slant height 3, \( r^2=9-h^2 \) and \( V=\frac13\pi(9-h^2)h=\frac\pi3(9h-h^3) \).
2. \( V'(h)=0\Rightarrow9-3h^2=0\Rightarrow h=\sqrt3 \), and \( V''\lt0 \) there, so it is a maximum.
3. \( V=\frac\pi3(9\sqrt3-3\sqrt3)=2\sqrt3\,\pi \). Answer: (b)

Teacher's Note:
1. Eliminate the radius using the slant height.
2. Maximise \( V(h) \).

Answer: (a) exists and equals \( \frac1{4\sqrt2} \)

Explanation:
1. Rationalise the numerator: it becomes \( \frac{(1+\sqrt{1+y^4})-2}{\sqrt{1+\sqrt{1+y^4}}+\sqrt2}=\frac{\sqrt{1+y^4}-1}{\sqrt{1+\sqrt{1+y^4}}+\sqrt2} \).
2. Rationalise again: \( \sqrt{1+y^4}-1=\frac{y^4}{\sqrt{1+y^4}+1} \), so the limit is \( \lim\frac1{(\sqrt{1+\sqrt{1+y^4}}+\sqrt2)(\sqrt{1+y^4}+1)}=\frac1{2\sqrt2\cdot2}=\frac1{4\sqrt2} \). Answer: (a)

Teacher's Note:
1. Rationalise the numerator twice.
2. Cancel \( y^4 \).

Answer: (c) 1

Explanation:
1. Divide by \( |x-y| \): \( \left|\frac{f(x+h)-f(x)}h\right|\le2|h|^{1/2} \), which tends to 0 as \( h\to0 \).
2. So \( |f'(x)|\le0 \), hence \( f'(x)=0 \) and \( f \) is constant. Since \( f(0)=1 \), \( f(x)=1 \) for all \( x \).
3. \( \int_0^1f^2dx=\int_0^11\,dx=1 \). Answer: (c)

Teacher's Note:
1. Form the difference quotient.
2. Show that the derivative is zero everywhere.