JEE Mathematics Limits Continuity and Differentiability MCQs Set 11

Answer: (c) \( -\sin1 \)

Explanation:
1. For \( x\lt0 \) near 0: \( [x]=-1 \) and \( |x|=-x \), so the expression is \( \frac{x(-1-x)\sin(-1)}{-x} \).
2. Since \( \frac{x(-1-x)}{-x}=1+x \), the expression is \( (1+x)\sin(-1) \).
3. As \( x\to0^- \) it tends to \( \sin(-1)=-\sin1 \). Answer: (c)

Teacher's Note:
1. Replace \( [x] \) and \( |x| \) by their values for small negative \( x \).
2. Cancel \( x \).

Answer: (b) \( \frac1{6\sqrt2} \)

Explanation:
1. \( \frac{dy}{dx}=\frac{3\sec t\tan t}{3\sec^2t}=\sin t \).
2. \( \frac{d^2y}{dx^2}=\frac{d}{dt}(\sin t)\cdot\frac{dt}{dx}=\frac{\cos t}{3\sec^2t}=\frac{\cos^3t}3 \).
3. At \( t=\frac\pi4 \): \( \frac{(1/\sqrt2)^3}3=\frac1{3\cdot2\sqrt2}=\frac1{6\sqrt2} \). Answer: (b)

Teacher's Note:
1. Find \( dy/dx \) parametrically.
2. Differentiate again with respect to \( t \) and divide by \( dx/dt \).

Answer: (a) equals \( \{-2,-1,0,1,2\} \)

Explanation:
1. For \( |x|\le2 \), \( f(x)=\max\{|x|,x^2\} \): it equals \( x^2 \) for \( |x|\ge1 \) and \( |x| \) for \( |x|\le1 \). The graphs meet at \( x=-1,0,1 \), giving corners at these three points.
2. For \( 2\lt|x|\le4 \), \( f(x)=8-2|x| \). At \( x=2 \): \( f \) is continuous (value 4) but the slope changes from \( 4 \) (of \( x^2 \)) to \( -2 \); similarly at \( x=-2 \).
3. So \( f \) is not differentiable at \( -2,-1,0,1,2 \). Answer: (a)

Teacher's Note:
1. Sketch \( \max\{|x|,x^2\} \) on \( [-2,2] \).
2. Look for corners and jumps.

Answer: (a) equals 0

Explanation:
1. Put \( x=1+h \), \( h\to0^+ \). Then \( |1-x|=h \), \( 1-|x|=-h \) and \( [1-x]=[-h]=-1 \).
2. The expression becomes \( \frac{(-h+\sin h)\sin(-\pi/2)}{h\cdot(-1)}=\frac{(-h+\sin h)(-1)}{-h}=\frac{\sin h-h}{h} \).
3. This equals \( \frac{\sin h}h-1\to1-1=0 \). Answer: (a)

Teacher's Note:
1. Substitute \( x=1+h \).
2. Evaluate the greatest integer function for small negative argument.

Answer: (c) \( \frac16\sqrt{\frac73} \)

Explanation:
1. For the point \( (x,\,x^{3/2}+7) \), \( d^2=\left(x-\frac12\right)^2+x^3 \).
2. \( \frac{d(d^2)}{dx}=2x-1+3x^2=0\Rightarrow(3x-1)(x+1)=0 \), so \( x=\frac13 \) (as \( x\ge0 \)).
3. \( d^2=\frac1{36}+\frac1{27}=\frac7{108} \), so \( d=\sqrt{\frac7{108}}=\frac16\sqrt{\frac73} \). Answer: (c)

Teacher's Note:
1. Write the squared distance as a function of \( x \).
2. Minimise it.

Answer: (a) \( \left(\frac43,2e\right) \)

Explanation:
1. \( \frac{dy}{dx}=e^{x^2}+2x^2e^{x^2} \), so the slope at \( (1,e) \) is \( e+2e=3e \).
2. The tangent is \( y-e=3e(x-1) \), i.e. \( y=3ex-2e \).
3. Testing the options: \( x=\frac43 \) gives \( y=4e-2e=2e \), so \( \left(\frac43,2e\right) \) lies on it; the other points do not. Answer: (a)

Teacher's Note:
1. Find the slope at the given point.
2. Write the tangent and test each option.

Answer: (d) does not exist

Explanation:
1. For \( x\to0^+ \): \( [x]=0 \), so the expression is \( \frac{\tan(\pi\sin^2x)+x^2}{x^2}\to\pi+1 \).
2. For \( x\to0^- \): \( [x]=-1 \) and \( |x|=-x \), so \( |x|-\sin(x[x])=-x+\sin x \), which is of order \( x^3 \); its square divided by \( x^2 \) tends to 0, and \( \frac{\tan(\pi\sin^2x)}{x^2}\to\pi \). The left limit is \( \pi \).
3. The two one-sided limits differ, so the limit does not exist. Answer: (d)

Teacher's Note:
1. Evaluate the left and right limits separately.
2. Use \( \tan t\approx t \) and \( \sin t\approx t \).

Answer: (a) 122

Explanation:
1. \( x^2-11x+30\le0\Rightarrow(x-5)(x-6)\le0 \), so \( S=[5,6] \).
2. \( f'(x)=9x^2-36x+27=9(x-1)(x-3)\gt0 \) on \( [5,6] \), so \( f \) is increasing there.
3. The maximum is \( f(6)=648-648+162-40=122 \). Answer: (a)

Teacher's Note:
1. Solve the inequality to get the interval.
2. Check the monotonicity of \( f \) on it.

Answer: (a) \( f \) is an increasing function of \( x \)

Explanation:
1. \( \frac d{dx}\frac x{\sqrt{a^2+x^2}}=\frac{a^2}{(a^2+x^2)^{3/2}} \) and \( \frac d{dx}\left[-\frac{d-x}{\sqrt{b^2+(d-x)^2}}\right]=\frac{b^2}{(b^2+(d-x)^2)^{3/2}} \).
2. So \( f'(x)=\frac{a^2}{(a^2+x^2)^{3/2}}+\frac{b^2}{(b^2+(d-x)^2)^{3/2}}\gt0 \) for all \( x \) (and it is continuous). Answer: (a)

Teacher's Note:
1. Differentiate each term with the quotient rule.
2. Look at the sign of \( f' \).

Answer: (b) an empty set

Explanation:
1. For \( x\lt0 \): \( f(x)=-\sin x+x+2(x-\pi)\cos x \), so \( f'(x)=-\cos x+1+2\cos x-2(x-\pi)\sin x \). For \( x\gt0 \): \( f'(x)=\cos x-1+2\cos x-2(x-\pi)\sin x \).
2. Only \( x=0 \) can be a problem. \( f'(0^-)=-1+1+2=2 \) and \( f'(0^+)=1-1+2=2 \), so they agree.
3. \( f \) is differentiable everywhere, so \( K \) is empty. Answer: (b)

Teacher's Note:
1. Remove the modulus on each side of 0.
2. Compare left and right derivatives at 0.

Answer: (b) 1

Explanation:
1. Write it as \( \frac{x}{\tan4x}\cdot\frac1{\sin^2x}\cdot\tan^22x \).
2. \( \frac x{\tan4x}\to\frac14 \), and \( \frac{\tan^22x}{\sin^2x}=\left(\frac{\tan2x}{2x}\right)^2\left(\frac{x}{\sin x}\right)^2\cdot4\to4 \).
3. The limit is \( \frac14\cdot4=1 \). Answer: (b)

Teacher's Note:
1. Replace \( \\cot \) by \( 1/\\tan \).
2. Use \( \lim\frac{\tan ax}{ax}=1 \) and \( \lim\frac{\sin x}x=1 \).

Answer: (c) 8

Explanation:
1. \( \cot^3x-\tan x=\frac{1-\tan^4x}{\tan^3x} \) and \( \cos\left(x+\frac\pi4\right)=\frac1{\sqrt2}(\cos x-\sin x) \).
2. \( 1-\tan^4x=(1-\tan^2x)(1+\tan^2x)=\frac{(\cos x-\sin x)(\cos x+\sin x)}{\cos^2x}\sec^2x \), so the factor \( \cos x-\sin x \) cancels.
3. The limit is \( \sqrt2\cdot\frac{\sec^4x(\cos x+\sin x)}{\tan^3x} \) at \( x=\frac\pi4 \): \( \sqrt2\cdot\frac{4\cdot\sqrt2}{1}=8 \). Answer: (c)

Teacher's Note:
1. Factorise the numerator.
2. Cancel \( \cos x-\sin x \) with the expanded denominator.

Answer: (b) \( \frac{x\log_e2x-\log_e2}x \)

Explanation:
1. Take logarithms: \( 2y\ln(2x)=2\ln2+2x-2y \), so \( y(1+\ln2x)=x+\ln2 \) and \( y=\frac{x+\ln2}{1+\ln2x} \).
2. \( \frac{dy}{dx}=\frac{(1+\ln2x)-(x+\ln2)\frac1x}{(1+\ln2x)^2} \).
3. So \( (1+\ln2x)^2\frac{dy}{dx}=1+\ln2x-1-\frac{\ln2}x=\frac{x\ln2x-\ln2}x \). Answer: (b)

Teacher's Note:
1. Take natural logarithm on both sides and solve for \( y \).
2. Apply the quotient rule.

Answer: (c) \( \left\{-\frac{3\pi}4,-\frac\pi4,\frac{3\pi}4,\frac\pi4\right\} \)

Explanation:
1. The graph of \( \min\{\sin x,\cos x\} \) has sharp corners where \( \sin x=\cos x \), i.e. \( \tan x=1 \).
2. In \( (-\pi,\pi) \) this gives \( x=\frac\pi4 \) and \( x=-\frac{3\pi}4 \), so \( S=\left\{-\frac{3\pi}4,\frac\pi4\right\} \).
3. Only option (c) contains both points. Answer: (c)

Teacher's Note:
1. Find where the two graphs cross.
2. Check which option contains every such point.

Answer: (b) \( \sqrt{\frac2\pi} \)

Explanation:
1. Rationalise: the limit is \( \lim\frac{\pi-2\sin^{-1}x}{\sqrt{1-x}}\cdot\frac1{\sqrt\pi+\sqrt{2\sin^{-1}x}} \), and the second factor tends to \( \frac1{2\sqrt\pi} \).
2. \( \pi-2\sin^{-1}x=2\cos^{-1}x \). Put \( x=\cos\theta \), \( \theta\to0^+ \): \( \frac{2\theta}{\sqrt{1-\cos\theta}}=\frac{2\theta}{\sqrt2\sin(\theta/2)}\to\frac{2\theta}{\sqrt2\,\theta/2}=2\sqrt2 \).
3. The limit is \( \frac{2\sqrt2}{2\sqrt\pi}=\sqrt{\frac2\pi} \). Answer: (b)

Teacher's Note:
1. Rationalise the numerator.
2. Substitute \( x=\cos\theta \).

Answer: (b) \( 4e \)

Explanation:
1. \( f'=f \) gives \( f(x)=Ce^x \), and \( f(1)=2 \) gives \( f(x)=2e^{x-1} \).
2. \( h'(1)=f'(f(1))f'(1)=f'(2)f'(1)=f(2)f(1) \). Here \( f(2)=2e \), so \( h'(1)=2e\cdot2=4e \). Answer: (b)

Teacher's Note:
1. Solve \( f'=f \) with the given condition.
2. Apply the chain rule at \( x=1 \).

Answer: (b) \( x=y\cot\theta+2\tan\theta \)

Explanation:
1. For \( x^2=8y \), \( \frac{dy}{dx}=\frac x4=m=\tan\theta \), so the point of contact is \( (4m,2m^2) \).
2. The tangent is \( y-2m^2=m(x-4m) \), i.e. \( y=mx-2m^2 \).
3. Dividing by \( m \): \( x=y\cot\theta+2\tan\theta \). Answer: (b)

Teacher's Note:
1. Use the slope to find the point of contact.
2. Rearrange the tangent equation.

Answer: (a) \( \varnothing \) (an empty set)

Explanation:
1. For \( x\lt0 \): \( |x-\pi|=\pi-x \) and \( \sin|x|=-\sin x \), so \( f=(x-\pi)(e^{-x}-1)\sin x \); for \( 0\le x\lt\pi \): \( f=-(x-\pi)(e^x-1)\sin x \); for \( x\ge\pi \): \( f=(x-\pi)(e^x-1)\sin x \).
2. At \( x=0 \): both sides give \( f'=0 \) because \( (e^{|x|}-1)\sin|x| \) is of order \( x^2 \). At \( x=\pi \): both sides give \( f'=0 \) because \( \sin x\to0 \) and \( |x-\pi|\sin x\sim(x-\pi)^2 \).
3. So \( f \) is differentiable everywhere and \( S \) is empty. Answer: (a)

Teacher's Note:
1. Remove the modulus signs piecewise.
2. Check the one-sided derivatives at the two suspect points.

Answer: (d) \( 2\sqrt2 \)

Explanation:
1. \( f(x)=\left(x-\frac1x\right)^2+2 \), so \( h(x)=t+\frac2t \) where \( t=x-\frac1x \).
2. For \( t\gt0 \), \( t+\frac2t\ge2\sqrt2 \) with equality at \( t=\sqrt2 \), a local minimum. (For \( t\lt0 \) the expression has a local maximum \( -2\sqrt2 \).)
3. The local minimum value is \( 2\sqrt2 \). Answer: (d)

Teacher's Note:
1. Write \( f \) in terms of \( g \).
2. Apply AM-GM for positive \( t \).

Answer: (c) 25

Explanation:
1. For radius \( r \) and angle \( \theta \): \( 2r+r\theta=20 \), so \( \theta=\frac{20-2r}r \).
2. Area \( A=\frac12r^2\theta=10r-r^2 \), and \( \frac{dA}{dr}=10-2r=0\Rightarrow r=5 \) (a maximum since \( A''\lt0 \)).
3. \( A_{\max}=50-25=25 \) sq m. Answer: (c)

Teacher's Note:
1. Use the perimeter condition to find \( \theta \).
2. Maximise the area as a function of \( r \).

Answer: (b) \( \frac1{16} \)

Explanation:
1. \( \cot x-\cos x=\frac{\cos x(1-\sin x)}{\sin x} \). Put \( x=\frac\pi2-h \): \( \cos x=\sin h \), \( \sin x=\cos h \), \( \pi-2x=2h \).
2. The limit is \( \lim_{h\to0}\frac{\sin h(1-\cos h)}{\cos h\cdot8h^3}=\frac18\lim\frac{\sin h\cdot2\sin^2(h/2)}{h^3\cos h} \).
3. \( =\frac18\cdot2\cdot1\cdot\frac14=\frac1{16} \). Answer: (b)

Teacher's Note:
1. Substitute \( x=\\pi/2-h \).
2. Use \( 1-\cos h=2\sin^2(h/2) \).

Answer: (b) \( \left(\frac12,\frac12\right) \)

Explanation:
1. At \( x=0 \): \( 6y=6\Rightarrow y=1 \). The point is \( (0,1) \).
2. \( y=\frac{x+6}{x^2-5x+6} \), so \( \frac{dy}{dx}=\frac{(x^2-5x+6)-(x+6)(2x-5)}{(x^2-5x+6)^2} \), which at \( x=0 \) is \( \frac{6+30}{36}=1 \).
3. The normal is \( y-1=-x \), i.e. \( x+y=1 \), which passes through \( \left(\frac12,\frac12\right) \). Answer: (b)

Teacher's Note:
1. Find the point on the Y-axis.
2. Slope of the normal is the negative reciprocal of the tangent slope.

Answer: (c) \( 4(\sqrt2-1) \)

Explanation:
1. By symmetry the centre is \( (0,4-r) \), with the circle touching the parabola at its vertex \( (0,4) \) and the lines \( y=\pm x \) in the first and second quadrants.
2. Distance from the centre to \( y=x \): \( \frac{|0-(4-r)|}{\sqrt2}=r\Rightarrow4-r=r\sqrt2 \), so \( r=\frac4{\sqrt2+1} \) (the other root is negative).
3. \( r=4(\sqrt2-1) \). Answer: (c)

Teacher's Note:
1. Place the centre on the Y-axis.
2. Equate the distance from the centre to the line with \( r \).

Answer: (c) \( \frac12 \)

Explanation:
1. The form is \( 1^\infty \), so \( p=e^{L} \) with \( L=\lim\frac{\tan^2\sqrt x}{2x}=\frac12\lim\left(\frac{\tan\sqrt x}{\sqrt x}\right)^2=\frac12 \).
2. So \( p=e^{1/2} \) and \( \log p=\frac12 \). Answer: (c)

Teacher's Note:
1. Recognise the \( 1^\infty \) form.
2. Use \( \lim\frac{\tan t}t=1 \).

Answer: (b) \( g'(0)=\cos(\log2) \)

Explanation:
1. Near \( x=0 \), \( \sin x \) is small and \( \log2\approx0.693\gt\sin x \), so \( f(x)=\log2-\sin x \) without the modulus.
2. Then \( g(x)=\log2-\sin(\log2-\sin x) \), which is differentiable, and \( g'(x)=-\cos(\log2-\sin x)\cdot(-\cos x)=\cos x\cos(\log2-\sin x) \).
3. \( g'(0)=\cos(\log2) \). Answer: (b)

Teacher's Note:
1. Remove the modulus near 0.
2. Apply the chain rule.

Answer: (c) \( x=2r \)

Explanation:
1. \( 4x+2\pi r=2 \), so \( r=\frac{1-2x}\pi \). The total area is \( A=x^2+\pi r^2=x^2+\frac{(1-2x)^2}\pi \).
2. \( A'(x)=2x-\frac{4(1-2x)}\pi=0\Rightarrow x=\frac2{\pi+4} \), and then \( r=\frac{1-2x}\pi=\frac1{\pi+4} \).
3. So \( x=2r \). Answer: (c)

Teacher's Note:
1. Use the wire length to relate \( x \) and \( r \).
2. Minimise the total area.

Answer: (c) 2

Explanation:
1. \( 1-\cos2x=2\sin^2x \), so the expression is \( \frac{2\sin^2x}{x^2}\cdot(3+\cos x)\cdot\frac{x}{\tan4x} \).
2. As \( x\to0 \): \( \frac{2\sin^2x}{x^2}\to2 \), \( 3+\cos x\to4 \) and \( \frac x{\tan4x}\to\frac14 \).
3. The limit is \( 2\cdot4\cdot\frac14=2 \). Answer: (c)

Teacher's Note:
1. Use \( 1-\\cos2x=2\\sin^2x \).
2. Split into standard limits.