JEE Mathematics Limits Continuity and Differentiability MCQs Set 04

Answer: (b) \( \frac32 \)

Explanation:
1. The limit has the form \( 1^\infty \), so it equals \( e^{\lim 2x\left(\frac ax-\frac4{x^2}\right)}=e^{2a} \).
2. \( e^{2a}=e^3 \Rightarrow a=\frac32 \).

Teacher's Note:
1. For a 1 to the power infinity form, use e to the limit of (base minus 1) times exponent.

Answer: (a) -2

Explanation:
1. Numerator: \( (2\sin^2x)^2\approx4x^4 \).
2. Denominator: \( 2x\left(x+\frac{x^3}3\right)-x\left(2x+\frac{8x^3}3\right)=-2x^4 \).
3. Limit \( =\frac4{-2}=-2 \).

Teacher's Note:
1. Expand tan x = x + x^3/3 and tan 2x = 2x + 8x^3/3 to get the x^4 terms of the denominator.

Answer: (a) \( (\sqrt2,1-\sqrt3) \)

Explanation:
1. At \( x=1 \): \( \frac2a=a\Rightarrow a^2=2 \).
2. At \( x=\sqrt2 \): \( a=\frac{2b^2-4b}{2\sqrt2} \), so \( 2\sqrt2a=2b^2-4b \).
3. For \( a=\sqrt2 \): \( b^2-2b-2=0\Rightarrow b=1\pm\sqrt3 \). For \( a=-\sqrt2 \): \( b^2-2b+2=0 \) has no real root.
4. So \( a=\sqrt2 \) and \( b=1\pm\sqrt3 \); the option listed is \( (\sqrt2,1-\sqrt3) \).

Teacher's Note:
1. Apply continuity at each junction separately and check the sign choices for a.

Answer: (b) \( \frac32 \)

Explanation:
1. \( e^{x^2}-\cos x\approx\left(1+x^2\right)-\left(1-\frac{x^2}2\right)=\frac{3x^2}2 \).
2. \( \sin^2x\approx x^2 \).
3. Limit \( =\frac32 \).

Teacher's Note:
1. Use series expansions to the order x^2 when numerator and denominator both vanish.

Answer: (c) 3

Explanation:
1. Near 0: \( (e^x-1)^2\approx x^2 \), \( \sin\frac xk\approx\frac xk \), \( \log\left(1+\frac x4\right)\approx\frac x4 \).
2. The limit is \( \frac{x^2}{\frac xk\cdot\frac x4}=4k \).
3. Continuity requires \( 4k=12 \), so \( k=3 \).

Teacher's Note:
1. Match the limit with f(0) for continuity.

Answer: (b) \( \frac29 \)

Explanation:
1. \( \frac{1-\cos3x}{x^2}=\frac{2\sin^2\frac{3x}2}{x^2}\to2\cdot\frac94=\frac92 \).
2. Since f is continuous, the limit is \( f\left(\frac92\right)=\frac29 \).

Teacher's Note:
1. For a continuous f, the limit can be moved inside: lim f(g(x)) = f(lim g(x)).

Answer: (d) \( -ne^{-nx} \)

Explanation:
1. \( \frac{d^2y}{dx^2}=n^2e^{nx} \).
2. \( x=\frac{\ln y}n \), so \( \frac{dx}{dy}=\frac1{ny} \) and \( \frac{d^2x}{dy^2}=-\frac1{ny^2}=-\frac{e^{-2nx}}n \).
3. Product \( =n^2e^{nx}\cdot\left(-\frac{e^{-2nx}}n\right)=-ne^{-nx} \).

Teacher's Note:
1. The second derivative of the inverse function is not the reciprocal of the second derivative.

Answer: (c) \( \frac13 \)

Explanation:
1. \( g'(7)=\frac1{f'(g(7))} \).
2. \( f(x)=7\Rightarrow x^2-x-2=0\Rightarrow x=2 \) (since \( x>\frac12 \)).
3. \( f'(2)=2\cdot2-1=3 \), so \( g'(7)=\frac13 \).

Teacher's Note:
1. Derivative of the inverse: reciprocal of f' at the corresponding point.

Answer: (b) Both Statement I and Statement II are true

Explanation:
1. Statement I: \( |x\sin\frac1x|\le|x|\to0 \), so f is continuous at 0.
2. Statement II: \( g(x)=x^2\sin\frac1x \) and \( g'(0)=\lim\frac{x^2\sin(1/x)}x=\lim x\sin\frac1x=0 \), so g is differentiable at 0.
3. Both statements are true.

Teacher's Note:
1. Multiplying by an extra factor of x improves smoothness at the origin.

Answer: (d) \( \frac14 \)

Explanation:
1. Put \( x=\pi+h \): \( \cos x=-\cos h\approx-1+\frac{h^2}2 \).
2. \( \sqrt{2+\cos x}\approx\sqrt{1+\frac{h^2}2}\approx1+\frac{h^2}4 \), so the numerator is about \( \frac{h^2}4 \).
3. The limit is \( \frac14 \), hence \( k=\frac14 \).

Teacher's Note:
1. Shift the variable so that the limit point is 0, then use series.

Answer: (b) continuous as well as differentiable

Explanation:
1. \( |f(0)|\le0 \Rightarrow f(0)=0 \).
2. \( \left|\frac{f(h)-f(0)}h\right|\le|h|\to0 \), so \( f'(0)=0 \).
3. f is differentiable, hence continuous, at 0.

Teacher's Note:
1. Squeeze the difference quotient between -|h| and |h|.

Answer: (b) \( -\frac12 \)

Explanation:
1. At \( x=0 \), \( \tan^{-1}1=\frac\pi4 \), so the limit is the derivative of \( \tan^{-1}u \) at 0, where \( u=\frac{x+1}{2x+1} \).
2. \( u'=\frac{(2x+1)-2(x+1)}{(2x+1)^2}=-\frac1{(2x+1)^2} \), so \( u'(0)=-1 \).
3. \( \frac{d}{dx}\tan^{-1}u=\frac{u'}{1+u^2}=\frac{-1}{2} \).

Teacher's Note:
1. A limit of the form [g(x) - g(0)]/x is just g'(0).

Answer: (d) \( \frac{x^2+y^2}{x^2} \)

Explanation:
1. \( \ln x=\frac12\sin^{-1}t\,\ln a \) and \( \ln y=\frac12\cos^{-1}t\,\ln a \).
2. Differentiating with respect to t: \( \frac{x'}x=\frac{\ln a}{2\sqrt{1-t^2}} \) and \( \frac{y'}y=-\frac{\ln a}{2\sqrt{1-t^2}} \).
3. So \( \frac{dy}{dx}=-\frac yx \), and \( 1+\left(\frac{dy}{dx}\right)^2=1+\frac{y^2}{x^2}=\frac{x^2+y^2}{x^2} \).

Teacher's Note:
1. Take logarithms before differentiating parametric exponent expressions.

Answer: (d) \( \cos^2x\sin(\sin x) \)

Explanation:
1. \( f'(x)=\cos(\sin x)\cos x \).
2. \( f''(x)=-\sin(\sin x)\cos^2x-\cos(\sin x)\sin x \).
3. \( \tan x\,f'(x)=\cos(\sin x)\sin x \).
4. So \( f''+\tan x\,f'=-\sin(\sin x)\cos^2x \), and \( g(x)=\sin(\sin x)\cos^2x \).

Teacher's Note:
1. The cos(sin x) sin x terms cancel, leaving only the first term.

Answer: (b) 3

Explanation:
1. The composite is undefined at \( x=1 \) (where u is undefined).
2. It is also undefined where \( u^2+u-2=0 \), i.e. \( u=1 \) or \( u=-2 \), which gives \( x=2 \) and \( x=\frac12 \).
3. So there are 3 points of discontinuity: \( \frac12,1,2 \).

Teacher's Note:
1. List both the points where u is undefined and where the outer denominator vanishes.

Answer: (c) 0

Explanation:
1. For x slightly greater than 1, \( \{x\}=x-1\to0^+ \).
2. The exponent \( \frac1{\ln x}\to+\infty \).
3. A small positive number raised to a large positive power tends to 0.

Teacher's Note:
1. Evaluate the base and exponent separately before deciding the form.

Answer: (c) 0

Explanation:
1. By the mean value theorem, the difference equals \( 2\cdot\frac23\xi^{-1/3} \) for some \( \xi\in(x-1,x+1) \).
2. As \( x\to\infty \), \( \xi^{-1/3}\to0 \), so the limit is 0.

Teacher's Note:
1. Differences of fractional powers with exponent less than 1 vanish at infinity.

Answer: (a) 1

Explanation:
1. Numerator \( \sim n^{3/2}+n^{4/3}\sim n^{3/2} \).
2. Denominator \( \sim n^{6/4}-n^{7/5}=n^{3/2}-n^{7/5}\sim n^{3/2} \), since \( \frac32>\frac75 \).
3. The ratio tends to 1.

Teacher's Note:
1. Compare the highest powers of n in the numerator and the denominator.

Answer: (a) 0

Explanation:
1. For \( x\to1^- \), \( [x]=0 \) and \( 1-x\to0^+ \).
2. \( 0^{\,t} \) with \( t>0 \) equals 0, so the value of the expression is 0 for every x just below 1, and the limit is 0.

Teacher's Note:
1. Do not treat this as 0 to the power 0; the exponent is strictly positive.

Answer: (a) even

Explanation:
1. \( f(x)=(x-3)(x^2-4) \), so \( h(x)=x^2-4 \) for \( x\ne3 \).
2. Continuity requires \( K=3^2-4=5 \), so \( h(x)=x^2-4 \) for all x.
3. This is an even function.

Teacher's Note:
1. Factor out (x - 3) to find the continuous extension.

Answer: (a) \( a=\frac12,\ b=\frac32 \)

Explanation:
1. Continuity at 1: \( a-b=-1 \).
2. Left derivative: \( 2a \); right derivative: \( \frac{d}{dx}\left(-\frac1x\right)=\frac1{x^2}=1 \). So \( 2a=1\Rightarrow a=\frac12 \).
3. Then \( b=a+1=\frac32 \).

Teacher's Note:
1. Derivable means continuous and with equal left and right derivatives.

Answer: (a) \( e^{x\,f'(0)} \)

Explanation:
1. \( f'(x)=\lim\frac{f(x+h)-f(x)}h=f(x)\lim\frac{f(h)-1}h=f(x)f'(0) \) (since \( f(0)=1 \)).
2. Solving \( \frac{f'}f=f'(0) \) with \( f(0)=1 \) gives \( f(x)=e^{xf'(0)} \).

Teacher's Note:
1. A function with f(x+y) = f(x)f(y) is an exponential.

Answer: (b) \( 3p(\ln4)^3 \)

Explanation:
1. \( (4^x-1)^3\approx(x\ln4)^3 \), \( \sin\frac xp\approx\frac xp \), \( \ln\left(1+\frac{x^2}3\right)\approx\frac{x^2}3 \).
2. Limit \( =\frac{x^3(\ln4)^3}{\frac{x}{p}\cdot\frac{x^2}3}=3p(\ln4)^3 \).

Teacher's Note:
1. Replace each factor by its first-order approximation.

Answer: (b) 5050

Explanation:
1. Since \( \sum_{k=1}^{100}1^k=100 \), the limit is the derivative of \( \sum x^k \) at x = 1.
2. \( \sum k x^{k-1} \) at x = 1 is \( \sum_{k=1}^{100}k=5050 \).

Teacher's Note:
1. A limit of the form (g(x) - g(1))/(x - 1) is g'(1).

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

Explanation:
1. Let \( s=\sin x\to1 \). Rationalise: the difference equals \( \frac{(2s^2+3s+4)-(s^2+6s+2)}{\sqrt{\cdot}+\sqrt{\cdot}}=\frac{(s-1)(s-2)}{\sqrt{\cdot}+\sqrt{\cdot}} \).
2. \( \tan^2x=\frac{s^2}{(1-s)(1+s)} \).
3. The product is \( \frac{s^2(s-1)(s-2)}{(1-s)(1+s)(\sqrt\cdot+\sqrt\cdot)}=\frac{-s^2(s-2)}{(1+s)(\sqrt\cdot+\sqrt\cdot)}\to\frac{1}{2\cdot6}=\frac1{12} \).

Teacher's Note:
1. Rationalise the difference of square roots, then cancel the factor (1 - s) with tan squared.

Answer: (d) zero

Explanation:
1. The numerator is dominated by \( 5^{n+1} \) (as \( 5>4>3 \)).
2. The denominator is dominated by \( 3^{2n+3}=27\cdot9^n \), since \( 9>5 \).
3. The ratio behaves like \( \frac{5\cdot5^n}{27\cdot9^n}\to0 \).

Teacher's Note:
1. Compare growth rates: the largest base dominates.

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

Explanation:
1. \( e^{-x^2/2}=1-\frac{x^2}2+\frac{x^4}8-\dots \) and \( \cos x=1-\frac{x^2}2+\frac{x^4}{24}-\dots \).
2. The difference is \( \left(\frac18-\frac1{24}\right)x^4=\frac{x^4}{12} \).
3. The denominator is about \( x^4 \), so the limit is \( \frac1{12} \).

Teacher's Note:
1. The x^2 terms cancel, so go to the fourth order terms of the series.