JEE Mathematics Limits Continuity and Differentiability MCQs Set 16

Answer: (d) \( \frac12 \)

Explanation:
1. \( \frac{2^{x+2}-16}{4^x-16}=\frac{4(2^x-4)}{(2^x-4)(2^x+4)}=\frac4{2^x+4} \).
2. At \( x=2 \): \( \frac48=\frac12 \), so \( k=\frac12 \).

Teacher's Note:
1. Factor \( 4^x-16 \) as a difference of squares.

Answer: (d) \( -\frac12 \)

Explanation:
1. Left limit: \( \frac{\sqrt{1+kx}-\sqrt{1-kx}}x\to\frac{kx}x=k \).
2. \( f(0)=\frac{1}{-2}=-\frac12 \), so \( k=-\frac12 \).

Teacher's Note:
1. Rationalise or use \( \sqrt{1+u}\approx1+\frac u2 \).

Answer: (d) no value

Explanation:
1. For \( x>0 \): \( \frac x{x+2x^2}=\frac1{1+2x}\to1 \).
2. For \( x<0 \): \( \frac x{-x+2x^2}=\frac{-1}{1-2x}\to-1 \). The one-sided limits differ, so no value of k works.

Teacher's Note:
1. Split into \( x>0 \) and \( x<0 \) because of the modulus.

Answer: (a) \( a=1,\ b=-1 \)

Explanation:
1. Left limit: \( -1+a \); value: \( a+b \); right limit: \( 1+b \).
2. Equating: \( -1+a=a+b \) gives \( b=-1 \); \( a+b=1+b \) gives \( a=1 \).

Teacher's Note:
1. Equate the three quantities pairwise.

Answer: (d) 7

Explanation:
1. \( x^3+x^2-16x+20=(x-2)^2(x+5) \).
2. So \( f(x)=x+5 \) for \( x\ne2 \) and \( k=7 \).

Teacher's Note:
1. Check that \( x=2 \) is a double root of the numerator.

Answer: (a) \( \frac12 \)

Explanation:
1. Both numerator and denominator vanish at \( \frac\pi4 \); by L'Hopital: \( \frac{-\sqrt2\sin x}{-\operatorname{cosec}^2x} \).
2. At \( \frac\pi4 \): \( \frac{-1}{-2}=\frac12 \).

Teacher's Note:
1. Apply L'Hopital's rule for the \( \frac00 \) form.

Answer: (d) 8

Explanation:
1. For \( x<0 \): \( \frac{2\sin^22x}{x^2}\to8 \).
2. For \( x>0 \): multiply by the conjugate: \( \frac{\sqrt x\left(\sqrt{16+\sqrt x}+4\right)}{\sqrt x}\to8 \).
3. Hence \( a=8 \).

Teacher's Note:
1. Evaluate both one-sided limits; they agree.

Answer: (d) none of these

Explanation:
1. \( x-[x]=0 \) exactly when x is an integer, so f is undefined at every integer.
2. There are infinitely many such points, so none of the first three options is correct.

Teacher's Note:
1. The fractional part vanishes at all integers.

Answer: (d) 2

Explanation:
1. \( \ln(1+x)\approx x \) and \( \ln(1-x)\approx-x \), so the numerator is \( \approx2x \).
2. The limit is 2, so \( f(0)=2 \).

Teacher's Note:
1. The assigned value must equal the limit.

Answer: (b) 1

Explanation:
1. Put \( x=\frac\pi2-h \): \( \ln\sin x=\ln\cos h\approx-\frac{h^2}2 \) and the exponent is \( \frac1{2h} \).
2. \( \frac{-h^2/2}{2h}\to0 \), so the limit is \( e^0=1 \).

Teacher's Note:
1. Use logarithms for the \( 1^\infty \) form.

Answer: (b) \( \frac13 \)

Explanation:
1. Near 0: \( \sin^{-1}x\approx x \) and \( \tan^{-1}x\approx x \).
2. \( f(x)\to\frac{2-1}{2+1}=\frac13 \).

Teacher's Note:
1. Divide numerator and denominator by x.

Answer: (c) \( \frac1{32} \)

Explanation:
1. Put \( x=\frac\pi2-h \). Then \( \tan\frac x2=\tan\left(\frac\pi4-\frac h2\right)=\frac{1-\tan\frac h2}{1+\tan\frac h2} \), so \( \frac{1-\tan\frac x2}{1+\tan\frac x2}=\tan\frac h2 \).
2. \( 1-\sin x=1-\cos h\approx\frac{h^2}2 \), \( \tan\frac h2\approx\frac h2 \) and \( (\pi-2x)^3=8h^3 \).
3. The limit is \( \frac{\frac h2\cdot\frac{h^2}2}{8h^3}=\frac1{32} \).

Teacher's Note:
1. Convert \( \frac{1-\tan(x/2)}{1+\tan(x/2)} \) with the tangent subtraction formula.

Answer: (a) \( A=-2,\ B=0 \) and \( f(0)=-1 \)

Explanation:
1. The numerator must vanish to order \( x^3 \): \( B=0 \) (constant term) and \( 2+A=0 \) (coefficient of x), so \( A=-2 \).
2. Then \( \sin2x-2\sin x\approx\left(2x-\frac{8x^3}6\right)-\left(2x-\frac{2x^3}6\right)=-x^3 \), so \( f(0)=-1 \).

Teacher's Note:
1. Match series coefficients so the numerator is \( O(x^3) \).

Answer: (c) \( -2 \)

Explanation:
1. Left limit: \( k \). Value at 0: \( -2 \).
2. So \( k=-2 \).

Teacher's Note:
1. Evaluate the polynomial piece at 0.

Answer: (a) 3

Explanation:
1. Near 0: \( (4^x-1)^3\approx x^3(\ln4)^3 \), \( \sin\frac xa\approx\frac xa \) and \( \ln\left(1+\frac{x^2}3\right)\approx\frac{x^2}3 \).
2. The limit is \( 3a(\ln4)^3 \), which must equal \( 9(\ln4)^3 \), so \( a=3 \).

Teacher's Note:
1. Replace each factor by its leading term.

Answer: (b) \( a=\frac35 \)

Explanation:
1. Near 0: \( \sin^3\sqrt x\approx x^{3/2} \), \( \log(1+3x)\approx3x \), \( (\tan^{-1}\sqrt x)^2\approx x \), \( e^{5\sqrt x}-1\approx5\sqrt x \).
2. The limit is \( \frac{3x\cdot x^{3/2}}{x\cdot5\sqrt x\cdot x}=\frac35 \).

Teacher's Note:
1. Count powers of \( \sqrt x \) in numerator and denominator.

Answer: (a) \( -8 \)

Explanation:
1. \( \cos^2x-\sin^2x-1=\cos2x-1=-2\sin^2x\approx-2x^2 \).
2. \( \sqrt{x^2+4}-2\approx\frac{x^2}4 \), so the limit is \( \frac{-2x^2}{x^2/4}=-8 \).

Teacher's Note:
1. Rationalise the denominator or expand the root.

Answer: (a) \( e^{15} \)

Explanation:
1. \( \ln(\cos x+3\sin x)\approx3x \) for small x, and \( \operatorname{cosec}x\approx\frac1x \).
2. So the exponent tends to \( 5\cdot3=15 \) and \( \lambda=e^{15} \).

Teacher's Note:
1. Take the logarithm of the \( 1^\infty \) form.

Answer: (b) \( a+b \)

Explanation:
1. \( \ln(1+ax)\approx ax \) and \( -\ln(1-bx)\approx bx \).
2. The limit is \( a+b \).

Teacher's Note:
1. The assigned value equals the limit at the point.

Answer: (d) \( x=1,2 \)

Explanation:
1. For \( 0\le x\le1 \): \( f(f(x))=f(1+x)=2+x \); for \( 1<x\le2 \): \( f(1+x)=3-(1+x)=2-x \). At \( x=1 \) the left value is 3 and the right value is 1: discontinuous.
2. At \( x=2 \): \( g(2)=f(3)=0 \), the left limit is 0 but for \( 2<x\le3 \), \( g(x)=f(3-x)=4-x\to2 \): discontinuous.
3. The points are \( x=1,2 \).

Teacher's Note:
1. Compute \( g \) piece by piece and test the break points of both stages.

Answer: (a) continuous at \( x=-3 \) and discontinuous at \( x=3 \)

Explanation:
1. At \( x=-3 \): \( f(-3)=6 \) and the right limit is \( -2(-3)=6 \): continuous.
2. At \( x=3 \): the left limit is \( -6 \) but \( f(3)=20 \): discontinuous.

Teacher's Note:
1. Compare both sides at each joining point.

Answer: (a) discontinuous at \( x=0 \)

Explanation:
1. As \( x\to0^+ \), \( e^{1/x}\to\infty \) and the fraction tends to 1.
2. As \( x\to0^- \), \( e^{1/x}\to0 \) and the fraction tends to \( -1 \). The limits differ, so f is discontinuous at 0.

Teacher's Note:
1. The two one-sided limits are \( +1 \) and \( -1 \).

Answer: (b) not continuous at \( x=0 \)

Explanation:
1. As \( x\to0^+ \), \( e^{1/x}\to\infty \) so \( f\to0 \).
2. As \( x\to0^- \), \( e^{1/x}\to0 \) so \( f\to1 \). The one-sided limits differ, so f is not continuous at 0.

Teacher's Note:
1. Check both sides of 0.

Answer: (c) discontinuous at the origin because \( \frac{|x|}x \) is discontinuous there

Explanation:
1. \( |x| \) is continuous everywhere.
2. \( \frac{|x|}x \) equals \( 1 \) for \( x>0 \) and \( -1 \) for \( x<0 \), so it is discontinuous at 0, which makes f discontinuous there.

Teacher's Note:
1. A sum is discontinuous when exactly one summand is.

Answer: (c) 0

Explanation:
1. \( f'(0)=\lim_{h\to0}\frac{h^2\sin\frac1h}h=\lim h\sin\frac1h=0 \).

Teacher's Note:
1. Use the definition of the derivative with the squeeze theorem.

Answer: (a) \( f'(0^+) \) and \( f'(0^-) \) do not exist

Explanation:
1. \( \frac{f(h)-f(0)}h=\sin\frac1h \), which oscillates between \( -1 \) and 1 as \( h\to0 \).
2. So neither one-sided derivative exists.

Teacher's Note:
1. The difference quotient reduces to \( \sin\frac1h \).

Answer: (b) 1

Explanation:
1. For \( x>0 \), \( f=\frac x{1+x} \) and \( f'(0^+)=1 \); for \( x<0 \), \( f=\frac x{1-x} \) and \( f'(0^-)=1 \).
2. So \( f'(0)=1 \).

Teacher's Note:
1. Differentiate each branch and check that the one-sided derivatives agree.

Answer: (d) 4

Explanation:
1. \( |x| \) is not differentiable at \( x=0 \); \( |\cos x| \) is not differentiable at \( \pm\frac\pi2 \) (both in \( (-2,2) \)).
2. \( \tan\left(x+\frac\pi4\right) \) is undefined at \( x=\frac\pi4 \) (and at \( -\frac{3\pi}4\approx-2.36 \), outside the interval).
3. The points are \( 0,\frac\pi2,-\frac\pi2,\frac\pi4 \): 4 points.

Teacher's Note:
1. List every point where any term fails to be differentiable inside the interval.

Answer: (a) \( \{0\} \)

Explanation:
1. At \( x=0 \) the term \( -|x| \) has a corner, so f is not differentiable there.
2. At \( x=1 \): \( \frac{f(x)-f(1)}{x-1}=(x-1)\sin\frac1{x-1}+\frac{1-|x|}{x-1}\to0-1=-1 \), so f is differentiable at 1.

Teacher's Note:
1. Check the special point \( x=1 \) with the definition.

Answer: (a) \( \{0,1\} \)

Explanation:
1. \( \frac{f(h)-f(0)}h=\frac1{1+e^{1/h}} \).
2. As \( h\to0^+ \), \( e^{1/h}\to\infty \) and the quotient tends to 0; as \( h\to0^- \), \( e^{1/h}\to0 \) and it tends to 1.

Teacher's Note:
1. The difference quotient equals \( \frac1{1+e^{1/h}} \).