JEE Mathematics Limits Continuity and Differentiability MCQs Set 03

Answer: (d) \( a=2 \) and \( L=\frac1{64} \)

Explanation:
1. \( \sqrt{a^2-x^2}=a-\frac{x^2}{2a}-\frac{x^4}{8a^3}-\dots \), so the numerator is \( x^2\left(\frac1{2a}-\frac14\right)+\frac{x^4}{8a^3}+\dots \).
2. Finiteness needs \( a=2 \); then \( L=\frac1{8\cdot8}=\frac1{64} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( e^{-5} \)

Explanation:
1. \( \left(1-\frac5{x+2}\right)^x\to e^{-5} \) since \( \frac{-5x}{x+2}\to-5 \).

Teacher's Note:
1. Use the 1-to-the-infinity form: limit equals e raised to the limit of (base - 1) times exponent.
2. (x - 3)/(x + 2) - 1 = -5/(x + 2).

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

Explanation:
1. The graphs of \( y=x \) and \( y=x^3 \) cross at \( x=-1,0,1 \).
2. At each crossing the slopes of the two graphs differ (\( 1 \) and \( 3x^2=3,0,3 \)), so there is a corner at \( -1,0,1 \); \( f \) is smooth elsewhere.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) \( e^2 \)

Explanation:
1. The expression is of the form \( 1^\infty \): the limit is \( \exp\left(\lim\frac{f(1+x)-f(1)}{x\,f(1)}\right)=\exp\left(\frac{f'(1)}{f(1)}\right) \).
2. This is \( e^{6/3}=e^2 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( (-1)^k(k-1)\pi \)

Explanation:
1. Just to the left of \( k \), \( [x]=k-1 \), so \( f(x)=(k-1)\sin\pi x \).
2. Its derivative is \( (k-1)\pi\cos\pi x\to(k-1)\pi\cos k\pi=(-1)^k(k-1)\pi \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) 1

Explanation:
1. As \( x\to0^+ \), \( 0<\sin x<1 \) and \( \frac1x\to\infty \), so \( (\sin x)^{1/x}\to0 \).
2. Also \( \left(\frac1x\right)^{\sin x}=e^{-\sin x\ln x} \) and \( \sin x\ln x\to0 \), so this term tends to 1.
3. The limit is \( 0+1=1 \).

Teacher's Note:
1. Handle the two terms separately.
2. x ln x tends to 0 as x tends to 0 from the right.

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

Explanation:
1. For small \( t \): \( \frac{t\ln(1+t)}{t^4+4}\approx\frac{t^2}{4} \).
2. Hence the integral \( \approx\frac{x^3}{12} \) and the limit is \( \frac1{12} \) (also by L'Hopital: \( \frac{x\ln(1+x)}{3x^2(x^4+4)}\to\frac1{12} \)).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) \( \frac8\pi f(2) \)

Explanation:
1. At \( x=\frac\pi4 \), \( \sec^2x=2 \), so the numerator and denominator both tend to 0; use L'Hopital's rule.
2. Numerator derivative: \( f(\sec^2x)\cdot2\sec^2x\tan x\to f(2)\cdot4 \). Denominator derivative: \( 2x\to\frac\pi2 \).
3. Limit \( =\frac{4f(2)}{\pi/2}=\frac8\pi f(2) \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) 2

Explanation:
1. Put \( y=x^2 \): \( \frac{x^2+\sin x^2}{x^4+\sin^2x} \).
2. Divide by \( x^2 \): \( \frac{1+\frac{\sin x^2}{x^2}}{x^2+\left(\frac{\sin x}{x}\right)^2}\to\frac{1+1}{0+1}=2 \).

Teacher's Note:
1. Reduce the expression to a standard form before reading off the answer.
2. Verify with a simple check of the final result.

Answer: (d) \( f \) is continuous everywhere but not differentiable at \( x=n\pi \)

Explanation:
1. \( \sin x \) is continuous, so is \( 1+|\sin x| \).
2. At \( x=n\pi \) the slope is \( -1 \) on one side and \( +1 \) on the other, so there is a corner: not differentiable there.

Teacher's Note:
1. Absolute value creates corners where the inside function crosses zero.
2. sin x = 0 at x = n pi.

Answer: (b) strictly increasing in (-1, 1) but neither continuous nor differentiable at 0

Explanation:
1. \( x+[x] \) equals \( x-1 \) on \( [-1,0) \) and \( x \) on \( [0,1) \), with a jump of 1 at 0.
2. It is strictly increasing but discontinuous (hence not differentiable) at 0.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) 0

Explanation:
1. Both numerator and denominator tend to infinity. By L'Hopital's rule: \( \frac{2\int_0^xe^{t^2}dt\cdot e^{x^2}}{e^{2x^2}}=\frac{2\int_0^xe^{t^2}dt}{e^{x^2}} \).
2. Applying it again: \( \frac{2e^{x^2}}{2xe^{x^2}}=\frac1x\to0 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( f \) is continuous at \( x=2 \)

Explanation:
1. At \( x=2 \): \( f(2)=2 \), and for \( x\to2^+ \) \( [x]=2 \), so the limit is 2: continuous at 2.
2. The left slope is 1 and the right slope is 0, so it is not differentiable at 2.
3. At \( x=3 \) and \( x=4 \) the greatest integer function jumps, so \( f \) is not continuous on \( [-2,4] \).

Teacher's Note:
1. Check one-sided limits and one-sided slopes separately.
2. The greatest integer function has jumps at integers greater than 2.

Answer: (a) \( \pi \)

Explanation:
1. \( \sin(\pi\cos^2x)=\sin(\pi-\pi\sin^2x)=\sin(\pi\sin^2x) \).
2. So the limit is \( \lim\frac{\sin(\pi\sin^2x)}{\pi\sin^2x}\cdot\frac{\pi\sin^2x}{x^2}=\pi \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) 0

Explanation:
1. The limit forces \( p(x)=ax^4+bx^3+x^2 \) (so \( p/x^2\to1 \)).
2. Then \( p'(x)=x(4ax^2+3bx+2) \) with roots 1, 2 gives \( 4ax^2+3bx+2=4a(x-1)(x-2) \): \( 8a=2 \), \( a=\frac14,\ b=-1 \).
3. So \( p(2)=4-8+4=0 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) continuous everywhere

Explanation:
1. \( \sin x \) is continuous, and so is its absolute value.
2. At \( x=0,\ \pi \) the function has corners (slopes change sign), so it is not differentiable everywhere.
3. It rises and falls repeatedly, so it is neither monotonic nor one-one.

Teacher's Note:
1. The modulus of a continuous function stays continuous.
2. Corners appear where sin x = 0.

Answer: (a) \( g(x) \) cannot be differentiable at \( x=2 \)

Explanation:
1. If \( g \) were differentiable at 2, then \( f=(f+g)-g \) would be differentiable there, a contradiction.
2. So \( g \) cannot be differentiable at 2; it may still be continuous (e.g. \( f=|x-2|,\ g=-|x-2| \)) so neither continuity nor discontinuity is forced.

Teacher's Note:
1. A difference of differentiable functions is differentiable.
2. Use a counterexample to see what is not forced.

Answer: (d) continuous, differentiable and strictly increasing in (-1, 1)

Explanation:
1. \( x|x| \) equals \( x^2 \) for \( x\ge0 \) and \( -x^2 \) for \( x<0 \); its derivative \( 2|x|\ge0 \) is zero only at 0.
2. So it is continuous, differentiable and strictly increasing.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( a_0=1,\ a_1=0 \)

Explanation:
1. Since \( \sin x=x-\frac{x^3}{6}+\dots \), we get \( \frac{\sin x}{x}=1-\frac{x^2}{6}+\dots \).
2. Comparing coefficients: \( a_0=1 \) and \( a_1=0 \).

Teacher's Note:
1. Divide the Maclaurin series of sin x by x term by term.
2. Only even powers of x occur.

Answer: (b) continuous and differentiable in (-1, 1) but not strictly increasing

Explanation:
1. \( |x-1|+|x+1|=(1-x)+(x+1)=2 \) on \( (-1,1) \), a constant function.
2. It is continuous and differentiable, but not strictly increasing.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( n=2,\ m=2 \)

Explanation:
1. The left derivative of \( |x-1| \) at 1 is \( -1 \), so \( p=-1 \).
2. With \( t=x-1\to0^+ \): \( g\approx\frac{t^n}{m\ln\cos t}\approx\frac{t^n}{-mt^2/2}=-\frac{2t^{n-2}}{m} \).
3. For a finite limit \( -1 \) we need \( n=2 \) and \( -\frac2m=-1 \), i.e. \( m=2 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.