JEE Mathematics Limits Continuity and Differentiability MCQs Set 15

Answer: (d) 1

Explanation:
1. \( \cos\frac xm=1-\frac{x^2}{2m^2}+\cdots \), so \( \left(\cos\frac xm\right)^m=\left(1-\frac{x^2}{2m^2}\right)^m\approx e^{-x^2/(2m)} \).
2. As \( m\to\infty \) the exponent tends to 0, so the limit is 1.

Teacher's Note:
1. The correction term is of order \( 1/m^2 \), which is too small to survive the power m.

Answer: (a) \( \frac{10}3 \)

Explanation:
1. \( 1-\cos2x=2\sin^2x\approx2x^2 \).
2. So the expression behaves like \( \frac{2x^2\cdot5x}{x^2\cdot3x}=\frac{10}3 \).

Teacher's Note:
1. Replace each trigonometric factor by its leading term as \( x\to0 \).

Answer: (c) \( a=1 \) and \( b=-2 \)

Explanation:
1. \( \frac{x^3+1}{x^2+1}=x+\frac{1-x}{x^2+1} \), so the limit is finite only if \( a=1 \).
2. Then the bracket tends to \( -b \), so \( -b=2 \) and \( b=-2 \).

Teacher's Note:
1. Divide the polynomials to isolate the linear part.

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

Explanation:
1. The base is \( 1+\frac2{n^2-n-1} \).
2. The limit is \( e^{\lim\frac{2n(n-1)}{n^2-n-1}}=e^2 \).

Teacher's Note:
1. This is a \( 1^\infty \) form: use \( e^{\lim g\cdot(f-1)} \).

Answer: (d) 100

Explanation:
1. Divide numerator and denominator by \( x^{10} \); each of the 100 terms in the numerator tends to 1.
2. The denominator tends to 1, so the limit is 100.

Teacher's Note:
1. Compare the leading powers of x.

Answer: (b) \( \pi \)

Explanation:
1. \( \sin(\pi\cos^2x)=\sin(\pi-\pi\sin^2x)=\sin(\pi\sin^2x) \).
2. \( \frac{\sin(\pi\sin^2x)}{x^2}\to\pi \) as \( x\to0 \).

Teacher's Note:
1. Use \( \sin(\pi-\theta)=\sin\theta \).

Answer: (d) \( \frac c\alpha\left(\frac1\alpha-\frac1\beta\right) \)

Explanation:
1. The roots of \( cx^2+bx+a=0 \) are \( \frac1\alpha \) and \( \frac1\beta \), so \( cx^2+bx+a=c\left(x-\frac1\alpha\right)\left(x-\frac1\beta\right) \).
2. As \( x\to\frac1\alpha \) the sine behaves like its argument, and \( x\alpha-1=\alpha\left(x-\frac1\alpha\right) \).
3. The limit is \( \frac c\alpha\left(\frac1\alpha-\frac1\beta\right) \).

Teacher's Note:
1. Reciprocal roots of the reversed quadratic let you factor the argument of the sine.

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

Explanation:
1. \( \cos(\sin x)-1\approx-\frac{\sin^2x}2\approx-\frac{x^2}2 \).
2. So the limit is \( -\frac12 \).

Teacher's Note:
1. Use \( \cos u-1\approx-\frac{u^2}2 \) with \( u=\sin x\approx x \).

Answer: (c) 3

Explanation:
1. \( \cos x-1\approx-\frac{x^2}2 \) and \( \cos x-e^x\approx-x \).
2. The numerator behaves like \( \frac{x^3}2 \), so \( n=3 \).

Teacher's Note:
1. Find the leading power of x in the numerator.

Answer: (d) is equal to 3

Explanation:
1. The numerator is \( \approx f'(2)(2h)=12h \) and the denominator is \( \approx f'(1)h=4h \).
2. So the limit is \( \frac{12}{4}=3 \).

Teacher's Note:
1. Replace each difference by derivative times the increment.

Answer: (c) \( -\log2 \)

Explanation:
1. \( \log_{x^2}2=\frac{\ln2}{2\ln x} \).
2. \( \frac{1-x^2}{2\ln x}\to\frac{-2(x-1)}{2(x-1)}=-1 \), so the limit is \( -\ln2 \).

Teacher's Note:
1. Convert the logarithm to the natural base.

Answer: (b) \( \frac14 \)

Explanation:
1. \( 1^3+\cdots+n^3=\frac{n^2(n+1)^2}4 \).
2. The ratio tends to \( \frac14 \).

Teacher's Note:
1. Use the formula for the sum of cubes.

Answer: (b) \( \frac14 \)

Explanation:
1. Put \( x=\cos t \) with \( t\to0^+ \): \( 1-\sqrt{\cos t}\approx\frac{t^2}4 \).
2. So the limit is \( \frac{t^2/4}{t^2}=\frac14 \).

Teacher's Note:
1. A substitution \( x=\cos t \) removes the inverse cosine.

Answer: (a) \( \frac12 \)

Explanation:
1. \( \tan2x-2\tan x\approx2x\cdot\frac{(2x)^2}{3}\ldots \); precisely \( \tan2x-2\tan x\approx\frac{8x^3}3-\frac{2x^3}3=2x^3 \), so the numerator is \( 2x^4 \).
2. \( (1-\cos2x)^2=4\sin^4x\approx4x^4 \), so the limit is \( \frac12 \).

Teacher's Note:
1. Expand \( \tan \) to the cubic term.

Answer: (d) does not exist

Explanation:
1. The expression is \( e^{\ln(1+\sin x)/x^2} \) with \( \frac{\ln(1+\sin x)}{x^2}\approx\frac1x \).
2. As \( x\to0^+ \) it tends to \( \infty \), and as \( x\to0^- \) the exponent tends to \( -\infty \) giving 0. The two sides differ, so the limit does not exist.

Teacher's Note:
1. Check both one-sided limits when the exponent changes sign.

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

Explanation:
1. The sum is \( \frac{1+2+\cdots+n}{1-n^2}=\frac{n(n+1)}{2(1-n^2)}=-\frac{n}{2(n-1)} \).
2. As \( n\to\infty \) this tends to \( -\frac12 \).

Teacher's Note:
1. Add the numerators first.

Answer: (d) \( \frac12 \)

Explanation:
1. With \( u=\sin x \), \( e^u-u-1\approx\frac{u^2}2 \).
2. So the limit is \( \frac{x^2/2}{x^2}=\frac12 \).

Teacher's Note:
1. Use the series of \( e^u \) up to \( u^2 \).

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

Explanation:
1. Put \( x=\frac\pi2-h \), so \( \pi-2x=2h \), \( \cot x=\tan h \) and \( \cos x=\sin h \).
2. \( \tan h-\sin h\approx\frac{h^3}2 \), so the limit is \( \frac{h^3/2}{8h^3}=\frac1{16} \).

Teacher's Note:
1. Shift the variable so that the limit point becomes 0.

Answer: (b) 1

Explanation:
1. For large x, \( x^4\sin\frac1x\approx x^3-\frac x6 \).
2. So the numerator is \( \approx x^3+x^2 \) and the denominator is \( \approx x^3 \), giving 1.

Teacher's Note:
1. Expand \( \sin\frac1x \) for large x.

Answer: (b) 4

Explanation:
1. The limit is the derivative of \( x^\lambda \) at 5: \( \lambda5^{\lambda-1} \).
2. \( \lambda5^{\lambda-1}=500 \) holds for \( \lambda=4 \) since \( 4\cdot125=500 \).

Teacher's Note:
1. The limit is \( f'(5) \) for \( f(x)=x^\lambda \).

Answer: (b) \( \frac9{2\log3} \)

Explanation:
1. \( 1-\cos3x\approx\frac{9x^2}2 \) and \( 3^x-1\approx x\ln3 \).
2. So the limit is \( \frac{9/2}{\ln3}=\frac9{2\log3} \).

Teacher's Note:
1. Use \( a^x-1\approx x\ln a \).

Answer: (d) does not exist

Explanation:
1. For \( 0<x<1 \), \( [x]=0 \) so \( f=0 \) and the right limit is 0.
2. For \( -1<x<0 \), \( [x]=-1 \) so \( f=\frac{\sin(-1)}{-1}=\sin1\ne0 \). The one-sided limits differ, so the limit does not exist.

Teacher's Note:
1. Evaluate the two one-sided limits separately.

Answer: (a) 1

Explanation:
1. \( 2\sin x-\sin2x=2\sin x(1-\cos x) \).
2. \( \frac{2\sin x(1-\cos x)}{x^3}\to2\cdot1\cdot\frac12=1 \).

Teacher's Note:
1. Factor the numerator using the double angle formula.

Answer: (b) \( e^{-1} \)

Explanation:
1. Put \( x=1+h \): \( \tan\left(\frac\pi4+\frac{\pi h}4\right)-1\approx\frac{\pi h}2 \) and \( \tan\left(\frac\pi2+\frac{\pi h}2\right)=-\cot\frac{\pi h}2\approx-\frac2{\pi h} \).
2. The \( 1^\infty \) form gives \( e^{\frac{\pi h}2\cdot\left(-\frac2{\pi h}\right)}=e^{-1} \).

Teacher's Note:
1. Apply \( 1^\infty\to e^{\lim(\text{base}-1)\cdot\text{exponent}} \).

Answer: (a) 0

Explanation:
1. \( \ln\cos x\approx-\frac{x^2}2 \), so \( \frac{\ln\cos x}{x^4}\approx-\frac1{2x^2}\to-\infty \).
2. Therefore \( (\cos x)^{1/x^4}\to e^{-\infty}=0 \).

Teacher's Note:
1. Take logarithms when the exponent blows up.

Answer: (c) \( n \)

Explanation:
1. As \( x\to0 \), the largest term \( n^{\operatorname{cosec}^2x} \) dominates the sum.
2. The expression tends to \( \left(n^{\operatorname{cosec}^2x}\right)^{\sin^2x}=n \).

Teacher's Note:
1. The largest term controls the sum of powers when the exponent tends to infinity.

Answer: (a) continuous everywhere except at \( x=1 \)

Explanation:
1. At \( x=0 \): left limit 0, \( f(0)=0 \) and right limit 0, so f is continuous.
2. At \( x=1 \): left limit 0 but right limit 4, so f is discontinuous at 1.

Teacher's Note:
1. Check the limits at each joining point.

Answer: (d) \( a=b+\frac23 \)

Explanation:
1. Continuity at 3 needs \( 3a+1=3b+3 \).
2. So \( a=b+\frac23 \).

Teacher's Note:
1. Equate the left and right values at the joining point.

Answer: (d) 6

Explanation:
1. With \( x=\frac\pi2-h \): \( \frac{k\sin h}{2h}\to\frac k2 \).
2. Continuity needs \( \frac k2=3 \), so \( k=6 \).

Teacher's Note:
1. Shift the variable to turn the limit into \( \frac{\sin h}h \).

Answer: (b) \( a=2,\ b=1 \)

Explanation:
1. At \( x=2 \): \( 2a+b=5 \). At \( x=10 \): \( 10a+b=21 \).
2. Subtracting: \( 8a=16 \), \( a=2 \), then \( b=1 \).

Teacher's Note:
1. Two joining points give two equations.

Answer: (b) continuous at \( x=0 \)

Explanation:
1. \( \left||x|\cos\frac1x\right|\le|x|\to0 \), so \( \lim_{x\to0}f(x)=0=f(0) \).
2. Hence f is continuous at 0.

Teacher's Note:
1. Squeeze the bounded cosine between \( \pm|x| \).

Answer: (c) continuous at \( x=1 \)

Explanation:
1. At \( x=1 \): \( f(1)=\frac12 \) and \( \lim_{x\to1^+}\left(2x^2-3x+\frac32\right)=2-3+\frac32=\frac12 \).
2. The two values agree, so f is continuous at 1.

Teacher's Note:
1. Evaluate both pieces at the joining point.