JEE Mathematics Limits Continuity and Differentiability MCQs Set 08

Answer: (a) \( e^4 \)

Explanation:
1. The form is \( 1^\infty \): \( \exp\left[\lim(x+1)\left(\frac{x+2}{x-2}-1\right)\right]=\exp\left[\lim\frac{4(x+1)}{x-2}\right] \).
2. The exponent tends to 4, so the limit is \( e^4 \).

Teacher's Note:
1. Use the 1^infinity rule: e^(lim g(x)(f(x) - 1)).
2. Compute the exponent first.

Answer: (b) \( e^5 \)

Explanation:
1. The form is \( 1^\infty \) with exponent \( \frac5x\tan^2\sqrt x\to5 \) (since \( \tan^2\sqrt x\sim x \)).
2. So the limit is \( e^5 \).

Teacher's Note:
1. Use e^(lim g (f - 1)).
2. tan^2(sqrt x) ~ x.

Answer: (a) \( 1 \)

Explanation:
1. Near pi/4, x is about 0.785, so [x] = 0.
2. The expression is \( 1^{1/\ln(\tan x)} \), and for x not equal to pi/4 the base is exactly 1.
3. So every value near pi/4 equals 1 and the limit is 1.

Teacher's Note:
1. Evaluate [x] near pi/4 first.
2. 1 raised to any power is 1.

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

Explanation:
1. The form is \( 1^\infty \): the exponent is \( \lim x\cdot\frac{2x-1}{x^2-4x+2}=2 \).
2. So the limit is \( e^2 \).

Teacher's Note:
1. Compute (base - 1) first: (x^2 - 2x + 1 - x^2 + 4x - 2) = 2x - 1.
2. Multiply by x.

Answer: (d) \( 1 \)

Explanation:
1. The form is \( 1^\infty \) with exponent \( \lim\frac{\cos x-1}{\sin x}=\lim\frac{-x^2/2}{x}=0 \).
2. So the limit is \( e^0=1 \).

Teacher's Note:
1. Use e^(lim (cos x - 1)/sin x).
2. (cos x - 1)/sin x ~ -x/2.

Answer: (b) \( e^{-x^2/2} \)

Explanation:
1. The form is \( 1^\infty \): exponent \( n\left(\cos\frac x{\sqrt n}-1\right)\approx n\cdot\left(-\frac{x^2}{2n}\right)=-\frac{x^2}2 \).
2. So the limit is \( e^{-x^2/2} \).

Teacher's Note:
1. cos u - 1 ~ -u^2/2 with u = x/sqrt n.
2. The factor n cancels the 1/n.

Answer: (c) 0

Explanation:
1. Use \( \cos(\sin x)-\cos x=2\sin\frac{x+\sin x}{2}\sin\frac{x-\sin x}{2} \).
2. For small x: \( \sin\frac{x+\sin x}2\approx x \) and \( \sin\frac{x-\sin x}2\approx\frac{x-\sin x}{2}\approx\frac{x^3}{12} \), so the numerator is about \( \frac{x^4}{6} \).
3. Hence \( f(x)\approx\frac{x^2}{6}\to0 \), and continuity requires a = 0.

Teacher's Note:
1. Use the product formula cos A - cos B = -2 sin((A+B)/2) sin((A-B)/2).
2. x - sin x ~ x^3/6.

Answer: (c) f is discontinuous at x = 0

Explanation:
1. At x = 0: for x slightly negative \( [x]=-1 \) and \( f=|x+\tfrac12|\to\tfrac12 \), whereas \( f(0)=0 \) and the right limit is 0. So f is discontinuous at 0.
2. At x = 1: left limit 0 (since [x] = 0), but f(1) = 1.5: discontinuous.
3. At x = -1: left limit \( |(-\tfrac12)(-2)|=1 \) but right limit \( \tfrac12 \): discontinuous.
4. At x = 2: left limit 2.5, f(2) = 5: discontinuous.

Teacher's Note:
1. Check the value of [x] on each side of the point.
2. Only one of the four statements holds.

Answer: (c) discontinuous at x = 1 since neither \( f(1^-) \) nor \( f(1^+) \) exists

Explanation:
1. At x = 1 the base \( 4x-3\to1 \) and the numerator \( \ln(x^2-2x+5)\to\ln4\ne0 \).
2. So \( f=\frac{\ln(x^2-2x+5)}{\ln(4x-3)} \) has a nonzero numerator over a denominator tending to 0.
3. From the right the value tends to +infinity and from the left to -infinity; neither one-sided limit is finite.

Teacher's Note:
1. Write the logarithm to base 4x - 3 as a ratio of natural logs.
2. The denominator vanishes at x = 1 while the numerator does not.

Answer: (d) discontinuous at x = 1

Explanation:
1. At x = 1: for x just below 1, [x] = 0 and \( \sin\frac{\pi}{2}(x)\to1 \) so f tends to 1; for x just above 1, [x] = 1 and \( \sin\frac\pi2(x+2)\to\sin\frac{3\pi}2=-1 \) so f tends to -1.
2. The one-sided limits differ, so f is discontinuous at x = 1.
3. (At x = 2 all values tend to 0, so f is continuous there; at x = 3 the limits are -3 and 3, so it is discontinuous.)

Teacher's Note:
1. Evaluate [x] on either side of the point.
2. Compute sin of the multiple of pi/2 carefully.

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

Explanation:
1. The limit at 0 from the left is \( \lim\frac{2px}{x(\sqrt{1+px}+\sqrt{1-px})}=p \).
2. The value at 0 is \( f(0)=\frac{1}{-2}=-\frac12 \).
3. Continuity at 0 requires \( p=-\frac12 \).

Teacher's Note:
1. Rationalise the numerator on the left.
2. Equate the left limit with f(0).

Answer: (d) exactly three points

Explanation:
1. The composite is \( \operatorname{sgn}(x(x-2)(x-3)) \).
2. The sign changes at the zeros of g, namely x = 0, 2, 3, and at each zero the function value is 0 while the neighbouring values are +1 or -1.
3. So it is discontinuous at exactly three points.

Teacher's Note:
1. sgn is discontinuous only where its argument is 0.
2. Factorise g(x).

Answer: (a) 3

Explanation:
1. The denominator \( t^2+t-2=(t+2)(t-1) \) vanishes at t = -2 and t = 1.
2. These correspond to \( x-1=-\frac12 \) (x = 1/2) and \( x-1=1 \) (x = 2).
3. Also t itself is undefined at x = 1. So there are three points: x = 1/2, 1, 2.

Teacher's Note:
1. Find where t is undefined and where the denominator is 0.
2. Convert the values of t back to x.

Answer: (b) at least one real solution in \( [0,\pi/4] \)

Explanation:
1. Let \( f(x)=2\tan x+5x-2 \): \( f(0)=-2<0 \) and \( f(\pi/4)=2+\frac{5\pi}{4}-2>0 \).
2. f is continuous, so by the intermediate value theorem it has a zero in the interval.
3. (f is also increasing, so the zero is unique, but only 'at least one' is guaranteed by the theorem.)

Teacher's Note:
1. Check the signs of f at the end points.
2. Apply the intermediate value theorem.

Answer: (b) continuous but not differentiable at x = 0

Explanation:
1. As \( x\to0^+ \), \( e^{1/x}\to\infty \), so \( \frac{3e^{1/x}+4}{2-e^{1/x}}\to-3 \); as \( x\to0^- \), \( e^{1/x}\to0 \) and the fraction tends to 2.
2. So \( f(x)\to0 \) from both sides: f is continuous at 0.
3. The right derivative is -3 and the left derivative is 2, so f is not differentiable at 0.

Teacher's Note:
1. Study the behaviour of e^(1/x) on each side of 0.
2. Compare the one-sided derivatives f(x)/x.

Answer: (a) continuous at x = 0 (but not differentiable there)

Explanation:
1. For x near 0, \( \sin^{-1}(\cos x)=\frac\pi2-|x| \).
2. This is continuous at 0 but has a corner there (slopes -1 and +1), so it is not differentiable.

Teacher's Note:
1. cos x = sin(pi/2 - |x|) near 0.
2. A graph with a corner is continuous but not differentiable.

Answer: (a) f is neither continuous nor differentiable at x = 0 and is not continuous at x = 2

Explanation:
1. As \( x\to0^- \), \( \{x\}=1+x\to1 \), so \( f\to0+1+0=1\ne f(0)=0 \): f is discontinuous at 0 (so not differentiable).
2. At x = 2: from the left \( \{x\}\to1 \) and \( f\to2+1+2\sin1 \); from the right \( f\to2 \): f is discontinuous at 2.

Teacher's Note:
1. The fractional part jumps from near 1 to 0 at each integer.
2. Compare left and right limits.

Answer: (d) f is not differentiable at x = 2 and x = 3

Explanation:
1. At x = 2: the left value is \( \frac35 \) and the right limit is \( \frac14(8-4)=1 \): f is discontinuous, hence not differentiable.
2. At x = 3: both sides give \( \frac{27-9}{4}=\frac92 \) (\( \frac94(1+1)=\frac92 \)), so f is continuous.
3. The left derivative at 3 is \( \frac{3x^2-2x}{4}=\frac{21}4 \); the right derivative is the slope of \( \frac94\cdot2 \) which is a constant, namely 0. They differ, so f is not differentiable at 3.

Teacher's Note:
1. Check continuity first.
2. For 3 < x < 4, |x - 4| + |2 - x| = (4 - x) + (x - 2) = 2, a constant.

Answer: (a) \( (-\infty,\infty) \)

Explanation:
1. For x > 0, \( f=\frac{x}{1+x} \) with derivative \( \frac1{(1+x)^2} \); for x < 0, \( f=\frac{x}{1-x} \) with derivative \( \frac{1}{(1-x)^2} \).
2. At x = 0 both one-sided derivatives equal 1, so f is differentiable everywhere.

Teacher's Note:
1. Treat x > 0 and x < 0 separately.
2. Compare the derivatives at 0.

Answer: (a) \( |f|^2 \) is differentiable everywhere

Explanation:
1. \( |f|^2=f^2 \) is a product of differentiable functions, so it is differentiable.
2. The functions \( |f| \) and \( f+|f|=2\max(f,0) \) can have corners where f changes sign, for example f(x) = x at x = 0.

Teacher's Note:
1. |f|^2 and f^2 are the same function.
2. Test with f(x) = x.