Let f 1 : R rightarrow R , f 2 : (-π/2, π/2 ) rightarrow R , f 3 : (-1, e π / 2 -2 ) rightarrow R and f 4 : R rightarrow R be functions defined by (i) f 1 (x)=sin (sqrt 1-e -x 2 ) , (ii) f 2 (x)= \ begin array ccc frac sin x tan -1 x & if & x neq 0 \\1 & if & x=0end array . , where the inverse trigonometric function tan -1 x assumes values in (-π/2, π/2 ) , (iii) f 3 (x)= [sin (log e (x+2) ) ] , where, for t in R ,[t] denotes the greatest integer less than or equal to t , (iv) f 4 (x)= \ begin array ccc x 2 sin (1/x ) & if & x neq 0 0 & if & x=0end array . . LIST-I LIST-II --- --- P. The function f 1 is 1. NOT continuous at x=0 Q. The function f 2 is 2. continuous at x=0 and NOT differentiable at x=0 R. The function f 3 is 3. differentiable at x=0 and its derivative is NOT continuous at x=0 S. The function f 4 is 4. differentiable at x=0 and its derivative is continuous at x =0

P rightarrow 2 ; Q rightarrow 1 ; R rightarrow 4 ; S rightarrow 3

P rightarrow 2 ; Q rightarrow 3 ; R rightarrow 1 ; S rightarrow 4

P rightarrow 4 ; Q rightarrow 1 ; R rightarrow 2 ; S rightarrow 3

P rightarrow 4 ; Q rightarrow 2 ; R rightarrow 1 ; S rightarrow 3

Mathematics · Limits, Continuity and Differentiability

JEE Advanced 2018 — Paper 2 — Question 45

Let $f_{1}: \mathrm{R} \rightarrow \mathrm{R}, f_{2}:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow \mathrm{R}, \mathrm{f}_{3}:\left(-1, \mathrm{e}^{\pi / 2}-2\right) \rightarrow \mathrm{R}$ and $\mathrm{f}_{4}: \mathrm{R} \rightarrow \mathrm{R}$ be functions defined by

(i) $f_{1}(x)=\sin \left(\sqrt{1-e^{-x^{2}}}\right)$ ,

(ii) $f_{2}(x)=\left\{\begin{array}{ccc}\frac{|\sin x|}{\tan ^{-1} x} & \text { if } & x \neq 0 \\1 & \text { if } & x=0\end{array}\right.$ , where the inverse trigonometric function $\tan ^{-1} x$ assumes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ ,

(iii) $f_{3}(x)=\left[\sin \left(\log _{\mathrm{e}}(x+2)\right)\right]$ , where, for $t \in \mathrm{R},[t]$ denotes the greatest integer less than or equal to $t$ ,

(iv) $f_{4}(x)=\left\{\begin{array}{ccc}x^{2} \sin \left(\frac{1}{x}\right) & \text { if } & x \neq 0 0 & \text { if } & x=0\end{array}\right.$ .

LIST-I	LIST-II
P. The function $f_{1}$ is	1. NOT continuous at $x=0$
Q. The function $f_{2}$ is	2. continuous at $x=0$ and NOT differentiable at $x=0$
R. The function $f_{3}$ is	3. differentiable at $x=0$ and its derivative is NOT continuous at $x=0$
S. The function $f_{4}$ is	4. differentiable at $x=0$ and its derivative is continuous at $x$ $=0$

Option A:
$\mathbf{P} \rightarrow \mathbf{2 ; Q} \rightarrow \mathbf{3 ; R} \rightarrow \mathbf{1 ; S \rightarrow 4}$
Option B:
$\mathbf{P} \rightarrow 4 ; Q \rightarrow 1 ; R \rightarrow 2 ; S \rightarrow 3$
Option C:
$\mathrm{P} \rightarrow 4$ ; Q $\rightarrow 2 ; \mathrm{R} \rightarrow 1 ; \mathrm{S} \rightarrow 3$
Option D:
$\mathrm{P} \rightarrow 2 ; \mathrm{Q} \rightarrow \mathbf{1} ; \mathrm{R} \rightarrow 4 ; \mathrm{S} \rightarrow \mathbf{3}$
Correct

Answer: D

Step-by-step solution

For $f_1(x)=\sin\left(\sqrt{1-e^{-x^2}}\right)$ , $f_1(0)=0$ . As $x\to0$ , $1-e^{-x^2}\sim x^2$ , so $\sqrt{1-e^{-x^2}}\sim |x|$ .

Hence $\lim_{x\to0}f_1(x)=0$ , continuity holds. Right derivative: $\lim_{h\to0^+}\frac{\sin\left(\sqrt{1-e^{-h^2}}\right)}{h}=1$ .

Left derivative: $\lim_{h\to0^-}\frac{\sin\left(\sqrt{1-e^{-h^2}}\right)}{h}=-1$ .

So $f_1$ is continuous but not differentiable at 0 (P→2). 2. For $f_2(x)=\frac{|\sin x|}{\tan^{-1}x}$ at $x=0$ , $f_2(0)=1$ .

RHL: $\lim_{x\to0^+}\frac{|\sin x|}{\tan^{-1}x}=1$ .

LHL: $\lim_{x\to0^-}\frac{|\sin x|}{\tan^{-1}x}=-1$ .

Limit does not exist, so $f_2$ is not continuous at 0 (Q→1). 3. For $f_3(x)=\left[\sin(\ln(x+2))\right]$ , at $x=0$ , $\sin(\ln2)\approx0.639$ , so $f_3(0)=0$ .

In a neighbourhood of 0, $\sin(\ln(x+2))$ lies strictly between 0 and 1, so $f_3(x)=0$ for all $x$ near 0.

Thus $f_3$ is constant near 0, hence differentiable with derivative 0, which is continuous (R→4). 4. For $f_4(x)=$$\begin{cases} x^2\sin(1/x) & x eq0\\ 0 & x=0\end{cases}$$$,$ |x^2\sin(1/x)|\le x^2\to0$, so continuity holds.

Derivative at 0: $\lim_{h\to0}\frac{h^2\sin(1/h)-0}{h}=0$ . For $x eq0$ , $f_4'(x)=2x\sin(1/x)-\cos(1/x)$ .

The limit $\lim_{x\to0}f_4'(x)$ does not exist due to oscillation of $-\cos(1/x)$ .

Therefore $f_4$ is differentiable at 0 but its derivative is not continuous at 0 (S→3). Thus the correct matching is P→2, Q→1, R→4, S→3, which corresponds to option D.

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2018
Paper: Paper 2
Subject: Mathematics
Chapter: Limits, Continuity and Differentiability
Topic: Continuity at a point & in an interval

Let f:(0, pi) arrow R be a twice differentiable function such that lim t arrow x f(x) sin t-f(t) sin x/t-x=sin ^2 x for all x in(0, pi) If…

Answer: D

Step-by-step solution

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