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JEE Advanced 2018 Paper 2

Section: Mathematics

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Question : 54 of 54

Marks: +1, -0

Let

f_1:\mathbb{R}\rightarrow\mathbb{R},f_2:\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\rightarrow\mathbb{R},f_3\left(-1,e^{\pi/2}-2\right)\rightarrow\mathbb{R}

f_4:\mathbb{R}\rightarrow\mathbb{R}

be functions defied by

f_1(x)=\sin\left(\sqrt{1-e^{-x^2}}\right)

f_2(x)=\begin{cases} \frac{|\sin x|}{\tan^{-1}x}, & \text{if } x\neq 0 \\ 1, & \text{if } x=0 \end{cases}

where the inverse trigonometric function

\tan^{-1}x

assumes values in

\left(-\frac{\pi}{2},\frac{\pi}{2}\right)

f_3(x)=\left[\sin\left(\log_e(x+2)\right)\right],

t\in\mathbb{R},[t]

denotes the greatest integer less than or equal to t,

f_4(x)=\begin{cases} x^2\sin\left(\frac{1}{x}\right), & \text{if } x\neq 0 \\ 0, & \text{if } x=0 \end{cases}

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 is derivative is NOT continuous at $x=0$
S. The function $f_4$ is	4. Differentiable at $x=0$ and its derivative is continuous at $x=0$

The correct options is:

	P	Q	R	S
A)	2	3	1	4
B)	4	1	2	3
C)	4	2	1	3
D)	2	1	4	3

[JEE Adv 2018 P2]

A
B
C
D

Solution:

(i)

f(x)=\sin\sqrt{1-e^{-x^2}}

f_1'(x)=\cos\sqrt{1-e^{-x^2}}\cdot\frac{1}{2\sqrt{1-e^{-x^2}}}\left(0-e^{-x^2}\cdot(-2x)\right)

at

x=0f_1'(x)

does not exist

So

P\rightarrow2

(ii)

f_2(x)=\begin{cases} \frac{|\sin x|}{\tan^{-1}x}, & x\neq 0 \\ 0, & x=0 \end{cases}

\lim\limits_{x\to 0^{+}} = \frac{\sin x}{x}\cdot\frac{x}{\tan^{-1}x}=1

\Rightarrow f_2(x)

does not continuous at

x=0

So

Q\rightarrow1

(iii)

f_3(x)=\left[\sin(\ln(x+2))\right]=0

1 < x+2 < e^{\pi/2}

\Rightarrow 0 < \ln(x+2) < \frac{\pi}{2}

\Rightarrow 0 < \sin(\ln(x+2)) < 1

\Rightarrow f_3(x)=0

So

R\rightarrow4

(iv)

f-4(x)=\begin{cases} x^2\sin\left(\frac{1}{x}\right), & x\neq 0 \\ 0, & x=0 \end{cases}

So

S\rightarrow3

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