JEE Advance - Mathematics (2018 - Paper 2 Offline - No. 18)
Let $${f_1}:R \to R,\,{f_2}:\left( { - {\pi \over 2},{\pi \over 2}} \right) \to R,\,{f_3}:( - 1,{e^{\pi /2}} - 2) \to R$$ and $${f_4}:R \to R$$ be functions defined by
(i) $${f_1}(x) = \sin (\sqrt {1 - {e^{ - {x^2}}}} )$$,
(ii) $${f_2}(x) = \left\{ \matrix{ {{|\sin x|} \over {\tan { - ^1}x}}if\,x \ne 0,\,where \hfill \cr 1\,if\,x = 0 \hfill \cr} \right.$$
the inverse trigonometric function tan$$-$$1x assumes values in $$\left( { - {\pi \over 2},{\pi \over 2}} \right)$$,
(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) $${f_4}(x) = \left\{ \matrix{ {x^2}\sin \left( {{1 \over x}} \right)\,if\,x \ne 0 \hfill \cr 0\,if\,x = 0 \hfill \cr} \right.$$
(i) $${f_1}(x) = \sin (\sqrt {1 - {e^{ - {x^2}}}} )$$,
(ii) $${f_2}(x) = \left\{ \matrix{ {{|\sin x|} \over {\tan { - ^1}x}}if\,x \ne 0,\,where \hfill \cr 1\,if\,x = 0 \hfill \cr} \right.$$
the inverse trigonometric function tan$$-$$1x assumes values in $$\left( { - {\pi \over 2},{\pi \over 2}} \right)$$,
(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) $${f_4}(x) = \left\{ \matrix{ {x^2}\sin \left( {{1 \over x}} \right)\,if\,x \ne 0 \hfill \cr 0\,if\,x = 0 \hfill \cr} \right.$$
| 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 $$ x = 0 $$ |
P $$ \to $$ 2 ; Q $$ \to $$ 3 ; R $$ \to $$ 1 ; S $$ \to $$ 4
P $$ \to $$ 4 ; Q $$ \to $$ 1 ; R $$ \to $$ 2 ; S $$ \to $$ 3
P $$ \to $$ 4 ; Q $$ \to $$ 2 ; R $$ \to $$ 1 ; S $$ \to $$ 3
P $$ \to $$ 2 ; Q $$ \to $$ 1 ; R $$ \to $$ 4 ; S $$ \to $$ 3
Explanation
(i) Given,
f1 : R $$ \to $$ R and f1(x) = sin$$(\sqrt {1 - {e^{ - {x^2}}}} )$$
$$ \therefore $$ f1(x) is continuous at x = 0
Now,
$${f_1}'(x) = \cos \sqrt {1 - {e^{ - {x^2}}}} .{1 \over {2\sqrt {1 - {e^{ - {x^2}}}} }}(2x{e^{ - {x^2}}})$$
At x = 0
f1'(x) does not exists.
$$ \therefore $$ f1(x) is not differential at x = 0
Hence, option (2) for P.
(ii) Given, $${f_2}(x) = \left\{ \matrix{ {{|\sin x|} \over {{{\tan }^{ - 1}}x'}}\,if\,x \ne 0 \hfill \cr 1,\,if\,x = 0 \hfill \cr} \right\}$$
$$ \Rightarrow {f_2}(x) = \left\{ \matrix{ {{ - \sin x} \over {{{\tan }^{ - 1}}x}}x < 0 \hfill \cr {{\sin x} \over {{{\tan }^{ - 1}}x}}x > 0 \hfill \cr 1\,x = 0 \hfill \cr} \right.$$
Clearly, f2(x) is not continuous at x = 0.
$$ \therefore $$ Option (1) for Q.
(iii) Given, f3(x) = [sin(loge(x + 2))], where [ ] is G.I.F.
and f3 : ($$-$$1, e$$\pi $$/2 $$-$$ 2) $$ \to $$ R
It is given,
$$ - 1 < x < {e^{\pi /2}} - 2$$
$$ \Rightarrow - 1 + 2 < x + 2 < {e^{\pi /2}} - 2 + 2$$
$$ \Rightarrow 1 < x + 2 < {e^{\pi /2}}$$
$$ \Rightarrow {\log _e}1 < {\log _e}(x + 2) < {\log _e}{e^{\pi /2}}$$
$$ \Rightarrow 0 < {\log _e}(x + 2) < {\pi \over 2}$$
$$ \Rightarrow \sin 0 < \sin {\log _e}(x + 2) < \sin {\pi \over 2}$$
$$ \Rightarrow 0 < \sin {\log _e}(x + 2) < 1$$
$$ \therefore $$ $$[\sin {\log _e}(x + 2)] = 0$$
$$ \therefore $$ $${f_3}(x) = 0,\,f{'_3}(x) = {f_3}''(x) = 0$$
It is differentiable and continuous at x = 0.
$$ \therefore $$ Option (4) for R
(iv) Given, $${f_4}(x) = \left\{ \matrix{ {x^2}\sin \left( {{1 \over x}} \right),\,if\,x \ne 0 \hfill \cr 0,\,if\,x = 0 \hfill \cr} \right.$$
Now, $$\mathop {\lim }\limits_{x \to 0} {f_4}(x) = \mathop {\lim }\limits_{x \to 0} {x^2}\sin \left( {{1 \over x}} \right) = 0$$
$${f_4}'(x) = 2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right)$$
For $$x = 0,\,{f_4}'(x) = \mathop {\lim }\limits_{h \to 0} {{f(0 + h) - f(0)} \over h}$$
$$ \Rightarrow {f_4}'(x) = \mathop {\lim }\limits_{h \to 0} {{{h^2}\sin \left( {{1 \over h}} \right) - 0} \over h}$$
$$ \Rightarrow {f_4}'(x) = \mathop {\lim }\limits_{h \to 0} h\sin \left( {{1 \over h}} \right) = 0$$
Thus,
$${f_4}'(x) = \left\{ \matrix{ 2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right),\,x \ne 0 \hfill \cr 0,\,x = 0 \hfill \cr} \right.$$
Again, $$\mathop {\lim }\limits_{x \to 0} $$
$$f'(x) = \mathop {\lim }\limits_{x \to 0} \left( {2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right)} \right)$$
does not exists.
Since, $$\mathop {\lim }\limits_{x \to 0} $$ cos$${\left( {{1 \over x}} \right)}$$ does not exists.
Hence, f'(x) is not continuous at x = 0.
$$ \therefore $$ Option (3) for S.
f1 : R $$ \to $$ R and f1(x) = sin$$(\sqrt {1 - {e^{ - {x^2}}}} )$$
$$ \therefore $$ f1(x) is continuous at x = 0
Now,
$${f_1}'(x) = \cos \sqrt {1 - {e^{ - {x^2}}}} .{1 \over {2\sqrt {1 - {e^{ - {x^2}}}} }}(2x{e^{ - {x^2}}})$$
At x = 0
f1'(x) does not exists.
$$ \therefore $$ f1(x) is not differential at x = 0
Hence, option (2) for P.
(ii) Given, $${f_2}(x) = \left\{ \matrix{ {{|\sin x|} \over {{{\tan }^{ - 1}}x'}}\,if\,x \ne 0 \hfill \cr 1,\,if\,x = 0 \hfill \cr} \right\}$$
$$ \Rightarrow {f_2}(x) = \left\{ \matrix{ {{ - \sin x} \over {{{\tan }^{ - 1}}x}}x < 0 \hfill \cr {{\sin x} \over {{{\tan }^{ - 1}}x}}x > 0 \hfill \cr 1\,x = 0 \hfill \cr} \right.$$
Clearly, f2(x) is not continuous at x = 0.
$$ \therefore $$ Option (1) for Q.
(iii) Given, f3(x) = [sin(loge(x + 2))], where [ ] is G.I.F.
and f3 : ($$-$$1, e$$\pi $$/2 $$-$$ 2) $$ \to $$ R
It is given,
$$ - 1 < x < {e^{\pi /2}} - 2$$
$$ \Rightarrow - 1 + 2 < x + 2 < {e^{\pi /2}} - 2 + 2$$
$$ \Rightarrow 1 < x + 2 < {e^{\pi /2}}$$
$$ \Rightarrow {\log _e}1 < {\log _e}(x + 2) < {\log _e}{e^{\pi /2}}$$
$$ \Rightarrow 0 < {\log _e}(x + 2) < {\pi \over 2}$$
$$ \Rightarrow \sin 0 < \sin {\log _e}(x + 2) < \sin {\pi \over 2}$$
$$ \Rightarrow 0 < \sin {\log _e}(x + 2) < 1$$
$$ \therefore $$ $$[\sin {\log _e}(x + 2)] = 0$$
$$ \therefore $$ $${f_3}(x) = 0,\,f{'_3}(x) = {f_3}''(x) = 0$$
It is differentiable and continuous at x = 0.
$$ \therefore $$ Option (4) for R
(iv) Given, $${f_4}(x) = \left\{ \matrix{ {x^2}\sin \left( {{1 \over x}} \right),\,if\,x \ne 0 \hfill \cr 0,\,if\,x = 0 \hfill \cr} \right.$$
Now, $$\mathop {\lim }\limits_{x \to 0} {f_4}(x) = \mathop {\lim }\limits_{x \to 0} {x^2}\sin \left( {{1 \over x}} \right) = 0$$
$${f_4}'(x) = 2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right)$$
For $$x = 0,\,{f_4}'(x) = \mathop {\lim }\limits_{h \to 0} {{f(0 + h) - f(0)} \over h}$$
$$ \Rightarrow {f_4}'(x) = \mathop {\lim }\limits_{h \to 0} {{{h^2}\sin \left( {{1 \over h}} \right) - 0} \over h}$$
$$ \Rightarrow {f_4}'(x) = \mathop {\lim }\limits_{h \to 0} h\sin \left( {{1 \over h}} \right) = 0$$
Thus,
$${f_4}'(x) = \left\{ \matrix{ 2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right),\,x \ne 0 \hfill \cr 0,\,x = 0 \hfill \cr} \right.$$
Again, $$\mathop {\lim }\limits_{x \to 0} $$
$$f'(x) = \mathop {\lim }\limits_{x \to 0} \left( {2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right)} \right)$$
does not exists.
Since, $$\mathop {\lim }\limits_{x \to 0} $$ cos$${\left( {{1 \over x}} \right)}$$ does not exists.
Hence, f'(x) is not continuous at x = 0.
$$ \therefore $$ Option (3) for S.
Comments (0)
