JEE Advance - Mathematics (2018 - Paper 2 Offline - No. 15)
Let $${E_1} = \left\{ {x \in R:x \ne 1\,and\,{x \over {x - 1}} > 0} \right\}$$ and
$${E_2} = \left\{ \matrix{ x \in {E_1}:{\sin ^{ - 1}}\left( {{{\log }_e}\left( {{x \over {x - 1}}} \right)} \right) \hfill \cr is\,a\,real\,number \hfill \cr} \right\}$$
(Here, the inverse trigonometric function $${\sin ^{ - 1}}$$ x assumes values in $$\left[ { - {\pi \over 2},{\pi \over 2}} \right]$$.).
Let f : E1 $$ \to $$ R be the function defined by f(x) = $${{{\log }_e}\left( {{x \over {x - 1}}} \right)}$$ and g : E2 $$ \to $$ R be the function defined by g(x) = $${\sin ^{ - 1}}\left( {{{\log }_e}\left( {{x \over {x - 1}}} \right)} \right)$$.
The correct option is :
$${E_2} = \left\{ \matrix{ x \in {E_1}:{\sin ^{ - 1}}\left( {{{\log }_e}\left( {{x \over {x - 1}}} \right)} \right) \hfill \cr is\,a\,real\,number \hfill \cr} \right\}$$
(Here, the inverse trigonometric function $${\sin ^{ - 1}}$$ x assumes values in $$\left[ { - {\pi \over 2},{\pi \over 2}} \right]$$.).
Let f : E1 $$ \to $$ R be the function defined by f(x) = $${{{\log }_e}\left( {{x \over {x - 1}}} \right)}$$ and g : E2 $$ \to $$ R be the function defined by g(x) = $${\sin ^{ - 1}}\left( {{{\log }_e}\left( {{x \over {x - 1}}} \right)} \right)$$.
| LIST-I | LIST-II |
|---|---|
| P. The range of $f$ is | 1. $\left( -\infty, \frac{1}{1-e} \right] \cup \left[ \frac{e}{e-1}, \infty \right)$ |
| Q. The range of $g$ contains | 2. $(0, 1)$ |
| R. The domain of $f$ contains | 3. $\left[ -\frac{1}{2}, \frac{1}{2} \right]$ |
| S. The domain of $g$ is | 4. $(-\infty, 0) \cup (0, \infty)$ |
| 5. $\left( -\infty, \frac{e}{e-1} \right)$ | |
| 6. $(-\infty, 0) \cup \left( \frac{1}{2}, \frac{e}{e-1} \right]$ |
P $$ \to $$ 4; Q $$ \to $$ 2; R $$ \to $$ 1 ; S $$ \to $$ 1
P $$ \to $$ 3; Q $$ \to $$ 3; R $$ \to $$ 6 ; S $$ \to $$ 5
P $$ \to $$ 4; Q $$ \to $$ 2; R $$ \to $$ 1 ; S $$ \to $$ 6
P $$ \to $$ 4; Q $$ \to $$ 3; R $$ \to $$ 6 ; S $$ \to $$ 5
Explanation
We have,
$${E_1} = \left\{ {x \in R:x \ne 1\,and\,{x \over {x - 1}} > 0} \right\}$$
$$ \therefore $$ $${E_1} = {x \over {x - 1}} > 0$$
$${E_1} = x \in ( - \infty ,0) \cup (1,\infty )$$
and $${E_2} = \left\{ \matrix{ x \in {E_1}:{\sin ^{ - 1}}\left( {{{\log }_e}\left( {{x \over {x - 1}}} \right)} \right) \hfill \cr is\,a\,real\,number \hfill \cr} \right\}$$
$${E_2} = - 1 \le {\log _e}{x \over {x - 1}} \le 1$$
$$ \Rightarrow {e^{ - 1}} \le {x \over {x - 1}} \le e$$
Now, $${x \over {x - 1}} \ge {e^{ - 1}} \Rightarrow {x \over {x - 1}} - {1 \over e} \ge 0$$
$$ \Rightarrow {{ex - x + 1} \over {e(x - 1)}} \ge 0 \Rightarrow {{x(e - 1) + 1} \over {(x - 1)e}} \ge 0$$
$$ \Rightarrow x \in \left( { - \infty ,\,{1 \over {1 - e}}} \right) \cup (1,\infty )$$
Also, $${x \over {x - 1}} \le e$$
$$ \Rightarrow {{(e - 1)x - e} \over {x - 1}} \ge 0$$
$$ \Rightarrow x \in ( - \infty ,\,1) \cup \left[ {{e \over {e - 1}},\infty } \right)$$
So, $${E_2} = \left( { - \infty ,\,{1 \over {1 - e}}} \right] \cup \left[ {{e \over {e - 1}},\,\infty } \right)$$
$$ \therefore $$ The domain of f and g are
$$\left( { - \infty ,\,{1 \over {1 - e}}} \right] \cup \left[ {{e \over {e - 1}},\,\infty } \right)$$
and Range of $${x \over {x - 1}}$$ is R+ $$-$$ {1}
$$ \Rightarrow $$ Range of f is R $$-$$ {0} or ($$-$$$$\infty $$, 0) $$ \cup $$ (0, $$\infty $$)
Range of g is $$\left[ { - {\pi \over 2},{\pi \over 2}} \right] - \{ 0\} $$ or $$\left[ { - {\pi \over 2},0} \right) \cup \left( {0,{\pi \over 2}} \right]$$
Now, P $$ \to $$ 4, Q $$ \to $$ 2, R $$ \to $$ 1, S $$ \to $$ 1
Hence, option (a) is correct answer.
$${E_1} = \left\{ {x \in R:x \ne 1\,and\,{x \over {x - 1}} > 0} \right\}$$
$$ \therefore $$ $${E_1} = {x \over {x - 1}} > 0$$
$${E_1} = x \in ( - \infty ,0) \cup (1,\infty )$$
and $${E_2} = \left\{ \matrix{ x \in {E_1}:{\sin ^{ - 1}}\left( {{{\log }_e}\left( {{x \over {x - 1}}} \right)} \right) \hfill \cr is\,a\,real\,number \hfill \cr} \right\}$$
$${E_2} = - 1 \le {\log _e}{x \over {x - 1}} \le 1$$
$$ \Rightarrow {e^{ - 1}} \le {x \over {x - 1}} \le e$$
Now, $${x \over {x - 1}} \ge {e^{ - 1}} \Rightarrow {x \over {x - 1}} - {1 \over e} \ge 0$$
$$ \Rightarrow {{ex - x + 1} \over {e(x - 1)}} \ge 0 \Rightarrow {{x(e - 1) + 1} \over {(x - 1)e}} \ge 0$$
$$ \Rightarrow x \in \left( { - \infty ,\,{1 \over {1 - e}}} \right) \cup (1,\infty )$$
Also, $${x \over {x - 1}} \le e$$
$$ \Rightarrow {{(e - 1)x - e} \over {x - 1}} \ge 0$$
$$ \Rightarrow x \in ( - \infty ,\,1) \cup \left[ {{e \over {e - 1}},\infty } \right)$$
So, $${E_2} = \left( { - \infty ,\,{1 \over {1 - e}}} \right] \cup \left[ {{e \over {e - 1}},\,\infty } \right)$$
$$ \therefore $$ The domain of f and g are
$$\left( { - \infty ,\,{1 \over {1 - e}}} \right] \cup \left[ {{e \over {e - 1}},\,\infty } \right)$$
and Range of $${x \over {x - 1}}$$ is R+ $$-$$ {1}
$$ \Rightarrow $$ Range of f is R $$-$$ {0} or ($$-$$$$\infty $$, 0) $$ \cup $$ (0, $$\infty $$)
Range of g is $$\left[ { - {\pi \over 2},{\pi \over 2}} \right] - \{ 0\} $$ or $$\left[ { - {\pi \over 2},0} \right) \cup \left( {0,{\pi \over 2}} \right]$$
Now, P $$ \to $$ 4, Q $$ \to $$ 2, R $$ \to $$ 1, S $$ \to $$ 1
Hence, option (a) is correct answer.
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