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Let $\quad E_1=\left\{x \in R : x \neq 1\right.$ and $\left.\frac{x}{x-1}>0\right\}$ and $\quad E_2=\left\{x \in E_1: \sin ^{-1}\left(\log _e\left(\frac{x}{x-1}\right)\right)\right.$ is a real number $\}$.
(Here, the inverse trigonometric function $\sin ^{-1} x$ assumes values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ )
Let $f : E _1 \rightarrow R$ be the function defined by $f(x)=\log _c\left(\frac{x}{x-1}\right)$ and $g: E_2 \rightarrow R$ be the function defined by $g(x)=\sin ^{-1}\left(\log _e\left(\frac{x}{x-1}\right)\right)$
| $LIST I$ | $LIST II$ |
| $P$ 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 | $4$ $(-\infty, 0) \cup(0, \infty)$ |
| $5$ $\left(-\infty, \frac{ e }{ e -1}\right]$ | |
| $6$ $(-\infty, 0) \cup\left(\frac{1}{2}, \frac{ e }{ e -1}\right]$ |
The correct option is:
$P \rightarrow 4 ; Q \rightarrow 2 ; R \rightarrow 1 ; S \rightarrow 1$
$P \rightarrow 3 ; Q \rightarrow 3 ; R \rightarrow 6 ; S \rightarrow 5$
$P \rightarrow 4 ; Q \rightarrow 2 ; R \rightarrow 1 ; S \rightarrow 6$
$P \rightarrow 4 ; Q \rightarrow 3 ; R \rightarrow 6 ; S \rightarrow 5$
Solution
$\begin{array}{l} P \rightarrow 4, Q \rightarrow 2, R \rightarrow 1, S \rightarrow 1 \\ E _1: \quad \frac{ x }{ x -1}>0 \quad \Rightarrow x \in(-\infty, 0) \cup(1, \infty) \\ E _2: \quad \frac{ x }{ x -1}>0 \text { and } \quad-1 \leq \log \left(\frac{ x }{ x -1}\right) \leq 1\end{array}$
$x \in(-\infty, 0) \cup(1, \infty) \frac{1}{e} \leq \frac{x}{x-1} \leq e $
$0 \leq \frac{x}{x-1}-\frac{1}{e} \frac{x}{x-1}-e \leq 0$
$0 \leq \frac{(e-1) x+1}{e(x-1)} \frac{x(e-1)-e}{x-1} \geq 0$
$x \in\left(-\infty, \frac{1}{1-e}\right] \cup(1, \infty) x \in(-\infty, 1) \cup\left[\frac{e}{e-1}, \infty\right)$
Intersection $x \in\left(-\infty, \frac{1}{1- e }\right] \cup\ [\frac{ e }{ e -1},$
So, domain of $g$ is $x \in\left(-\infty, \frac{1}{1- e }\right] \cup\left[\frac{ e }{ e -1}, \infty\right)$
Range of $\frac{x}{x-1}$ is $R ^{+}-\{1\}$
Range of $f$ is $R-\{0\}$ or $(-\infty, 0) \cup(0, \infty)$
Range of $g$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] /\{0\}$
