Inverse Circular Functions Ques 5
- $E _1=x \in R: x \neq 1$ and $\frac{x}{x-1}> 0 $ and $E _2=x \in E _1: \sin ^{-1} \log _e \frac{x}{x-1}$ is a real number
(Here, the inverse trigonometric function $\sin ^{-1} x$ assumes values in $-\frac{\pi}{2}, \frac{\pi}{2}$ ). Let $f: E _1 \rightarrow R$ be the function defined by $f(x)=\log _e \frac{x}{x-1}$ and $g: E _2 \rightarrow R$ be the function defined by $g(x)=\sin ^{-1} \log _e \frac{x}{x-1}$.
(2018 Adv.)
| List I | List II | ||
|---|---|---|---|
| P. The range of $f$ is | 1. | $-\infty, \frac{1}{1-e} \cup \frac{e}{e-1}, \infty$ | |
| Q. | The range of $g$ contains | 2. | $(0,1]$ |
| The domain of $f$ contains |
3. | $-\frac{1}{2}, \frac{1}{2}$ | |
| S. The domain of $g$ is | 4. | $(-\infty, 0) \cup(0, \infty)$ | |
| 5. | $-\infty, \frac{e}{e-1}$ |
The correct answer is
(a) $P \rightarrow 4 ; Q \rightarrow 2 ; R \rightarrow 1 ; S \rightarrow 1$
(b) $P \rightarrow 3 ; Q \rightarrow 3 ; R \rightarrow 6 ; S \rightarrow 5$
(c) $P \rightarrow 4 ; Q \rightarrow 2 ; R \rightarrow 1 ; S \rightarrow 6$
(d) $P \rightarrow 4 ; Q \rightarrow 3 ; R \rightarrow 6 ; S \rightarrow 5$
Numerical Value Based
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Answer:
Correct Answer: 5.(a)
Solution:
Formula:
Domains and Ranges of Inverse Trigonometric:
- We have,
$ \begin{alignedat} & E _1= \{ x \in R: x \neq 1 \text { and } \frac{x}{x-1}>0 \} \\ \therefore & E _1=\frac{x}{x-1}>0 \\ & E _1=x \in(-\infty, 0) \cup(1, \infty) \end{aligned} $
and
$E _2= \{ x \in E _1: \sin ^{-1} (\log _e (\frac{x}{x-1}))$ is a real number $\}$
$ E _2=-1 \leq \log _e \frac{x}{x-1} \leq 1 \Rightarrow $
$ e^{-1} \leq \frac{x}{x-1} \leq e $
Now, $\quad \frac{x}{x-1} \geq e^{-1} \Rightarrow \frac{x}{x-1}-\frac{1}{e} \geq 0$
$\Rightarrow \frac{e x-x+1}{e(x-1)} \geq 0 \Rightarrow \frac{x(e-1)+1}{(x-1) e} \geq 0$
$\Rightarrow x \in ( -\infty, \frac{1}{1-e}] \cup(1, \infty)$
Also,
$ \Rightarrow \quad \frac{(e-1) x-e}{x-1} \geq 0 $
$\Rightarrow x \in(-\infty, 1) \cup \left(\frac{e}{e-1}, \infty\right)$
So, $\quad E _2=(-\infty, \frac{1}{1-e}] \cup [\frac{e}{e-1}, \infty )$
$\therefore$ The domains of $f$ and $g$ are
$ (-\infty, \frac{1}{1-e}] \cup [\frac{e}{e-1}, \infty) $
and Range of $\frac{x}{x}{x-1}$ is $R^{+}-{1}$
$\Rightarrow$ Range of $f$ is $R-{0}$ or $(-\infty, 0) \cup(0, \infty)$
Range of $g$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]-{0}$ or $[-\frac{\pi}{2}, 0 ) \cup ( 0, \frac{\pi}{2}]$
Now, $P \rightarrow 4, Q \rightarrow 2, R \rightarrow 1, S \rightarrow 1$
BETA
