MCQ
If $\sin \left( {{{\sin }^{ - 1}}\frac{1}{5} + {{\cos }^{ - 1}}x} \right) = 1$, then $x$ is equal to
- A$1$
- B$0$
- C$\frac{4}{5}$
- ✓$\frac{1}{5}$
$\therefore \,\,{\sin ^{ - 1}}\frac{1}{5} = \frac{\pi }{2} - {\cos ^{ - 1}}x = {\sin ^{ - 1}}x$
$\therefore \,\,\,x = \frac{1}{5}$.
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(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: