- Home
- Standard 12
- Mathematics
If $f(x) = \cos (\log x)$, then $f(x)f(y) - \frac{1}{2}[f(x/y) + f(xy)] = $
$ - 1$
$\frac{1}{2}$
$ - 2$
None of these
Solution
(d) Given $f(x) = \cos \,(\log x)\,\, \Rightarrow \,f(y) = \cos \,(\log y)$
Then $f(x).f(y) – \frac{1}{2}\left[ {f\left( {\frac{x}{y}} \right) + f(xy)} \right]$
$ = \cos \,(\log x)\cos \,(\log y) – \frac{1}{2}\left[ {\cos \left( {\log \frac{x}{y}} \right) + \cos \,(\log xy)} \right]$
$ = \cos \,(\log x)\,\cos \,(\log y) – \frac{1}{2}\,[2\cos \,(\log x)\cos \,(\log y)]= 0.$
Similar Questions
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:
