If $R \subset A \times B$ and $S \subset B \times C$ be two relations,then $(S \circ R)^{-1} = $
- A$S^{-1} \circ R^{-1}$
- B$R^{-1} \circ S^{-1}$
- C$S \circ R$
- D$R \circ S$
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DifficultAP EAMCET 2021
View SolutionMatch the items of List-$I$ with those of the items of List-$II$:
List-$I$ List-$II$ $A$. Range of $\sec ^{-1}\left[1+\cos ^2 x\right]$, where $[.]$ denotes the greatest integer function $I$. Odd function $B$. Domain of $f(x)$ where $f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}$ $II$. $\left\{0, \frac{1}{2}\right\}$ $C$. $f(x+y)=f(x)+f(y) ; f(1)=5$ $III$. $\left\{\sec ^{-1} 5, \sec ^{-1} 4\right\}$ $D$. $\sin ^{-1} x-\cos ^{-1} x+\sin ^{-1}(1-x)=0 \Rightarrow x \in$ $IV$. $R$ $V$. $\left\{\sec ^{-1} 1, \sec ^{-1} 2\right\}$
| List-$I$ | List-$II$ |
| $A$. Range of $\sec ^{-1}\left[1+\cos ^2 x\right]$, where $[.]$ denotes the greatest integer function | $I$. Odd function |
| $B$. Domain of $f(x)$ where $f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}$ | $II$. $\left\{0, \frac{1}{2}\right\}$ |
| $C$. $f(x+y)=f(x)+f(y) ; f(1)=5$ | $III$. $\left\{\sec ^{-1} 5, \sec ^{-1} 4\right\}$ |
| $D$. $\sin ^{-1} x-\cos ^{-1} x+\sin ^{-1}(1-x)=0 \Rightarrow x \in$ | $IV$. $R$ |
| $V$. $\left\{\sec ^{-1} 1, \sec ^{-1} 2\right\}$ |
DifficultTS EAMCET 2020
View SolutionIf $e^{f(x)}=\frac{10+x}{10-x}, x \in(-10,10)$ and $f(x)=k f\left(\frac{200 x}{100+x^2}\right)$, then $k$ is equal to
DifficultTS EAMCET 2003
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