Let \( f: R \rightarrow[-1,1] \) is defined by \( f(x)=\sin (2 x+1) \). If domain is restricted ....
0Let \( f: R \rightarrow[-1,1] \) is defined by \( f(x)=\sin (2 x+1) \).
\( \mathrm{P} \)
If domain is restricted to:
\begin{tabular}{|l|l|l|l|}
\hline \multicolumn{1}{|c|}{ Column-I } & & \begin{tabular}{l}
Column- \\
II
\end{tabular} \\
\hline A & {\( [-3 \pi / 4-1 / 2,-\pi / 2-1 / 2] \)} & P. & \begin{tabular}{l}
\( f \) is one- \\
one and \\
onto
\end{tabular} \\
\hline B. & {\( [-3 \pi / 4-1 / 2,-1 / 2] \)} & Q. & \begin{tabular}{l}
\( f \) is one- \\
one but \\
not onto
\end{tabular} \\
\hline C. & {\( [\pi / 4-1 / 2,3 \pi / 4-1 / 2] \)} & R. & \begin{tabular}{l}
\( f \) is onto \\
but not \\
one-one
\end{tabular} \\
\hline D & {\( \left[-\frac{3 \pi}{4}-\frac{1}{2}, \frac{\pi}{2}-\frac{1}{2}\right] \cup \)} & S. & \begin{tabular}{l}
fis \\
neither \\
one-one \\
nor onto
\end{tabular} \\
\hline
\end{tabular}
A
B
C D
(1) \( \begin{array}{lllll}\mathrm{Q} & \mathrm{R} & \mathrm{P} & \mathrm{S}\end{array} \)
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