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Q.
Let $f: R \rightarrow\left(0, \frac{\pi}{6}\right]$ be defined as $f(x)=\sin ^{-1}\left(\frac{4}{4 x^2-12 x+17}\right)$ then $f(x)$ is
Inverse Trigonometric Functions
Solution:
We have $f(x)=\sin ^{-1}\left(\frac{4}{4 x^2-12 x+17}\right)$
$=\sin ^{-1}\left(\frac{1}{x^2-3 x+\frac{17}{4}}\right)=\sin ^{-1}\left(\frac{1}{\left(x-\frac{3}{2}\right)^2+\frac{17}{4}-\frac{9}{4}}\right)=\sin ^{-1}\left(\frac{1}{\left(x-\frac{3}{2}\right)^2+2}\right)$
Hence $f(x) \in\left(0, \frac{\pi}{6}\right]$, so co-domain $=$ Range $\left(\right.$ As $\left.x^2-3 x+\frac{17}{4}=\left(x-\frac{3}{2}\right)^2+2 \in[2, \infty)\right)$
Also $y=4 x^2-12 x+17$ is many one function.
Hence $f(x)$ is surjective but not injective.