Question:
Let $f: \mathbf{R}-\left\{\frac{\alpha}{6}\right\} \rightarrow \mathbf{R}$ be defined by $f(x)=\frac{5 x+3}{6 x-\alpha}$
Then the value of $\alpha$ for which $(f o f)(x)=x$, for all $x \in \mathbf{R}-\left\{\frac{\alpha}{6}\right\}$, is :
Correct Option: , 2
Solution:
$f(x)=\frac{5 x+3}{6 x-\alpha}=y$..(1)
$5 x+3=6 x y-\alpha y$
$x(6 y-5)=\alpha y+3$
$x=\frac{\alpha y+3}{6 y-5}$
$\mathrm{f}^{-1}(\mathrm{x})=\frac{\alpha \mathrm{x}+3}{6 \mathrm{x}-5}$...(2)
fo $f(x)=x$
$f(x)=f^{-1}(x)$
From eq $^{\mathrm{n}}$ (i) \& (ii)
Clearly $(\alpha=5)$
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