Solve the Following Questions


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 :

  1. No such $\alpha$ exists

  2. 5

  3. 8

  4. 6

Correct Option: , 2


$f(x)=\frac{5 x+3}{6 x-\alpha}=y$..(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 $f(x)=x$


From eq $^{\mathrm{n}}$ (i) \& (ii)

Clearly $(\alpha=5)$

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