Let $f: R-\left\{\frac{\alpha}{6}\right\} \rightarrow R$ be defined by $f(x)=\frac{5 x+3}{6 x-\alpha}$. Then the value of $\alpha$ for which (fof) $(x)=x$, for all $x \in R-\left\{\frac{\alpha}{6}\right\}$, is :

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Sets Relations and Functions

Let

f : R - {\frac{α}{6}} \to R

be defined by

f (x) = \frac{5 x + 3}{6 x - α}

. Then the value of

α

for which (fof)

(x) = x

, for all

x \in R - {\frac{α}{6}}

, is :

Answer: (D)

Composing the function with itself,

f (f (x)) = \frac{5 \cdot \frac{5 x + 3}{6 x - α} + 3}{6 \cdot \frac{5 x + 3}{6 x - α} - α} = \frac{5 ( 5 x + 3 ) + 3 ( 6 x - α )}{6 ( 5 x + 3 ) - α ( 6 x - α )} = \frac{43 x + 15 - 3 α}{( 30 - 6 α ) x + 18 + α ^{2}}

Requiring this to equal

x

for every admissible

x

gives

(30 - 6 α) x^{2} + (18 + α^{2}) x = 43 x + 15 - 3 α

The coefficient of

x^{2}

must vanish, so

α = 5

. Then

18 + α^{2} = 43

and

15 - 3 α = 0

, so the identity holds, and the required value is

α = 5

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Single correct option