JEE MAIN - Mathematics (2021 - 20th July Evening Shift - No. 5)
Let $$f:R - \left\{ {{\alpha \over 6}} \right\} \to R$$ be defined by $$f(x) = {{5x + 3} \over {6x - \alpha }}$$. Then the value of $$\alpha$$ for which (fof)(x) = x, for all $$x \in R - \left\{ {{\alpha \over 6}} \right\}$$, is :
No such $$\alpha$$ exists
5
8
6
Explanation
$$f(x) = {{5x + 3} \over {6x - \alpha }} = y$$ ..... (i)
$$5x + 3 = 6xy - \alpha y$$
$$x(6y - 5) = \alpha y + 3$$
$$x = {{\alpha y + 3} \over {6y - 5}}$$
$${f^{ - 1}}(x) = {{\alpha x + 3} \over {6x - 5}}$$ ...... (ii)
fo $$f(x) = x$$
$$f(x) = {f^{ - 1}}(x)$$
From eqn (i) & (ii)
Clearly $$(\alpha = 5)$$
$$5x + 3 = 6xy - \alpha y$$
$$x(6y - 5) = \alpha y + 3$$
$$x = {{\alpha y + 3} \over {6y - 5}}$$
$${f^{ - 1}}(x) = {{\alpha x + 3} \over {6x - 5}}$$ ...... (ii)
fo $$f(x) = x$$
$$f(x) = {f^{ - 1}}(x)$$
From eqn (i) & (ii)
Clearly $$(\alpha = 5)$$
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