Let f : R - α 6 → R be defined by f x = 5 x + 3 6 x - α . Then the value of α for which f o f x = x , for all x ∈ R - α 6 , is

We have, f x = y = 5 x + 3 6 x - α … i ⇒ 5 x + 3 = 6 x y - α y ⇒ x ( 6 y - 5 ) = α y + 3 ⇒ x = α y + 3 6 y - 5 Hence, f - 1 x = α x + 3 6 x - 5 … ii Now, f o f x = x f ( x ) = f - 1 ( x ) From equation i and i i , we have 5 x + 3 6 x - α = α x + 3 6 x - 5 Clearly, α = 5

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MathematicsFunctionsCompositeEasy2 minPYQ_2021

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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 $(f o f) (x) = x$ , for all $x \in R - \{\frac{α}{6}\},$ is

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Answer:

Solution:

We have,

$f (x) = y = (\frac{5 x + 3}{6 x - α}) \dots (i)$

$\Rightarrow 5 x + 3 = 6 x y - α y$

$\Rightarrow x (6 y - 5) = α y + 3$

$\Rightarrow x = \frac{α y + 3}{6 y - 5}$

Hence,

$f^{- 1} (x) = (\frac{α x + 3}{6 x - 5}) \dots (ii)$

Now,

$(f o f) (x) = x$

$f (x) = f^{- 1} (x)$

From equation $(i)$ and $(i i)$ , we have

$(\frac{5 x + 3}{6 x - α}) = (\frac{α x + 3}{6 x - 5})$

Clearly, $α = 5$

Stream:JEESubject:MathematicsTopic:FunctionsSubtopic:Composite

⏱ 2mℹ️ Source: PYQ_2021

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