Let f: R−{α6}→R be definced by f(x)= 5x+36x−α. Then the value of α for which (fof)(x)=x, for all x∈R−{α6}, is:

Given f(x)=5x+36x−αlet f−1(x)=y⇒ x=f(y)⇒ x= 5y+36y−α⇒ 6xy−αx=5y+3⇒ 6xy−5y=αx+3⇒ (6x−5)y=αx+3⇒ y=αx+36x−5∴f−1(x)=αx+36x−5and fof(x)=x⇒ f(x)=f−1(x)⇒ 5x+36x−α=αx+36x−5⇒ (5x+3)(6x−5)=(αx+3)(6x−α)⇒ 30x2−25x+18x−15=6αx2−α2x+18x−3α⇒ 30x2−7x−15=6αx2+(18−α2)x−3α∴ 6α=30 ⇒ α=5

Let f: R−{α6}→R be definced by f(x)= 5x+36x−α. Then the value of α for which (fof)(x)=x, for all x∈R−{α6}, is:

Given f(x)=5x+36x−αlet f−1(x)=y⇒ x=f(y)⇒ x= 5y+36y−α⇒ 6xy−αx=5y+3⇒ 6xy−5y=αx+3⇒ (6x−5)y=αx+3⇒ y=αx+36x−5∴f−1(x)=αx+36x−5and fof(x)=x⇒ f(x)=f−1(x)⇒ 5x+36x−α=αx+36x−5⇒ (5x+3)(6x−5)=(αx+3)(6x−α)⇒ 30x2−25x+18x−15=6αx2−α2x+18x−3α⇒ 30x2−7x−15=6αx2+(18−α2)x−3α∴ 6α=30 ⇒ α=5

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$L e t f : R - {\frac{α}{6}} \to R b e d e f i n c e d b y f (x) = \frac{5 x + 3}{6 x - α} . T h e n t h e v a l u e o f α f o r w h i c h (f o f) (x) = x, f o r a l l x \in R - {\frac{α}{6}}, i s :$

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Detailed Solution

$\begin{array}{l} G i v e n f (x) = \frac{5 x + 3}{6 x - α} \\ l e t f^{- 1} (x) = y \Rightarrow x = f (y) \\ \Rightarrow x = \frac{5 y + 3}{6 y - α} \\ \Rightarrow 6 x y - α x = 5 y + 3 \\ \Rightarrow 6 x y - 5 y = α x + 3 \\ \Rightarrow (6 x - 5) y = α x + 3 \\ \Rightarrow y = \frac{α x + 3}{6 x - 5} \\ ∴ f^{- 1} (x) = \frac{α x + 3}{6 x - 5} \end{array}$

$\begin{array}{l} a n d f o f (x) = x \\ \Rightarrow f (x) = f^{- 1} (x) \\ \Rightarrow \frac{5 x + 3}{6 x - α} = \frac{α x + 3}{6 x - 5} \\ \Rightarrow (5 x + 3) (6 x - 5) = (α x + 3) (6 x - α) \\ \Rightarrow 30 x^{2} - 25 x + 18 x - 15 = 6 α x^{2} - α^{2} x + 18 x - 3 α \\ \Rightarrow 30 x^{2} - 7 x - 15 = 6 α x^{2} + (18 - α^{2}) x - 3 α \\ ∴ 6 α = 30 \Rightarrow α = 5 \end{array}$