Let ⨍x=αxx+1,x≠-1 then for what value of α is ⨍⨍x=x?

⨍x=αxx+1,x≠1⨍⨍x=x (or) ffx=αfxfx+1=ααxx+1αxx+1+1=x or α2xα+1x+1=x⇒α+1x2+x+-α2x=0....(1) ⇒α+1=0 and 1-α2=0 As it is true ∀x≠-1, equ-(1) is an identity∴α=-1

Let ⨍x=αxx+1,x≠-1 then for what value of α is ⨍⨍x=x?

⨍x=αxx+1,x≠1⨍⨍x=x (or) ffx=αfxfx+1=ααxx+1αxx+1+1=x or α2xα+1x+1=x⇒α+1x2+x+-α2x=0....(1) ⇒α+1=0 and 1-α2=0 As it is true ∀x≠-1, equ-(1) is an identity∴α=-1

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Let $⨍ (x) = \frac{αx}{x + 1},x≠-1$ then for what value of $α$ is $⨍ (⨍ (x)) =x?$

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-1

$- \sqrt{2}$

$\sqrt{2}$

answer is D.

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Detailed Solution

$⨍ (x) = \frac{αx}{x + 1},x≠1⨍ (⨍ (x)) =x (or) f (f (x)) = \frac{α f (x)}{f (x) + 1} = \frac{α (\frac{αx}{x + 1})}{\frac{αx}{x + 1} +1} =x or \frac{α^{2} x}{(α + 1) x+1} =x$ $\Rightarrow (α + 1) (x^{2}) + x + (- α^{2}) x = 0 . . . . (1) \Rightarrow α + 1 = 0 and 1 - α^{2} = 0 [As it is true \forall x \neq - 1, equ - (1) is an identity]$ $∴ α = - 1$