The value of the parameter -for which the function f x -1- x - -0 is the inverse of itself- isA- -2B- 1C- -1D- 2

The correct option is C 1y=1+α x ⇒ x =y 1/α f^ 1(x)=x 1/α=f(x)=1+α x ⇒x 1/α=1+α x ⇒ x 1=α +α ^2 x Equating the coefficient of x α ^2=1 amp; α = 1 α = 1 ...

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Standard XI

Mathematics

Definition of Functions

The value of ...

Question

The value of the parameter $α$ ,for which the function $f (x) = 1 + α x, α \neq 0$ is the inverse of itself, is

-2

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-1

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Solution

The correct option is B -1
$y = 1 + α x \Rightarrow x = \frac{y - 1}{α}$
$f^{- 1} (x) = \frac{x - 1}{α} = f (x) = 1 + α x \Rightarrow \frac{x - 1}{α} = 1 + α x \Rightarrow x - 1 = α + α^{2} x$
Equating the coefficient of x
$α^{2} = 1$ & $α = - 1$
$α = - 1$

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The value of the parameter α ,for which the function f(x)=1+αx,α≠0 is the inverse of itself, is

The correct option is B -1y=1+αx⇒x=y−1α f−1(x)=x−1α=f(x)=1+αx⇒x−1α=1+αx⇒x−1=α+α2x Equating the coefficient of x α2=1 & α=−1 α=−1

The value of the parameter $α$ ,for which the function $f (x) = 1 + α x, α \neq 0$ is the inverse of itself, is

The correct option is B -1
$y = 1 + α x \Rightarrow x = \frac{y - 1}{α}$
$f^{- 1} (x) = \frac{x - 1}{α} = f (x) = 1 + α x \Rightarrow \frac{x - 1}{α} = 1 + α x \Rightarrow x - 1 = α + α^{2} x$
Equating the coefficient of x
$α^{2} = 1$ & $α = - 1$
$α = - 1$