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The value of $\alpha(\neq 0)$ for which the function $f(x)=1+\alpha x$ is the inverse of itself is
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Verified Answer
The correct answer is:
$-1$
Let $y=1+\alpha x$
$$
[f(x)=y]
$$
$$
\begin{array}{ll}
\Rightarrow & x=\frac{y-1}{\alpha} \\
\Rightarrow & f^{-1}(x)=\frac{x-1}{\alpha}
\end{array}
$$
It is given that, $f(x)=f^{-1}(x)$
$$
\begin{aligned}
&\Rightarrow \quad 1+\alpha x=\frac{x-1}{\alpha} \\
&\Rightarrow \quad \alpha+\alpha^{2} x=x-1
\end{aligned}
$$
On compare the coefficients, we get
$$
\begin{array}{ll}
& \alpha^{2}=1 \text { and } \alpha=-1 \\
\Rightarrow \quad & \alpha=-1
\end{array}
$$
$$
[f(x)=y]
$$
$$
\begin{array}{ll}
\Rightarrow & x=\frac{y-1}{\alpha} \\
\Rightarrow & f^{-1}(x)=\frac{x-1}{\alpha}
\end{array}
$$
It is given that, $f(x)=f^{-1}(x)$
$$
\begin{aligned}
&\Rightarrow \quad 1+\alpha x=\frac{x-1}{\alpha} \\
&\Rightarrow \quad \alpha+\alpha^{2} x=x-1
\end{aligned}
$$
On compare the coefficients, we get
$$
\begin{array}{ll}
& \alpha^{2}=1 \text { and } \alpha=-1 \\
\Rightarrow \quad & \alpha=-1
\end{array}
$$
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