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If a function $f:(-1,1) \rightarrow B(\subseteq R)$ is defined as $f(x)=x+x^2+x^3+\ldots \infty$, then in order to have the inverse function of $f, B$ is equal to
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Verified Answer
The correct answer is:
$\left(\frac{-1}{2}, \infty\right)$
Given, $f(x)=x+x^2+x^3+\ldots \infty$
Let $y=f(x)$
[sum of infinite GP series]
$\Rightarrow y=\frac{x}{1-x}=\frac{1}{1-x}-1$
Given, Domain of $f=(-1,1)$
$\begin{array}{ll}
\text { i.e. } & -1 < x < 1 \Rightarrow-1 < -x < 1 \\
\Rightarrow & 0 < 1-x < 2 \\
& \quad \frac{1}{2} < \frac{1}{1-x} < \infty \\
& \frac{1}{2}-1 < \frac{1}{1-x}-1 < \infty-1 \\
\Rightarrow & -\frac{1}{2} < y < \infty
\end{array}$
So, range $=y \in\left(\frac{-1}{2}, \infty\right)=B$ (according to question $)$.
Hence, $\rho([A: B])=2$
for this $-c+2 a-b=0$
$\Rightarrow \quad a=\frac{b+c}{2}$
Let $y=f(x)$
[sum of infinite GP series]
$\Rightarrow y=\frac{x}{1-x}=\frac{1}{1-x}-1$
Given, Domain of $f=(-1,1)$
$\begin{array}{ll}
\text { i.e. } & -1 < x < 1 \Rightarrow-1 < -x < 1 \\
\Rightarrow & 0 < 1-x < 2 \\
& \quad \frac{1}{2} < \frac{1}{1-x} < \infty \\
& \frac{1}{2}-1 < \frac{1}{1-x}-1 < \infty-1 \\
\Rightarrow & -\frac{1}{2} < y < \infty
\end{array}$
So, range $=y \in\left(\frac{-1}{2}, \infty\right)=B$ (according to question $)$.
Hence, $\rho([A: B])=2$
for this $-c+2 a-b=0$
$\Rightarrow \quad a=\frac{b+c}{2}$
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