Q42 / 160 · TS EAMCET 2021 (06 Aug Shift 2) · Mathematics

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

  • A
    $\left(-\infty, \frac{1}{2}\right)$
  • Correct answer
    B
    $\left(\frac{-1}{2}, \infty\right)$
  • C
    ( -1,1 )
  • D
    R

Solution

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 $-c+2 a-b=0$ $\Rightarrow \quad a=\frac{b+c}{2}$

Hence, the correct answer is option: 'B'.

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