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Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsFunctions1 Mark
MCQ
If $\text{F}:[1,\infty)\rightarrow[2,\infty)$ is given by $\text{f(x)}=\text{x}+\frac{1}{\text{x}},$ then $f^{-1}(x)$ equals:
  • ✓
    $\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$
  • B
    $\frac{\text{x}}{1+\text{x}^2}$
  • C
    $\frac{\text{x}-\sqrt{\text{x}^2-4}}{2}$
  • D
    $1+\sqrt{\text{x}^2-4}$
✓

Answer

Correct option: A.
$\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$
Let $f^{-1}(x) = y$
$\Rightarrow\ \text{f(y)} = \text{x}$
$\Rightarrow\ \text{y}+\frac{1}{\text{y}}=\text{x}$
$\Rightarrow\ \text{y}^2 + 1 = \text{xy}$
$\Rightarrow\ \text{y}^2 - \text{xy} + 1 = 0$
$\Rightarrow\ \text{y}^2-2\times\text{y}\times\frac{\text{x}}{2}+\big(\frac{\text{x}}{2}\big)^2-\big(\frac{\text{x}}{2}\big)^2+1=0$
$\Rightarrow\ \text{y}^2-2\times\text{y}\times\frac{\text{x}}{2}+\big(\frac{\text{x}}{2}\big)^2=\frac{\text{x}^2-1}{4}$
$\Rightarrow\ \Big(\text{y}-\frac{\text{x}}{2}\Big)^2=\frac{\text{x}^2-1}{4}$
$\Rightarrow\ \text{y}-\frac{\text{x}}{2}=\frac{\sqrt{\text{x}^2-4}}{2}$
$\Rightarrow\ \text{y}=\frac{\text{x}}{2}+\frac{\sqrt{\text{x}^2-4}}{2}$
$\Rightarrow\ \text{y}=\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$

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