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Relations and Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsRelations and Functions1 Mark
MCQ
Let $\text{A}=\{\text{x}\in\text{R}:\text{x}\geq1\}.$ The inverse of the function $f : A \rightarrow A$ given by $\text{f(x)}=2^{\text{x}(\text{x}-1)},$ is:
  • A
    $\big(\frac{1}{2}\big)^{\text{x}(\text{x}-1)}$
  • ✓
    $\frac{1}{2}\big\{1+\sqrt{1+4\log_2\text{x}}\big\}$
  • C
    $\frac{1}{2}\big\{1-\sqrt{1+4\log_2\text{x}}\big\}$
  • D
    Not defined
✓

Answer

Correct option: B.
$\frac{1}{2}\big\{1+\sqrt{1+4\log_2\text{x}}\big\}$
Given function is $\text{A}=\{\text{x}\in\text{R}:\text{x}\geq1\}.$
The inverse of the function $f : A \rightarrow A$ given by $\text{f(x)}=2^{\text{x}(\text{x}-1)}$
$\text{f(x)}=\text{y}$
$2^{\text{x}(\text{x}-1)}=\text{y}$
$\text{x}(\text{x}-1)=\log_2\text{y}$
$\text{x}^2+\text{x}=\log_2\text{y}$
$\text{x}^2+\text{x}+\frac{1}{4}=\log_2\text{y}+\frac{1}{4}$
$\Big(\text{x}-\frac{1}{2}\Big)^2=\frac{4\log_2\text{y}+1}{4}$
$\text{x}-\frac{1}{2}=\pm\sqrt{\frac{4\log_2\text{y}+1}{4}}$
$\text{x}=\frac{1}{2}\pm\sqrt{\frac{4\log_2\text{y}+1}{4}}$
$\text{x}=\frac{1}{2}+\sqrt{\frac{4\log_2\text{y}+1}{4}}$
$\text{f}^{-1}(\text{x})=\frac{1+\sqrt{4\log_2\text{y}+1}}{2}$

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