MCQ
Let $\text{A}=\{\text{x}\in\text{R}:\text{x}\geq1\}.$ The inverse of the function f : A → 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
    $\text{Not defined}$

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 → 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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