Let A={x ∈ R : x ≥ 1}. The inverse of the function, f : A→A given by


Let $A=\{x \in R: x \geq 1\}$. The inverse of the function, $f: A \rightarrow A$ given by $f(x)=2^{x(x-1)}$, is

(a) $\left(\frac{1}{2}\right)^{x(x-1)}$

(b) $\frac{1}{2}\left\{1+\sqrt{1+4 \log _{2} x}\right\}$

(c) $\frac{1}{2}\left\{1-\sqrt{1+4 \log _{2} x}\right\}$

(d) not defined


Let $f^{-1}(x)=y \ldots(1)$

$\Rightarrow f(y)=x$

$\Rightarrow 2^{y(y-1)}=x$

$\Rightarrow 2^{y^{2}-y}=x$

$\Rightarrow y^{2}-y=\log _{2} x$

$\Rightarrow y^{2}-y+\frac{1}{4}=\log _{2} x+\frac{1}{4}$

$\Rightarrow\left(y-\frac{1}{2}\right)^{2}=\frac{4 \log _{2} x+1}{4}$

$\Rightarrow y-\frac{1}{2}=\pm \frac{\sqrt{4 \log _{2} x+1}}{2}$

$\Rightarrow y=\frac{1}{2} \pm \frac{\sqrt{4 \log _{2} x+1}}{2}$

$\Rightarrow y=\frac{1}{2}+\frac{\sqrt{4 \log _{2} x+1}}{2}$                            $(\because \mathrm{y} \geq 1)$

So, $f^{-1}(x)=\frac{1}{2}\left(1+\sqrt{1+4 \log _{2} x}\right)$

So, the answer is (b).

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