Question

Let

 \[A = \left\{ x \in R : x \geq 1 \right\}\] The inverse of the function, 

\[f : A \to A\] given by

\[f\left( x \right) = 2^{x \left( x - 1 \right)} , is\]

 

पर्याय

• \[\left( \frac{1}{2} \right)^{x \left( x - 1 \right)}\]

• \[\frac{1}{2} \left\{ 1 + \sqrt{1 + 4 \log_2 x} \right\}\]

•  \[\frac{1}{2} \left\{ 1 - \sqrt{1 + 4 \log_2 x} \right\}\]

• not defined

Answer 1

\[\text{Let} f^{- 1} \left( x \right) = y . . . \left( 1 \right)\] 
\[ \Rightarrow f\left( y \right) = x\] 
\[ \Rightarrow 2^{y\left( y - 1 \right)} = x\] 
\[ \Rightarrow 2^{y^2 - y} = x\] 
\[ \Rightarrow y^2 - y = \log_2 x\] 
\[ \Rightarrow y^2 - y + \frac{1}{4} = \log_2 x + \frac{1}{4}\] 
\[ \Rightarrow \left( y - \frac{1}{2} \right)^2 = \frac{4 \log_2 x + 1}{4}\] 
\[ \Rightarrow y - \frac{1}{2} = \pm \frac{\sqrt{4 \log_2 x + 1}}{2}\] 
\[ \Rightarrow y = \frac{1}{2} \pm \frac{\sqrt{4 \log_2 x + 1}}{2}\] 
\[ \Rightarrow y = \frac{1}{2} + \frac{\sqrt{4 \log_2 x + 1}}{2} \left( \because y \geq1 \right)\] 
\[So, f^{- 1} \left( x \right) = \frac{1}{2}(1 + \sqrt{1 + 4 \log_2 x} ) [\text{from}\left( 1 \right)]\] 

So, the answer is (b).