Question

Let $f :\left(-\infty, \frac{1}{2}\right] \rightarrow\left[\frac{1}{2}, \infty\right)$ be a function defined as $f ( x )=16^{ x ^2- x }$ then which one of the following must be CORRECT?

• A. $f^{-1}(x)$ is not possible because $f(x)$ is not a bijective function
• B. $f^{-1}(x)=\frac{1}{2}-\sqrt{\log _{16} x+\frac{1}{4}}$
• C. $f ^{-1}( x )=\frac{1}{2}+\sqrt{\log _{16} x +\frac{1}{4}}$
• D. $f ^{-1}( x )=\frac{1}{2}+\sqrt{\log _{16} x -\frac{1}{4}}$

Answer 1

Answer: (B)

$f(x)=16^{x^2-x}=16^{\left(x-\frac{1}{2}\right)^2-\frac{1}{4}} \geq \frac{1}{2} \Rightarrow f(x)$ is onto
$f ^{\prime}( x )=(2 x -1) 16^{ x ^2- x } \ln 16<0 \forall x <\frac{1}{2}$
$\Rightarrow f ( x )$ is one-one
$y=16^{x^2-x} \Rightarrow\left(x-\frac{1}{2}\right)^2-\frac{1}{4}=\log _{16} y$
$x=\frac{1}{2}-\sqrt{\log _{16} y +\frac{1}{4}}$ OR $x =\frac{1}{2}+\sqrt{\log _{16} y +\frac{1}{4}}$ (rejective)
$\left.f ^{-1}( x )=\frac{1}{2}-\sqrt{\log _{16} x +\frac{1}{4}} . \quad\right]$