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1.Relation and Function
hard
If the function $f:[1,\;\infty ) \to [1,\;\infty )$ is defined by $f(x) = {2^{x(x - 1)}},$ then ${f^{ - 1}} (x)$ is
A
${\left( {\frac{1}{2}} \right)^{x(x - 1)}}$
B
$\frac{1}{2}(1 + \sqrt {1 + 4{{\log }_2}x} )$
C
$\frac{1}{2}(1 - \sqrt {1 + 4{{\log }_2}x} )$
D
Not defined
(IIT-1999)
Solution
(b) Given $f(x) = {2^{x(x – 1)}}\,$
$\Rightarrow \,\,x\,(x – 1) = {\log _2}f(x)$
$ \Rightarrow \,\,{x^2} – x – {\log _2}f(x) = 0\,\, $
$\Rightarrow \,\,x = \frac{{1 \pm \sqrt {1 + 4{{\log }_2}f(x)} }}{2}$
Only $x = \frac{{1 + \sqrt {1 + 4{{\log }_2}f(x)} }}{2}$ lies in the domain
$\therefore \,\,{f^{ – 1}}(x) = \frac{1}{2}\,[1 + \sqrt {1 + 4\,{{\log }_2}x} ]$.
Standard 12
Mathematics
