Aspire Faculty ID #16202 · Topic: NIMCET 2011 · Just now
NIMCET 2011

If the function $f:[1,\infty)\to[1,\infty)$ is defined by $f(x)=2^{x(x-1)}$, then $f^{-1}(x)$ is:

Solution

Given $2^{x(x-1)} = y$ Take $\log_2$: $x(x-1) = \log_2 y$ Quadratic: $x^{2}-x-\log_2 y = 0$ So $x = \dfrac{1 \pm \sqrt{1+4\log_2 y}}{2}$ Since $x \ge 1$, choose positive sign: $f^{-1}(x)=\dfrac12\left(1+\sqrt{1+4\log_2 x}\right)$
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