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Question: If \(f:\lbrack 1,\infty) \rightarrow \lbrack 1,\infty)\) is defined as \(f(x) = 2^{x(x - 1)}\) then ...

If f:[1,∞)→[1,∞)f:\lbrack 1,\infty) \rightarrow \lbrack 1,\infty) is defined as f(x)=2x(x−1)f(x) = 2^{x(x - 1)} then f−1(x)f^{- 1}(x) is equal to

A

(12)x(x−1)\left( \frac{1}{2} \right)^{x(x - 1)}

B

12(1+1+4log⁡2x)\frac{1}{2}\left( 1 + \sqrt{1 + 4\log_{2}x} \right)

C

12(1−1+4log⁡2x)\frac{1}{2}\left( 1 - \sqrt{1 + 4\log_{2}x} \right)

D

Not defined

Answer

12(1+1+4log⁡2x)\frac{1}{2}\left( 1 + \sqrt{1 + 4\log_{2}x} \right)

Explanation

Solution

Given f(x)=2x(x−1)⇒x(x−1)=log⁡2f(x)f(x) = 2^{x(x - 1)} \Rightarrow x(x - 1) = \log_{2}f(x)

⇒x2−x−log⁡2f(x)=0⇒x=1±1+4log⁡2f(x)2x^{2} - x - \log_{2}f(x) = 0 \Rightarrow x = \frac{1 \pm \sqrt{1 + 4\log_{2}f(x)}}{2}Only x=1+1+4log⁡2f(x)2x = \frac{1 + \sqrt{1 + 4\log_{2}f(x)}}{2} lies in the domain

∴\therefore f−1(x)=12[1+1+4log⁡2x]f^{- 1}(x) = \frac{1}{2}\lbrack 1 + \sqrt{1 + 4\log_{2}x\rbrack}