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

Correct Answer is : 12{1+1+4log2x}

\frac{1}{2}\{1+\sqrt{1+4 \log _{2} x}\}

\frac{1}{2}\{1-\sqrt{1+4 \log _{2} x}\}

NIMCET 2011 Question Paper with solutions

Question : 20 of 120

Marks: +1, -0

If the function

f:[1, ∞) →[1, ∞)

is defined by

f(x)=2^{x(x-1)}

, then

f^{-1}(x)

Solution:

Let

Y=f(x)=2^{x\{x-1)}

On taking log, we get

\log _{2} y=x(x-1) \log _{2} 2

⇒ x^{2}-x-\log _{2} y=0

⇒ x=\ {1 ± √{1+4 \log _{2} Y}}/{2}

x= {1+√{1+4 \log _{2} y}}/{2}

f^{-1}(x)= {1+√{1+4 \log _{2} x}}/{2}

Since, for

y ∈[1, ∞), 4 \log _{2} y ≥ 0

⇒ √{1+4 \log _{2} y} ≥ 1

⇒ √{1+4 \ln x}>1

∴ x= {1-√{1+4 \log _{2} y}}/{2} ≤ 0 ∉[1, ∞)

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