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Question: If : (-∞, 1\] →\(\left[ \frac { 1 } { 2 } , \infty \right)\), where f(x) = 2<sup>x(x-2)</sup> , f<su...

If : (-∞, 1] →[12,∞)\left[ \frac { 1 } { 2 } , \infty \right), where f(x) = 2x(x-2) , f-1(x) is given by

A

1−1+log⁡2x1 - \sqrt { 1 + \log _ { 2 } x }

B

1−1−log⁡2x1 - \sqrt { 1 - \log _ { 2 } x }

C

1+1+log⁡2x1 + \sqrt { 1 + \log _ { 2 } x }

D

1+1−log⁡2x1 + \sqrt { 1 - \log _ { 2 } x }

Answer

1−1+log⁡2x1 - \sqrt { 1 + \log _ { 2 } x }

Explanation

Solution

f(f−1(x))=2f−1(x)(f−1(x)−2)f \left( f ^ { - 1 } ( x ) \right) = 2 ^ { f ^ { - 1 } ( x ) \left( f ^ { - 1 } ( x ) - 2 \right) }

⇒ (f−1(x))2 – 2f-1(x) – log2x = 0

⇒ f−1(x) = 2±4+4log⁡2x2=1±1+log⁡2x\frac { 2 \pm \sqrt { 4 + 4 \log _ { 2 } x } } { 2 } = 1 \pm \sqrt { 1 + \log _ { 2 } x }

But range of f−1(x) is (-∞, 1] thus