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Question: If f:\[1, ∞) → \[2, ∞) is given by f(x) = \(x + \frac{1}{x}\)then f<sup>−1</sup>(x) equals...

If f:[1, ∞) → [2, ∞) is given by f(x) = x+1xx + \frac{1}{x}then f−1(x) equals

A

(x+x2−4)2\frac{\left( x + \sqrt{x^{2} - 4} \right)}{2}

B

x(1+x2)\frac{x}{\left( 1 + x^{2} \right)}

C

(x−x2−4)2\frac{\left( x - \sqrt{x^{2} - 4} \right)}{2}

D

(1+x2−4)\left( 1 + \sqrt{x^{2} - 4} \right)

Answer

(x+x2−4)2\frac{\left( x + \sqrt{x^{2} - 4} \right)}{2}

Explanation

Solution

f: [1, ∞) → [2, ∞)

F(x) x+1x=yx + \frac{1}{x} = y ⇒ x2- yx + 1 = 0 ⇒ x y±y2−42\frac{y \pm \sqrt{y^{2} - 4}}{2}

But {x>1y>2 \left\{ \begin{matrix} x > 1 \\ y > 2 \end{matrix} \right.\ ∴ x=y+y2−42x = \frac{y + \sqrt{y^{2} - 4}}{2}

∴ f−1(x)=x+x2−42f^{- 1}(x) = \frac{x + \sqrt{x^{2} - 4}}{2}

∴ ‘A’ is the correct alternative