If {f}:[1, ∞) arrow[2, ∞) is given by {f}(x)=x+1/x then {f}^{-1}(x) equals

Correct Answer is : x+x2−42{x+{x²-4}}{2}2x+x2−4

Correct Answer is : x+x2−42{x+{x²-4}}{2}2x+x2−4

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MHTCET 11 May 2024 Shift 2 Paper

Question : 106 of 150

Marks: +1, -0

\mathrm{f}:[1, \infty) \rightarrow[2, \infty)

\mathrm{f}(x)=x+\frac{1}{x}

\mathrm{f}^{-1}(x)

equals

Solution:

Concept:

To find the inverse, set

y = f(x)

, solve for

x

in terms of

y

, and then replace

y

x

Explanation:

Given

f(x) = x + \frac{1}{x}

Let

y = x + \frac{1}{x}

Multiply throughout by

x

xy = x^2 + 1

Rearrange as a quadratic in

x

x^2 - xy + 1 = 0

Using the quadratic formula:

x = \frac{y \pm \sqrt{y^2 - 4}}{2}

Since

x \in [1, \infty)

, choose the positive sign because the negative sign gives

x < 1

for

y > 2

Thus:

x = \frac{y + \sqrt{y^2 - 4}}{2}

Replacing

y

x

, we get:

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

Answer:

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

Hence option A is correct.

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