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JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Application +4 | -1

Let

f\left( {x + \frac{1}{x}} \right) = {x^2} + \frac{1}{{{x^2}}},\;x \ne 0

, then

f\left( x \right)

is equal to

View answer and explanation

Correct option: C

{x^2} - 2

Option C (

x^2 - 2

) is correct because substituting

x + \frac{1}{x}

into the function results in

x^2 + \frac{1}{x^2}

.

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