JEE MATHEMATICS MCQ: FUNCTIONS

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Application +4 | -1

Let

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

then

f\left( x \right)

equals

View answer and explanation

Correct option: B

{x^2} - 2\;for\;\mathrm{ }all\;\left| x \right| \ge 2

Option B (

{x^2} - 2\;for\;\mathrm{ }all\;\left| x \right| \ge 2

) is correct because it correctly expresses f(x) in terms of x, considering the domain restriction.

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