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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}, x \in R-\{0\}

, then

f(x)

is equal to

View answer and explanation

Correct option: C

x^2-2

, when

|x| \geq 2

Option C (

x^2-2

, when

|x| \geq 2

) is correct because by substituting

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

into the equation

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

, we find that

f(z) = z^2 - 2

, which translates to

f(x) = x^2 - 2

when

|x| \geq 2

.

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S=\left\{x: f(x)=f^{-1}(x)\right\}