JEE MATHEMATICS MCQ: FUNCTIONS

1 Medium

1 of 1

Application +4 | -1

If a function

f\left( x \right)

satisfies the condition

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

View answer and explanation

Correct option: B

{x^2} - 2

for all

x

satisfying

\left| x \right| \ge 2

Option B (

{x^2} - 2

for all

x

satisfying

\left| x \right| \ge 2

) is correct because the function f(x) is defined as x² - 2 only when |x| >= 2, which is the range of x + 1/x.

More Questions from FUNCTIONS

Practice All Chapter MCQs →

f:(-\infty, 2] \longrightarrow(-\infty, 4]

f(x)=x(4-x)

f^{-1}(x)

g(f(x))=|\sin x|

f(g(x))=(\sin \sqrt{x})^2

f:\left( {2,\;3} \right) \to \left( {0,\;1} \right)

f\left( x \right) = x - \left[ x \right]

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

f: R \rightarrow R

f(x)=x^4

f(x)=(x+1)^2-1(x \geq-1)

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