JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Application +4 | -1

2f\left( {{x^2}} \right) + 3f\left( {\frac{1}{{{x^2}}}} \right) = {x^2} - 1\;and\,x \in R - \left\{ 0 \right\}

, then

f\left( {{x^4}} \right)

View answer and explanation

Correct option: A

\frac{{\left( {1 - {x^4}} \right)\left( {2{x^4} + 3} \right)}}{{5{x^4}}}

Option A (

\frac{{\left( {1 - {x^4}} \right)\left( {2{x^4} + 3} \right)}}{{5{x^4}}}

) is correct because it is the result of solving the system of equations for f(x²) and then substituting x⁴.

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\}