JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Application +4 | -1

f\left( x \right)

satisfies the relation

2f\left( x \right) + f\left( {1 - x} \right) = {x^2}

for all real x, then

f\left( x \right)

View answer and explanation

Correct option: B

\frac{{{x^2} + 2x - 1}}{3}

The correct answer is B (

\frac{{{x^2} + 2x - 1}}{3}

) because it is the solution obtained by substituting (1-x) into the original equation and solving the resulting system of equations.

More Questions from FUNCTIONS

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

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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\}