JEE MATHEMATICS MCQ: FUNCTIONS

1 Medium

1 of 1

Application +4 | -1

Let

f: \mathbb{R}-\{0,1\} \rightarrow \mathbb{R}

be a function such that

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

. then

f(2)

is equal to

View answer and explanation

Correct option: B

\frac{9}{4}

Option B (

\frac{9}{4}

) is correct because it is the value of f(2) obtained by solving the system of equations derived from the given functional equation.

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\}