JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Application +4 | -1

Let

f: Q-\{0,1\} \rightarrow Q

be a function such that

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

for all rational numbers

x \neq 0,1

. Then the value of

f(2)

View answer and explanation

Correct option: B

\frac{7}{4}

Option B (7/4) is correct because by substituting different values of x, we can create a system of equations to solve for f(2).

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

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