Q32 / 60 · KVPY 2014 (SA) · Mathematics

Let $R$ be the set of all real numbers and let $f$ be a fucntion $R$ to $R$ such that $f(x)+\left(x+\frac{1}{2}\right) f(1-x)=1$, for all $x \in R .$ Then $2 f(0)+3 f(1)$ is equal to.

  • A
    2
  • B
    0
  • Correct answer
    C
    $-2$
  • D
    $-4$

Solution

Given $f(x)+\left(x+\frac{1}{2}\right) f(1-x)=1 \quad ......\text {(1)}$ but $x=0$ $\begin{array}{l} f(0)+\frac{1}{2} f(1)=1 \\ \Rightarrow 2 f(0)+f(1)=2 \quad ......\text {(2)}\end{array}$ $\begin{array}{l} \text { put } x=1 \text { in }(1) \\ \Rightarrow f(1)+\frac{3}{2} f(0)=1 \\ \Rightarrow 2 f(1)+3 f(0)=2 \quad ......\text {(3)} \end{array}$ Solving $(2) \&(3)$ we have $\begin{array}{l} F(0)=2 \& f(1)=-2 \\ \therefore 2 f(0)+f(1)=4-6=-2 \end{array}$

Hence, the correct answer is option: 'C'.

More Functions Questions