Q106 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Mathematics
Let $f: R \rightarrow R$ be defined by $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}$ for all $x$ and $y$. If $f^{\prime}(0)$ exists and equals -1 and $f(0)=1$, then $f(2)=$
- Correct answerA-1
- B0
- C1/2
- D1
Solution
We have, $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}$
On differentiating w.r.t. ' $x$ ', $y$ as constant
$f^{\prime}\left(\frac{x+y}{2}\right)=\frac{f^{\prime}(x)}{2}$
Put, $x=0, y=2 x$ we get
$\begin{aligned}
& & f^{\prime}(x) & =f^{\prime}(0)=-1 \quad\left[\because f^{\prime}(0)=-1\right] \\
\therefore & & f(x) & =-x+c \Rightarrow f(0)=0+c \Rightarrow 1=c \\
\therefore & & f(x) & =-x+1, f(2)=-2+1=-1
\end{aligned}$
Hence, the correct answer is option: 'A'.