A differentiable function f(x) has a local maximum at x=0 . Let y=2...

Given, $\quad y=2 f(x)+a x-b\qquad....(i)$ Differentiating both side w.r. to $x$$\begin{aligned} \Rightarrow \quad \frac{d y}{d x} & =2 f^{\prime}(x)+a(1) \qquad....(ii) \\ \frac{d^2 y}{d x^2} & =2 f^{\prime \prime}(x) \end{aligned}$$\qquad....(iii)\because y$ has a relative maxima at $x=0$$\begin{aligned} & \therefore & \left(\frac{d y}{d x}\right)_{x=0} & =0 \\ & & & \\ \Rightarrow & & 2 f^{\prime}(0)+a & =0 \\ \Rightarrow & & 2(0)+a & =0 \\ & & a & =0 \end{aligned}$Also, $\quad\left(\frac{d^2 y}{d x^2}\right)_{x=0} < 0\begin{aligned} \Rightarrow & 2 f^{\prime \prime}(0) & < 0 \\ \Rightarrow & f^{\prime \prime}(0) & =0\end{aligned}$$$Clearly,$y$has relative maxima for$a=0$and all values of$b$.