Consider the function f(x)=(x-1)^2(x+1)(x-2)^3 . Let f(x) and g(x)...

$f^{\prime \prime}(x)=g^{\prime \prime}(x)$ integrate b|s w.r.t. (x) $f^{\prime}(x)+C_1=g^{\prime}(x)+C_2$...(1) Putting $x=1$$\begin{aligned} & f^{\prime}(1)+C_1=g^{\prime}(1)+C_2 \\ & \Rightarrow C_1-C_2=g^{\prime}(1)-f^{\prime}(1) \\ & \Rightarrow C_1-C_2=6-4=2 \end{aligned}$again integrating equation (1) $f(x)+C_1 x+a_1=g(x)+C_2 x+a_2$...(3) Putting $x=2\begin{aligned} & f(2)+C_1 \times 2+a_1=g(2)+2 C_2+a_2 \\ & \Rightarrow a_1-a_2=g(2)+2 C_2-2 C_1-f(2) \\ & \quad=9-3-2\left(C_1-C_2\right) \\ & \quad=6-2 \times 2=2 \\ & \Rightarrow \mathrm{a}_1-\mathrm{a}_2=2 \end{aligned}$ Rearranging equation (3) again, we get $f(x)-g(x)=-x $\left(C_1-C_2\right)- \left(a_1-a_2\right)$ Putting $x=4$, we get\begin{aligned} {[f(x)-g(x)]_{x=4} } =-4\left(C_1-C_2\right)-\left(a_1-a_2\right) \\ & =-4 \times(2)-2 \\ {[f(x)-g(x)]_{x=4} } & =-10 \end{aligned}$$$