$f(x)$ and $g(x)$ are two differentiable functions on $[0,2]$ such that $f^{\prime \prime}(x)-g^{\prime \prime}(x)=0$ $f^{\prime}(1)=2 g^{\prime}(1)=4 f(2)=3 g(2)=9$ then $f(x)-g(x)$ at $x=3 / 2$ is

Question

Aicharya · Accepted Answer

$\because f^{\prime \prime}(x)-g^{\prime \prime}(x)=0$ Integrating, $f^{\prime}(x)-g^{\prime}(x)=c \Rightarrow f^{\prime}(1)-g^{\prime}(1)=c \Rightarrow 4-2=c \Rightarrow c=2$ $	herefore f^{\prime}(x)-g^{\prime}(x)=2 ;$ Integrating, $f(x)-g(x)=2 x+c_1$ $\Rightarrow f(2)-g(2)=4+c_1 \Rightarrow 9-3=4+c_1 \Rightarrow c_1=2 \quad 	herefore f(x)-g(x)=2 x+2$ At $x=3 / 2, f(x)-g(x)=3+2=5$

$f(x)$ and $g(x)$ are two differentiable functions on $[0,2]$ such that $f^{\prime \prime}(x)-g^{\prime…

Solution