Question

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

• A. 0
• B. 2
• C. 10
• D. 5

Answer 1

Answer: (D)

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