Question

If $f\left(x\right),g\left(x\right)$ be twice differentiable functions on $[0,2]$ satisfying$f"\left(x\right)=g"\left(x\right),f^{\prime }\left(1\right)=2g^{\prime }\left(1\right)=4$ and $f\left(2\right)=3g\left(2\right)=9$, then $f\left(x\right)-g\left(x\right)$ at $x=4equals$

• A. $0$
• B. $10$
• C. $8$
• D. $2$

Answer 1

Answer: (B)

$10$

Given, $f^{{}^{\prime \prime }}=g^{{}^{\prime \prime }}$
$f^{\prime }\left(1\right)=2g^{\prime }\left(1\right)=4$
$f\left(x\right)=3g\left(2\right)=9$
$f\left(x\right)-g\left(x\right)=h\left(x\right)$
$h^{\prime }\left(x\right)=f^{\prime }\left(x\right)-g^{\prime }\left(x\right)$
$h^{\prime }\left(1\right)=2$
$h^{{}^{\prime \prime }}\left(x\right)=f^{{}^{\prime \prime }}\left(x\right)-g^{{}^{\prime \prime }}\left(x\right)=0$
$h\left(2\right)=f\left(2\right)-g\left(2\right)$
$=9-3=6$
$h\left(2\right)=6$
$h\left(x\right)=a{x}^2+bx+c$
$4a+2b+c=6$
$2a+b=2$
$a=0\therefore b=2$
$\therefore 4+c=6$
$c=2$
Therefore, $e2x+2=h\left(x\right)$
$\Rightarrow h\left(4\right)=10$