Question

Let f be twice differentiable function such that $f^{\prime \prime }\left(x\right)=-f\left(x\right)$ and $f^{\prime }\left(x\right)=g\left(x\right)$,$h\left(x\right)={f\left(x\right)}^2+{g\left(x\right)}^2$. If $h\left(5\right)=11$ then $h\left(10\right)$ is equal to

• A. $22$
• B. $11$
• C. $0$
• D. $1$

Answer 1

Answer: (B)

$11$

Given $h\left(x\right)={f\left(x\right)}^2+{g\left(x\right)}^2$

$h^{\prime }\left(x\right)=2f^{\prime }\left(x\right)f\left(x\right)+2g^{\prime }\left(x\right)g\left(x\right)$

$=2g\left(x\right)f\left(x\right)+2f^{\prime \prime }\left(x\right)g\left(x\right),\left(\because f^{\prime \prime }\left(x\right)=g^{\prime }\left(x\right)\right)$

$=0$ $\left(\because -f^{\prime \prime }\left(x\right)=f\left(x\right)\right)$

$h\left(x\right)$ is are constant function

$h\left(5\right)=h\left(10\right)=11$