Question

Let $f$ be the twice differentiable function such that $f^{\prime \prime }\left(x\right)=-f\left(x\right)$ and $f^{\prime }\left(x\right)=g\left(x\right)$. If $h^{\prime }\left(x\right)=[f{\left(x\right)}^2+g{\left(x\right)}^2],h\left(1\right)=8,h\left(0\right)=2$, then $h\left(2\right)$ equals to

• A. $1$
• B. $2$
• C. $3$
• D. None of these

Answer 1

Answer: (D)

None of these

We have $h^{\prime }\left(x\right)=[f{\left(x\right)}^2+g{\left(x\right)}^2]$
$\Rightarrow h^{\prime \prime }\left(x\right)=2f\left(x\right)f^{\prime }\left(x\right)+2g\left(x\right)g^{\prime }\left(x\right)$
$\Rightarrow h^{\prime \prime }\left(x\right)=2f\left(x\right)g\left(x\right)+2g\left(x\right)f^{\prime \prime }\left(x\right)[\because g(x)=f^{\prime }(x);g(x)=f^{\prime \prime }(x\left)\right]$
$\Rightarrow h^{\prime \prime }\left(x\right)=0$
$\Rightarrow h^{\prime }\left(x\right)=C$, constant for all $x\in R$
$\Rightarrow h\left(x\right)=xC+C_1$
$\Rightarrow h\left(0\right)=C_1;h\left(1\right)=C+C_1$
$\Rightarrow C_1=2;C=6$
Therefore, $h\left(x\right)=6x+2$
$\Rightarrow h\left(2\right)=14$