Question

Let $f$ be a twice differentiable function such that $f^{{}^{\prime \prime }}\left(x\right)=-f\left(x\right)$, and $f^{{}^{\prime }}\left(x\right)=g\left(x\right)$, $h\left(x\right)=\left[f\right(x{)}^2]+\left[g\right(x{)}^2]$. Find $h\left(10\right)$ if $h\left(5\right)=11$. (IIT-JEE, 1982)

Answer 1

Given that $f$ is twice differentiable function such that $f^{{}^{\prime \prime }}\left(x\right)$

$f^{{}^{\prime \prime }}\left(x\right)=-f\left(x\right)$, and $f^{{}^{\prime }}\left(x\right)=g\left(x\right)$, $h\left(x\right)=\left[f\right(x{)}^2]+\left[g\right(x{)}^2]$

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

$=2f\left(x\right)g\left(x\right)+2g\left(x\right)f^{{}^{\prime \prime }}\left(x\right)$ $[\therefore g(x)-f^{{}^{\prime }}(x)\Rightarrow g^{{}^{\prime }}(x)=f^{{}^{\prime \prime }}(x\left)\right]$

$=2f\left(x\right)g\left(x\right)+2g\left(x\right)(-f(x\left)\right)$

$=2f\left(x\right)g\left(x\right)-2f\left(x\right)g\left(x\right)$

$=0$

$\therefore h^{{}^{\prime }}\left(x\right)=0,\forall x$

$\Rightarrow h$ is a constant function.

$\therefore h(5=h(10)=11$