Let $f$ be twice differentiable function such that $\mathrm{f}^{\prime \prime}(x)=-\mathrm{f}(x), \mathrm{f}^{\prime}(x)=\mathrm{g}(x)$ and $\mathrm{h}(x)=(\mathrm{f}(x))^2+(\mathrm{g}(x))^2$. If $\mathrm{h}(5)=1$, then the value of $h(10)$ is

Question

Aicharya · Accepted Answer

$\begin{aligned} \quad \mathrm{h}(x) & =[\mathrm{f}(x)]^2+[\mathrm{g}(x)]^2 \ 	herefore \quad \mathrm{~h}^{\prime}(x) & \left.=2 \mathrm{f}(x)] \mathrm{f}^{\prime}(x)+2 \mathrm{~g}(x)ight] \mathrm{g}^{\prime}(x) \ & =2\left[-\mathrm{f}^{\prime \prime}(x)ight] \mathrm{f}^{\prime}(x)+2\left[\mathrm{f}^{\prime}(x)ight] \cdot \mathrm{f}^{\prime \prime}(x) \ & =0 \end{aligned}$ $	herefore \quad \mathrm{h}(x)$ is a constant function. $	herefore \quad h(5)=1 \Rightarrow h(10)=1$

Let $f$ be twice differentiable function such that $\mathrm{f}^{\prime \prime}(x)=-\mathrm{f}(x),…

Solution