Question

If $f\text{'}\left(x\right)=\varphi \left(x\right)$ and $\varphi \text{'}\left(x\right)=f\left(x\right)$ for all $x$. Also $f\left(3\right)=5$ and $f\text{'}\left(3\right)=4$. Then the value of $\left[f\right(10){]}^2-[\varphi \left(10\right){]}^2$ is .

Answer 1

$\left[f\right(x){]}^2-[\varphi \left(x\right){]}^2$
Differentiating w.r.t. $x$
$\frac{d}{dx}\left(\right[f\left(x\right){]}^2-\left[\varphi \right(x\left){]}^2\right)=2\left[f\right(x\left)f^{\prime }\right(x)-\varphi (x\left){\varphi }^{\prime }\right(x\left)\right]=2\left[f\right(x\left)\varphi \right(x)-\varphi (x\left)f\right(x\left)\right]=0$
So,
$\left[f\right(x){]}^2-[\varphi \left(x\right){]}^2=\text{constant}\Rightarrow \left[f\right(10){]}^2-[\varphi \left(10\right){]}^2=\left[f\right(3){]}^2-[\varphi \left(3\right){]}^2\Rightarrow \left[f\right(10){]}^2-[\varphi \left(10\right){]}^2=5^2-4^2=9$