Question

$f^{{}^{\prime }}\left(x\right)=g\left(x\right)$ and $g^{{}^{\prime }}\left(x\right)=-f\left(x\right)$ for all real x and $f\left(5\right)=2=f^{{}^{\prime }}\left(5\right)$ then $f^2\left(10\right)+g^2\left(10\right)$ is -

• A. $2$
• B. $4$
• C. $8$
• D. None of these

Answer 1

Answer: (C)

$8$

Given, $f^{{}^{\prime }}\left(x\right)=g\left(x\right)$ and $g^{{}^{\prime }}\left(x\right)=-f\left(x\right)$

Now $\frac{d}{dx}[f^2\left(x\right)+g^2\left(x\right)]=2f\left(x\right)f^{{}^{\prime }}\left(x\right)+2g\left(x\right)g^{{}^{\prime }}\left(x\right)$

$=2f\left(x\right)g\left(x\right)-2g\left(x\right)f\left(x\right)=0$

$\therefore f^2\left(x\right)+g^2\left(x\right)$ =constant

$f^2\left(5\right)+g^2\left(5\right)=4+4=8$

$\therefore f^2\left(10\right)+g^2\left(10\right)$ = 8