Question

lf $f^{\prime }\left(x\right)=g\left(x\right)$ and $g^{\prime }\left(x\right)=-f\left(x\right)$ for all $x$ and $f\left(2\right)=4=g\left(2\right)$, then $f^2\left(24\right)+g^2\left(24\right)$ is

• A. $32$
• B. $24$
• C. $64$
• D. $48$

Answer 1

Answer: (A)

$32$

$f\left(x\right)=-g^{\prime }\left(x\right)$ (given)
By multiply both sides with $f^{\prime }\left(x\right)$, we get
$f\left(x\right)f^{\prime }\left(x\right)=-g^{\prime }\left(x\right)f^{\prime }\left(x\right)$ ...(1)
$\because f^{\prime }\left(x\right)=g\left(x\right)$
$\therefore$ Eqn (1) becomes
$f\left(x\right)f^{\prime }\left(x\right)=-g\left(x\right)g^{\prime }\left(x\right)$
Now integrating both sides, we get
$f^2\left(x\right)+g^2\left(x\right)=c$
Where, $c$ is the constant of integration.
Now, it is given that $f\left(2\right)=g\left(2\right)=4$
$\therefore c=32$
$⟹f^2\left(24\right)+g^2\left(24\right)=32$
Hence, option A is correct.