Q118 / 180 · KCET 2013 · Mathematics
If $f(x)$ is a function such that $f^{\prime \prime}(x)+f(x)=0$ and $g(x)=[f(x)]^{2}+\left[f^{\prime}(x)\right]^{2}$ and $g(3)=8$, then $g(8)$ is equal to
- A0
- B3
- C5
- Correct answerD8
Solution
Given,
$f^{\prime \prime}(x)+f(x)=0...(i)$
and $g(x)=[f(x)]^{2}+\left[f^{\prime}(x)\right]^{2}$
Now, on differentiating w.r.t. $x$, we get
$g^{\prime}(x)=2 f(x) \cdot f^{\prime}(x)+2 f^{\prime}(x) \cdot f^{\prime \prime}(x)$
$=2 f(x) \cdot f^{\prime}(x)+2 f^{\prime}(x)\{-f(x)\} \quad$ [from Eq. (i)]
$=2 f(x) \cdot f^{\prime}(x)-2 f^{\prime}(x) \cdot f(x)$
$=0$
$\begin{array}{ll}
\Rightarrow & g(x)=\text { constant } \\
\because & g(3)=8 \\
\therefore & g(8)=8 (given)
\end{array}$
Hence, the correct answer is option: 'D'.