KCET · Maths · Differential Equations
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
- A 0
- B 3
- C 5
- D 8
Answer & Solution
Correct Answer
(D) 8
Step-by-step Solution
Detailed explanation
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)]
\(=2 f(x) \cdot f^{\prime}(x)-2 f^{\prime}(x) \cdot f(x)\)
\(=0\)
\(\begin{array}{ll}
\Rightarrow & g(x)=\text { constant } \\
\because & g(3)=8 \\
\therefore & g(8)=8 (given)
\end{array}\)
\(f^{\prime \prime}(x)+f(x)=0...(i)\)
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)]
\(=2 f(x) \cdot f^{\prime}(x)-2 f^{\prime}(x) \cdot f(x)\)
\(=0\)
\(\begin{array}{ll}
\Rightarrow & g(x)=\text { constant } \\
\because & g(3)=8 \\
\therefore & g(8)=8 (given)
\end{array}\)
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