Question

Let \(f(x)\) be a twice differentiable function such that \(f^{''}(x)=-f(x)\), and \(f'(x)=g(x)\).
Let \(h(x)=\{f(x)\}^{2}+\{g(x)\}^{2}\). If \(h(2)=7\), then \(h(8)\) is equal to:

• A. \(7\)
• B. \(14\)
• C. \(0\)
• D. not defined

Hint:

Differentiate h and use f'' = -f to show that h is constant.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
We have \(g=f'\), so \(g'=f''=-f\).

Step 2: Differentiate h
\(h'(x)=2f(x)f'(x)+2g(x)g'(x)=2fg+2g(-f)=0\).

Step 3: Conclusion
Since \(h'(x)=0\) for all x, h is a constant function.

Step 4: Calculation
\(h(8)=h(2)=7\).

Final Answer:
h(8) equals 7. \[ \boxed{\text{(A) }7} \]