Question

Let f(x) be a twice differentiable function such that \(f^{''}(x) = -f(x)\), \(f^'(x) = g(x)\) and \(h(x) = \{f(x)\}^2+\{g(x)\}^2\). If \(h(5) = 11\), then \(h(10)\) is equal to ...

• A. \(11\)
• B. \(22\)
• C. \(0\)
• D. Not defined

Hint:

Differentiate h and show it is zero, so h is constant.

Answer 1

The Correct Option is A

Step 1: Differentiate h
\(h(x) = f^2+g^2\), so \(h'(x) = 2ff' + 2gg'\). Since \(f' = g\) and \(g' = f'' = -f\), we get \(h'(x) = 2fg + 2g(-f)\).

Step 2: Simplify
\(h'(x) = 2fg - 2fg = 0\).

Step 3: Conclude
A function with zero derivative everywhere is constant, so \(h(10) = h(5) = 11\). Option (A).

Step 4: Why not the others
Nothing in the data lets h change, so \(22\) or \(0\) are impossible. The function exists since it is twice differentiable, so \(h\) is defined.

Final Answer:
h is constant and h(10) = 11. \[ \boxed{\text{(A)}\ 11} \]