Question

Let $f : \mathbb{R} \to \mathbb{R}$ be a twice differentiable function such that $f''(x) = -f(x)$. Let $h : \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $h'(x) = f(x)^2 + (f'(x))^2$ for all $x \in \mathbb{R}$. If $h(0) = 1$ and $h\left(\frac{1}{3}\right) = \frac{5}{3}$, then the value of $h(100)$ is

• A. 201
• B. 101
• C. 100
• D. 202

Hint:

The expression $f(x)^2 + (f'(x))^2$ is analogous to $\sin^2 x + \cos^2 x$, which is constant when $f''(x) = -f(x)$.
Recognizing this conservative-like quantity simplifies the derivative of $h(x)$ to a constant immediately.

Answer 1

The Correct Option is A

Step 1: Look at what $h'(x)$ really is.
We are told $h'(x) = f(x)^2 + (f'(x))^2$. The trick is to check whether this expression is secretly a constant.

Step 2: Differentiate that expression.
Let $E(x) = f(x)^2 + (f'(x))^2$. Then $E'(x) = 2 f(x) f'(x) + 2 f'(x) f''(x)$.

Step 3: Use the given relation $f''(x) = -f(x)$.
Replace $f''(x)$ with $-f(x)$: $E'(x) = 2 f(x) f'(x) + 2 f'(x)(-f(x)) = 2 f(x) f'(x) - 2 f(x) f'(x) = 0$.

Step 4: Conclude $E(x)$ is constant.
Since its derivative is zero everywhere, $E(x) = f(x)^2 + (f'(x))^2$ is the same number for all $x$. So $h'(x)$ is a constant, call it $k$. That means $h(x) = kx + c$, a straight line.

Step 5: Pin down $k$ and $c$ from the two data points.
From $h(0) = 1$ we get $c = 1$. From $h\left(\frac{1}{3}\right) = \frac{5}{3}$ we get $\frac{k}{3} + 1 = \frac{5}{3}$, so $\frac{k}{3} = \frac{2}{3}$ and $k = 2$.

Step 6: Evaluate at $x = 100$.
So $h(x) = 2x + 1$, and $h(100) = 2(100) + 1 = 201$.
\[ \boxed{201} \]