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
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
Show 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.
Recognizing this conservative-like quantity simplifies the derivative of $h(x)$ to a constant immediately.
Updated On: Jun 16, 2026
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
We are given a twice-differentiable function $f(x)$ satisfying a second-order differential relation.
Another function $h(x)$ has its derivative defined in terms of $f(x)$ and $f'(x)$.
We need to analyze the expression $f(x)^2 + (f'(x))^2$ to simplify $h'(x)$ and subsequently find $h(100)$.
Step 2: Key Formula or Approach:
We use the differentiation rules:
\[ \frac{d}{dx}[u^2] = 2u \frac{du}{dx} \]
And the fundamental theorem of calculus.
Step 3: Detailed Explanation:
• Let us define a auxiliary function $g(x) = f(x)^2 + (f'(x))^2$.
• Let us find the derivative of $g(x)$ with respect to $x$:
\[ g'(x) = \frac{d}{dx} \left[ f(x)^2 + (f'(x))^2 \right] \]
Using the chain rule:
\[ g'(x) = 2f(x)f'(x) + 2f'(x)f''(x) \]
• Factor out $2f'(x)$:
\[ g'(x) = 2f'(x) \left[ f(x) + f''(x) \right] \]
• We are given that $f''(x) = -f(x)$, which means $f(x) + f''(x) = 0$.
Substituting this into our equation:
\[ g'(x) = 2f'(x) \cdot 0 = 0 \]
• Since the derivative of $g(x)$ is zero for all real numbers, $g(x)$ must be a constant:
\[ g(x) = C \implies f(x)^2 + (f'(x))^2 = C \]
• Substituting this back into the expression for $h'(x)$:
\[ h'(x) = C \]
• Integrating both sides with respect to $x$:
\[ h(x) = Cx + D \]
where $D$ is the constant of integration.
• We are given $h(0) = 1$:
\[ C(0) + D = 1 \implies D = 1 \]
• We are also given $h\left(\frac{1}{3}\right) = \frac{5}{3}$:
\[ C\left(\frac{1}{3}\right) + 1 = \frac{5}{3} \implies \frac{C}{3} = \frac{2}{3} \implies C = 2 \]
• So, the function $h(x)$ is uniquely determined as:
\[ h(x) = 2x + 1 \]
• Now, we can find the value of $h(100)$:
\[ h(100) = 2(100) + 1 = 201 \]
Step 4: Final Answer:
The value of $h(100)$ is 201.
We are given a twice-differentiable function $f(x)$ satisfying a second-order differential relation.
Another function $h(x)$ has its derivative defined in terms of $f(x)$ and $f'(x)$.
We need to analyze the expression $f(x)^2 + (f'(x))^2$ to simplify $h'(x)$ and subsequently find $h(100)$.
Step 2: Key Formula or Approach:
We use the differentiation rules:
\[ \frac{d}{dx}[u^2] = 2u \frac{du}{dx} \]
And the fundamental theorem of calculus.
Step 3: Detailed Explanation:
• Let us define a auxiliary function $g(x) = f(x)^2 + (f'(x))^2$.
• Let us find the derivative of $g(x)$ with respect to $x$:
\[ g'(x) = \frac{d}{dx} \left[ f(x)^2 + (f'(x))^2 \right] \]
Using the chain rule:
\[ g'(x) = 2f(x)f'(x) + 2f'(x)f''(x) \]
• Factor out $2f'(x)$:
\[ g'(x) = 2f'(x) \left[ f(x) + f''(x) \right] \]
• We are given that $f''(x) = -f(x)$, which means $f(x) + f''(x) = 0$.
Substituting this into our equation:
\[ g'(x) = 2f'(x) \cdot 0 = 0 \]
• Since the derivative of $g(x)$ is zero for all real numbers, $g(x)$ must be a constant:
\[ g(x) = C \implies f(x)^2 + (f'(x))^2 = C \]
• Substituting this back into the expression for $h'(x)$:
\[ h'(x) = C \]
• Integrating both sides with respect to $x$:
\[ h(x) = Cx + D \]
where $D$ is the constant of integration.
• We are given $h(0) = 1$:
\[ C(0) + D = 1 \implies D = 1 \]
• We are also given $h\left(\frac{1}{3}\right) = \frac{5}{3}$:
\[ C\left(\frac{1}{3}\right) + 1 = \frac{5}{3} \implies \frac{C}{3} = \frac{2}{3} \implies C = 2 \]
• So, the function $h(x)$ is uniquely determined as:
\[ h(x) = 2x + 1 \]
• Now, we can find the value of $h(100)$:
\[ h(100) = 2(100) + 1 = 201 \]
Step 4: Final Answer:
The value of $h(100)$ is 201.
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