Q44 / 79 · NEST (NISER) 2026 · Mathematics
Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a twice differentiable function such that $f''(x) = -f(x)$. Let $h : \mathbb{R} \rightarrow \mathbb{R}$ 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 $h\left(\dfrac{1}{3}\right) = \dfrac{5}{3}$, then the value of $h(100)$ is
- A$100$
- Correct answerB$201$
- C$202$
- D$101$
Solution
Given $f''(x) = -f(x)$
Multiplying both sides by $2f'(x)$:
$2f'(x)f''(x) + 2f(x)f'(x) = 0$
Integrating with respect to $x$:
$(f'(x))^2 + (f(x))^2 = C$, where $C$ is a constant.
Given $h'(x) = (f(x))^2 + (f'(x))^2$
$h'(x) = C$
Integrating with respect to $x$:
$h(x) = Cx + D$, where $D$ is a constant.
Given $h(0) = 1$, we get $D = 1$.
Given $h\left(\dfrac{1}{3}\right) = \dfrac{5}{3}$, we get:
$C\left(\dfrac{1}{3}\right) + 1 = \dfrac{5}{3} \Rightarrow C = 2$
Thus, $h(x) = 2x + 1$.
For $x = 100$:
$h(100) = 2(100) + 1 = 201$
Answer: $201$
Multiplying both sides by $2f'(x)$:
$2f'(x)f''(x) + 2f(x)f'(x) = 0$
Integrating with respect to $x$:
$(f'(x))^2 + (f(x))^2 = C$, where $C$ is a constant.
Given $h'(x) = (f(x))^2 + (f'(x))^2$
$h'(x) = C$
Integrating with respect to $x$:
$h(x) = Cx + D$, where $D$ is a constant.
Given $h(0) = 1$, we get $D = 1$.
Given $h\left(\dfrac{1}{3}\right) = \dfrac{5}{3}$, we get:
$C\left(\dfrac{1}{3}\right) + 1 = \dfrac{5}{3} \Rightarrow C = 2$
Thus, $h(x) = 2x + 1$.
For $x = 100$:
$h(100) = 2(100) + 1 = 201$
Answer: $201$
Hence, the correct answer is option: 'B'.