Definite Integration
Definite Integration
nta_pyq_2025_apr
Grade 12
Question:
Let $f:(0,\infty)\to\mathbb{R}$ be a twice differentiable function. If for some $a \neq 0$, $\displaystyle\int_0^1 f(\lambda x)\,d\lambda = af(x)$, $f(1) = 1$ and $f(16) = \dfrac{1}{8}$, then $16 - f'\!\left(\dfrac{1}{16}\right)$ is equal to ____.
Step-by-Step Solution
Key Concept: Substitute $u = \lambda x$ to convert $\int_0^1 f(\lambda x)d\lambda = \tfrac{1}{x}\int_0^x f(u)du = af(x)$; differentiate both sides to get a separable ODE in $f$.
Let $u = \lambda x$: $\int_0^1 f(\lambda x)d\lambda = \dfrac{1}{x}\int_0^x f(u)du = af(x) \Rightarrow \int_0^x f(u)du = axf(x)$.
Differentiating: $f(x) = af(x)+axf'(x) \Rightarrow \dfrac{f'(x)}{f(x)} = \dfrac{1-a}{a}\cdot\dfrac{1}{x}$.
Integrating: $\ln f(x) = \dfrac{1-a}{a}\ln x + C$. Using $f(1)=1 \Rightarrow C=0$, so $f(x) = x^{(1-a)/a}$.
Using $f(16) = 1/8$: $16^{(1-a)/a} = 2^{-3}$, so $\dfrac{4(1-a)}{a} = -3 \Rightarrow a = 4$.
$f(x) = x^{-3/4}$, $f'(x) = -\dfrac{3}{4}x^{-7/4}$.
$f'\!\left(\dfrac{1}{16}\right) = -\dfrac{3}{4}\left(\dfrac{1}{16}\right)^{-7/4} = -\dfrac{3}{4}\cdot 2^7 = -96$.
$$16 - f'\!\left(\frac{1}{16}\right) = 16-(-96) = 112.$$
Correct Answer: 112