Question ID: #447
Let for some function $y=f(x)$, $\int_{0}^{x} t f(t) dt = x^2 f(x)$, $x > 0$ and $f(2)=3$. Then $f(6)$ is equal to:
- (1) 1
- (2) 2
- (3) 6
- (4) 3
Solution:
Given equation: $\int_{0}^{x} t f(t) dt = x^2 f(x)$.
Differentiate both sides with respect to $x$ using Leibniz rule:
$$ \frac{d}{dx} \left( \int_{0}^{x} t f(t) dt \right) = \frac{d}{dx} (x^2 f(x)) $$
$$ x f(x) = x^2 f'(x) + 2x f(x) $$
Since $x > 0$, we can divide by $x$:
$$ f(x) = x f'(x) + 2 f(x) $$
$$ x f'(x) = -f(x) $$
$$ \frac{f'(x)}{f(x)} = -\frac{1}{x} $$
Integrate both sides:
$$ \int \frac{f'(x)}{f(x)} dx = – \int \frac{1}{x} dx $$
$$ \ln |f(x)| = -\ln |x| + \ln C $$
$$ f(x) = \frac{C}{x} $$
Use the initial condition $f(2) = 3$:
$$ 3 = \frac{C}{2} \Rightarrow C = 6 $$
Thus, $f(x) = \frac{6}{x}$.
We need to find $f(6)$:
$$ f(6) = \frac{6}{6} = 1 $$
Ans. (1)
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