Question ID: #391
Let $f:(0,\infty)\rightarrow R$ be a function which is differentiable at all points of its domain and satisfies the condition $x^{2}f^{\prime}(x)=2xf(x)+3$ with $f(1)=4$. Then $2f(2)$ is equal to:
- (1) 29
- (2) 19
- (3) 39
- (4) 23
Solution:
We are given the differential equation:
$$ x^2 f'(x) – 2x f(x) = 3 $$
Dividing the entire equation by $x^4$ to put it in the form of the quotient rule derivative:
$$ \frac{x^2 f'(x) – 2x f(x)}{x^4} = \frac{3}{x^4} $$
Recognizing the left side as the derivative of $\frac{f(x)}{x^2}$:
$$ \frac{d}{dx}\left( \frac{f(x)}{x^2} \right) = 3x^{-4} $$
Integrating both sides with respect to $x$:
$$ \frac{f(x)}{x^2} = \int 3x^{-4} dx $$
$$ \frac{f(x)}{x^2} = 3 \left( \frac{x^{-3}}{-3} \right) + C $$
$$ \frac{f(x)}{x^2} = -\frac{1}{x^3} + C $$
Multiplying by $x^2$ to isolate $f(x)$:
$$ f(x) = -\frac{1}{x} + Cx^2 $$
Using the initial condition $f(1) = 4$:
$$ 4 = -\frac{1}{1} + C(1)^2 $$
$$ 4 = -1 + C \Rightarrow C = 5 $$
The function is therefore:
$$ f(x) = -\frac{1}{x} + 5x^2 $$
We need to find the value of $2f(2)$. First, calculate $f(2)$:
$$ f(2) = -\frac{1}{2} + 5(2)^2 = -\frac{1}{2} + 20 = \frac{39}{2} $$
$$ 2f(2) = 2 \times \frac{39}{2} = 39 $$
Ans. (3)
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