MCQ
Let $\mathrm{f}:(0, \infty) \rightarrow \mathbf{R}$ be a function which is differentiable at all points of its domain and satisfies the condition $x^{2} f^{\prime}(x)=2 x f(x)+3$, with $f(1)=4$. Then $2 f(2)$ is equal to:
- A29
- B19
- ✓39
- D23
✓
Answer
Correct option: C.
39
(C)
Sol. $x^{2} f^{\prime}(x)-2 x f(x)=3$
$\left(\frac{x^{2} f^{\prime}(x)-2 x f(x)}{\left(x^{2}\right)^{2}}\right)=\frac{3}{\left(x^{2}\right)^{2}}$
$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{f}(\mathrm{x})}{\mathrm{x}^{2}}\right)=\frac{3}{\mathrm{x}^{4}}$
Integrating both sides
$\frac{f(x)}{x^{2}}=-\frac{1}{x^{3}}+C$
$f(x)=-\frac{1}{x}+C x^{2}$
put $\mathrm{x}=1$
$4=-1+C \Rightarrow C=5$
$f(x)=-\frac{1}{x}+5 x^{2}$
Now $2 \times f(2)=2 \times\left[-\frac{1}{2}+5 \times 2^{2}\right]$
$=39$
Sol. $x^{2} f^{\prime}(x)-2 x f(x)=3$
$\left(\frac{x^{2} f^{\prime}(x)-2 x f(x)}{\left(x^{2}\right)^{2}}\right)=\frac{3}{\left(x^{2}\right)^{2}}$
$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{f}(\mathrm{x})}{\mathrm{x}^{2}}\right)=\frac{3}{\mathrm{x}^{4}}$
Integrating both sides
$\frac{f(x)}{x^{2}}=-\frac{1}{x^{3}}+C$
$f(x)=-\frac{1}{x}+C x^{2}$
put $\mathrm{x}=1$
$4=-1+C \Rightarrow C=5$
$f(x)=-\frac{1}{x}+5 x^{2}$
Now $2 \times f(2)=2 \times\left[-\frac{1}{2}+5 \times 2^{2}\right]$
$=39$
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