Q61 / 170 · Manipal (MET) 2020 · Mathematics
Let $f(x)$ be differentiable on the interval $(0, \infty)$ such that $f(1)=1$ and
$\lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$ for each $x\gt0$
Then, $f(x)$ is equal to
- Correct answerA$\frac{1}{3 x}+\frac{2}{3} x^2$
- B$-\frac{x}{3}+\frac{4 x^2}{3}$
- C$-\frac{1}{x}$
- D$-\frac{1}{x}+\frac{2}{x^2}$
Solution
Given, $\lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$
Using L' Hospital's rule, we get
$\lim _{t \rightarrow x} \frac{t^2 f^{\prime}(x)-2 x f(t)}{-1}=1$
$\Rightarrow \quad x^2 f^{\prime}(x)-2 x f(x)+1=0$
$\Rightarrow \quad \frac{x^2 f^{\prime}(x)-2 x f(x)}{\left(x^2\right)^2}+\frac{1}{x^4}=0$
$\Rightarrow \quad \frac{d}{d x}\left(\frac{f(x)}{x^2}\right)=-\frac{1}{x^4}$
On integrating both sides, we get
$\frac{f(x)}{x^2}=+\frac{1}{3 x^3}+c \Rightarrow f(x)=\frac{1}{3 x}+c x^2$
Also, $\quad f(1)=1 \Rightarrow 1=\frac{1}{3 \times 1}+c(1)^2$
$\Rightarrow \quad 1=\frac{1}{3}+c \Rightarrow \frac{2}{3}=c$
$\therefore \quad f(x)=\frac{1}{3 x}+\frac{2 x^2}{3}$
Hence, the correct answer is option: 'A'.