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JEE Advanced · Mathematics · 16. Limits

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 \(x>0\). Then, \(f(x)\) is

  1. A
    \(\frac{1}{3 x}+\frac{2 x^2}{3}\)
  2. B
    \(-\frac{1}{3 x}+\frac{4 x^2}{3}\)
  3. C
    \(-\frac{1}{x}+\frac{2}{x^2}\)
  4. D
    \(\frac{1}{x}\)
Verified Solution

Answer & Solution

Correct Answer

(A)
\(\frac{1}{3 x}+\frac{2 x^2}{3}\)

Step-by-step Solution

Detailed explanation

\[
\text { } \lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1
\]

\[
\begin{aligned}
& \Rightarrow & x^2 f^{\prime}(x)-2 x f(x)+1 & =0 \\
& \Rightarrow & \frac{x^2 f^{\prime}(x)-2 x f(x)}{\left(x^2\right)^2}+\frac{1}{x^4} & =0 \\
& \Rightarrow & \frac{d}{d x}\left(\frac{f(x)}{x^2}\right) & =-\frac{1}{x^4} \\
\Rightarrow & & f(x) & =c x^2+\frac{1}{3 x} \text { also } f(1)=1 \Rightarrow c=\frac{2}{3} . \\
& \text { Hence, } & f(x) & =\frac{2}{3} x^2+\frac{1}{3 x}
\end{aligned}
\]
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