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
- A
\(\frac{1}{3 x}+\frac{2 x^2}{3}\)
- B
\(-\frac{1}{3 x}+\frac{4 x^2}{3}\)
- C
\(-\frac{1}{x}+\frac{2}{x^2}\)
- D
\(\frac{1}{x}\)
Answer & Solution
Correct Answer
(A)
\(\frac{1}{3 x}+\frac{2 x^2}{3}\)
Step-by-step Solution
Detailed explanation
\[
\text { } \lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1
\]
\[
\begin{aligned}
& \Rightarrow & x^2 f^{\prime}(x)-2 x f(x)+1 & =0 \\
& \Rightarrow & \frac{x^2 f^{\prime}(x)-2 x f(x)}{\left(x^2\right)^2}+\frac{1}{x^4} & =0 \\
& \Rightarrow & \frac{d}{d x}\left(\frac{f(x)}{x^2}\right) & =-\frac{1}{x^4} \\
\Rightarrow & & f(x) & =c x^2+\frac{1}{3 x} \text { also } f(1)=1 \Rightarrow c=\frac{2}{3} . \\
& \text { Hence, } & f(x) & =\frac{2}{3} x^2+\frac{1}{3 x}
\end{aligned}
\]
\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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