Let f(x) be differentiable on the interval (0, ∞) such that f(1)=1 , and _ t arrow x (t^2 f(x)-x^2 f(t))/(t-x)=1 for each x>0 . Then, f(x) is

Correct answer: A. (1)/(3 x)+(2 x^2)/(3) _ t arrow x (t^2 f(x)-x^2 f(t))/(t-x)=1 aligned & & x² f^ (x)-2 x f(x)+1 & =0 & & x^2 f^ (x)-2 x f(x) (x²)^2 +(1)/(x^4)& =0 & & (d)/(d x)((f(x))/(x^2))& =-(1)/(x^4) & & f(x) & =c x^2+(1)/(3 x) also f(1)=1 c=(2)/(3) . & Hence, & f(x) & = 2 3 x^2+(1)/(3 x) aligned

JEE Advanced 2007Paper 1

Exam: JEE Advanced
Year: 2007
Paper: Paper 1
Subject: Mathematics
Chapter: Limits

JEE Advanced 2007 · Paper 1 · Mathematics · Limits

Let f(x) be differentiable on the interval (0, ∞) such that f(1)=1...

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)

Options

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 $\frac{1}{x}$

Correct answer

A. $\frac{1}{3 x}+\frac{2 x^2}{3}$

Step-by-step solution

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

\begin{aligned}& \Rightarrow & x² f^{\prime}(x)-2 x f(x)+1 & =0 \\ & \Rightarrow & \frac{x^2 f^{\prime}(x)-2 x f(x)}{ \left(x²\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) & =\ffrac{2}{3}x^2+\frac{1}{3 x} \end{aligned}

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