Question

Assertion :Let $f\left(x\right)$ be a differential on the interval $(0,\infty )$ such that $f\left(1\right)=1$ and $\underset{t\to x}{lim}\frac{t^{2}f\left(x\right)-x^{2}f\left(t\right)}{t-x}=1$ for each $x>0.$ Then $f\left(x\right)=\frac{1}{3x}+\frac{2x^2}{3}$ Reason: Differential equation of $f\left(x\right)$ is linear.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

$\underset{t\to x}{lim}\frac{t^{2}f\left(x\right)-x^{2}f\left(t\right)}{t-x}=1$

Now applying L-Hospital's rule

$\underset{t\to x}{lim}\frac{2tf\left(x\right)-x^2{f}^{\prime }\left(t\right)}{1}=1\Rightarrow 2xf\left(x\right)-x^2{f}^{\prime }\left(x\right)=1$

$f^{\prime }\left(x\right)-\frac{2}{x}f\left(x\right)=-\frac{1}{x^2},$ which is linear

Reason is true

$I.F.=e^{\int -\frac{2}{x}dx}=e^{-2\log x}=\frac{1}{x^2}$

Solution is $\frac{y}{x^2}=-\int \frac{1}{x^4}dx+c\Rightarrow \frac{y}{x^2}=\frac{x^{-3}}{3}+c$

$\frac{y}{x^2}=\frac{1}{3x^3}+c\Rightarrow \frac{y}{x^2}-\frac{1}{3x^3}=c\Rightarrow 1-\frac{1}{3}=c(\because f\left(1\right)=1)$

$\Rightarrow c=\frac{2}{3}\Rightarrow \frac{y}{x^2}=\frac{2}{3}+\frac{1}{3x^3}\Rightarrow y=\frac{2x^2}{3}+\frac{1}{3x}$

Assertion is also true and is correctly explained by reason.