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If $f(x)$ is a differentiable function in the interval (0, ∞) 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 $f(3/2)$ is equal to :

Option 1)
$\frac{13}{6}$

Option 2)
$\frac{23}{18}$

Option 3)
$\frac{25}{9}$

Option 4)
$\frac{31}{18}$

Answers (1)

best_answer

As we learnt in

L - Hospital Rule -

$In \:the\:form\:of\:\:\;\frac{0}{0}\:\:and\:\:\frac{\infty }{\infty }\:\:\:we\:differentiate\:\:\frac{N^{r}}{D^{r}}\:\:separately.$

$\Rightarrow \lim_{x\rightarrow a}\:\:\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\:\:\frac{f'(x)}{g'(x)}$

- wherein

$\lim_{x\rightarrow a}\:\:\frac{\frac{d}{dx}\:f(x)}{\frac{d}{dx}\:g(x)}$

$Where \:\:f(x)\:\:and\:\:g(x)=0$

$\lim_{t\rightarrow x}\:\frac{t^{2}f\left ( x \right )-x^{2}f\left ( t \right )}{t-x}=1$

$\lim_{t\rightarrow x}\:\frac{2tf\left ( x \right )-x^{2}f'\left ( t \right )}{1}=1$

$\therefore 2xf\left ( x \right )-x^{2}f'\left ( x \right )=1$

Now, let y=f(x)

$2xy-x^{2}\frac{dy}{dx}=1$

$\Rightarrow x^{2}\frac{dy}{dx}-2xy=-1$

$\Rightarrow \frac{dy}{dx}-\frac{2}{x}y=-\frac{1}{x^{2}}$

$P=-\frac{2}{x}\:and \:Q=-\frac{1}{x^{2}}$

$\therefore \int Pdx=-2\int \frac{dx}{x}$ $=-2logx = log \frac{1}{x^{2}}$

$\therefore$ If $e^{log\frac{1}{x^{2}}}=\frac{1}{x^{2}}$

$\therefore \:Solution\: is$

$y\cdot \frac{1}{x^{2}}=\int -\frac{1}{x^{2}}\times \frac{1}{x^{2}}\:\:dx=\int \frac{1}{x^{4}}dx$

$=-\int x^{-4}dx=\frac{-x^{-4+1}}{-4+1}+C$

$\frac{y}{x^{2}}=\frac{-x^{-3}}{-3}+C$ $=C+\frac{1}{3x^{3}}$

Put, x=1, y=1

$\frac{1}{1}=C+\frac{1}{3}$

$\therefore C=1-\frac{1}{3}=\frac{2}{3}$

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

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

Put, $x=\frac{3}{2}$

$y=\frac{1}{3\times \frac{3}{2}}+\frac{2}{3}\times \frac{9}{4}$

$=\frac{2}{9}+\frac{3}{2}$ $=\frac{4+27}{18}=\frac{31}{18}$

Option 1)

$\frac{13}{6}$

This option is incorrect.

Option 2)

$\frac{23}{18}$

This option is incorrect.

Option 3)

$\frac{25}{9}$

This option is incorrect.

Option 4)

$\frac{31}{18}$

This option is correct.

Posted by

divya.saini

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