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)

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.

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