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Stuck here, help me understand: - Let be a differentiablefunction such thatand . Then: - Limit , continuity and differentiability - JEE Main

 Let f be a differentiable function such that f'(x) = 7 - \frac{3}{4}\frac{f(x)}{x}, (x>0) and f(1)\neq 4 . Then \lim_{x\rightarrow 0^{+}} xf\left(\frac{1}{x} \right ):

  • Option 1)

    exists and equals 0

  • Option 2)

    does not exist

  • Option 3)

    exist and equals \frac{4}{7}

  • Option 4)

    exists and equals 4

Answers (1)
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A admin

 

Indeterminate forms -

The form

\frac{0}{0},\:\frac{\infty}{\infty },0\times \infty ,1^{\infty },0^{\infty} ,\infty ^{0}, \infty-\infty\ \:  are known as indeterminate form means

they do not exist directly

-

 

 

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

Given f{}'(x)=7-\frac{3}{4}\frac{f(x)}{x}, \: \: x>0

and   f(1)\neq 4

\frac{dy}{dx}+\frac{3}{4}\frac{y}{x}=7

This is in the form of L.D.E

I.F. = e^{\int P dx}=e^{\frac{3}{4}\int \frac{1}{x}dx}

      = x^{\frac{3}{4}}

Now,

y\cdot x^{\frac{3}{4}}=\int 7\cdot x^{\frac{3}{4}} dx

y\cdot x^{\frac{3}{4}}=7\frac{x\frac{7}{4}}{\frac{7}{4}}+C

f(x)=4x+Cx^{}\frac{-3}{4}

So,

f(\frac{1}{x})=\frac{4}{x}+Cx^{\frac{3}{4}}

\lim_{x\rightarrow 0^{+}}\: x\: f(\frac{1}{x})=\lim_{x\rightarrow 0^{+}}(4+C\cdot x^{\frac{7}{4}})=4

 


Option 1)

exists and equals 0

Option 2)

does not exist

Option 3)

exist and equals \frac{4}{7}

Option 4)

exists and equals 4

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