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  • Solve! - Limit , continuity and differentiability - JEE Main-2

If the function f defined as  f(x) = \frac{1}{x}- \frac{k-1}{e^{2x}-1} , x\neq 0

is continuous at  x=0, then the ordered pair (k, f (0)) is equal
to :

  • Option 1)

    (3, 2)

  • Option 2)

    (3, 1)

  • Option 3)

    (2, 1)

  • Option 4)

    ( 1/3,2)

 

Answers (2)

best_answer

As we have learned

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

 

 

Continuity at a point -

A function f(x)  is said to be continuous at  x = a in its domain if 

1.  f(a) is defined  : at  x = a.

2. \lim_{x\rightarrow a}\:f(x)\:exists\:means\:limit\:x\rightarrow a

of  f(x) at  x = a exists from left and right.

3. \lim_{x\rightarrow a}\:f(x)=f(a)\:then\:the\:limit\:equals \:the\:value\:at\:x=a

-

\lim_{x\rightarrow 0}\frac{1}{x}-\frac{k-1}{e^{2x}-1}

 

 \lim_{x\rightarrow 0}\frac{\left ( (e^{2x}-1)-x (k-1)\right )}{x(e^{2x}-1)}

By using L' hospital rule 

\frac{2e^{2x}-1(k-1)}{e^{2x}-1+2xe^{2x}}

put k=3; 

f(0)=1

 

 

 

 

 


Option 1)

(3, 2)

This is incorrect

Option 2)

(3, 1)

This is correct

Option 3)

(2, 1)

This is incorrect

Option 4)

( 1/3,2)

This is incorrect

Posted by

Himanshu

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