JEE Main On 16 April 2018 Question 24
Question: If the function $ f $ defined as $ f(x)=\frac{1}{x}-\frac{k-1}{e^{2x}-1}, $ $ x\ne 0, $ is continuous at $ x=0, $ then the ordered pair $ (k,f(0)) $ is equal to? [JEE Main 16-4-2018]
Options:
A) (3, 1)
B) (3, 2)
C) $ ( \frac{1}{3},2 ) $
D) (2, 1)
Show Answer
Answer:
Correct Answer: A
Solution:
If the function is continuous at $ x=0. $ $ {\lim _{x\to 0}}f(x) $ exists and is equal
to $ f(0) $ $ {\lim _{x\to 0}}f(x) $ $ ={\lim _{x\to 0}}\frac{1}{x}-\frac{k-1}{e^{2x}-1} $
$ ={\lim _{x\to 0}}\frac{e^{2x}-1-kx+x}{(x)(e^{2x}-1)} $ Using Taylor?s expansion for $ e^{2x}, $ we get
$ ={\lim _{x\to 0}}\frac{(1+2x+\frac{{{(2x)}^{2}}}{2!})+\frac{{{(2x)}^{2}}}{3!}+…)-1-kx+x}{(x)((1+2x+\frac{{{(2x)}^{2}}}{2!}+\frac{{{(2x)}^{3}}}{3!}+…)-1)} $ $ ={\lim _{x\to 0}}\frac{(3-k)x+\frac{4x^{2}}{2!}+\frac{8x^{3}}{3!}+…}{(2x^{2}+\frac{4x^{3}}{2!}+\frac{8x^{4}}{3!}+…)} $
For the limit to exist, power of $ x $ in the numerator should be greator than or equal to the power of $ x $ in the denominator.
Therefore, coefficient of in numerator is equal to zero
$ \Rightarrow $ $ 3-k=0 $
$ \Rightarrow $ $ k=3 $ So the limit reduces to $ {\lim _{x\to 0}}\frac{(x^{2})(\frac{4}{2!}+\frac{8x}{3!}+…)}{(x^{2})(2+\frac{4x}{2!}+\frac{8x2}{3!}+…)} $
$ ={\lim _{x\to 0}}\frac{\frac{4}{2!}+\frac{8x}{3!}+…}{2+\frac{4x}{2!}+\frac{8x2}{3!}+…} $ =1 Hence, $ f(0)=1 $
Therefore, answer is option A.
