If f(x)=`{{:(,(1)/(x)-(2)/(e^(2x)-1),x ne 0),(,1,x=0):}`

Options

A. f(x) is differentiable at x=0
B. f(x) is not differentiable at x=0
C. `f'(0)=(1)/(3)`
D. f(x) is continuous but not differenitable at x=0

Answer

A

Explanation

We have
`underset(x to 0)lim f(x)=underset(x to 0)lim ((1)/(x)-(2)/(e^(2x)-1))`
`Rightarrow underset(x to 0)lim f(x)=underset(x to 0)lim (e^(2x)-1-2x)/(x(e^(2x)-1))`
`Rightarrow underset(x to 0)lim f(x)=underset(x to 0)lim ((1+2x+(2x)^(2)/(2!)+(2x)^(3)/(3!)+....)-1 2x)/(x(1+2x+((2x)^(2))/(2!)+......-1))`
`Rightarrow underset(x to 0)lim f(x)=underset(x to 0)lim (((2x)^(2))/(2!)+((2x)^(3))/(3!)+....)/(x(2x+((2x)^(2))/(2!)+....))`
`Rightarrow underset(x to 0)lim f(x)=((2^(2))/(2!))/(2)=1=f(0)`
So, f(x) is continuous at x=0
Now, `("LHD at x=0")=underset(h to 0)lim (f(0-h)-f(0))/(-h)`
`Rightarrow ("LHD at x=0")=underset(h to 0)lim (f(-h)-1)/(-h)`
`Rightarrow ("LHD at x=0")=underset(h to 0)lim (-(1)/(h)-(2)/(e^(-2h)-1)-1)/h`
`Rightarrow ("LHD at x=0")=underset(h to 0)lim (e^(-2h)-1+2h+h(e^(-2h-1)))/(h^(2)(e^(-2h-1)))`
`{1-2h+((2h)^(2))/(2!)-(2h)^(3)/(3!)+.....}`
`=underset(h to 0)lim (+h{-2h+((2h)^(2))/(2!)-((2h)^(3))/(3!)}+....-1+2h)/(h^(2)(-2h+(2h)^(2)/(2!)-(2h)^(3)/(3!)+.....))`
`=underset(h to 0)lim ((-8)/(3!)+(4)/(2!)h^(3)+.....)/(h^(3)(-2+(4h)/(2!).....))=-(1)/(3)`
Similarly, we have
`("RHD at x=0")=(1)/(3)`
`therefore ("LHD at x=0")=("RHD at x=0")`
Thus, f(x) is differentiable at x=0