Let `f (x)= (e ^(tan x) -e ^(x) +ln (sec x+ tan x)-x)/(tan x-x)` be a continous function at `x=0.` The value of `f (0)` equals:

Options

A. `(1)/(2)`
B. `(2)/(3)`
C. `(3)/(2)`
D. 2

Answer

C

Explanation

For f(x) to be continuous at x=0, we must have
`underset(x to 0)lim (e^(tan x)-e^(x)+In (sec x+tan x)-x)/(tan x-x)=f(0)`
`Rightarrow f(0)=underset(x to 0)lim e^(x)((e^(tan x-x)-1))/(tan x-x)+underset(x to 0)lim ("In"(sec x+tan x))/(tan x-x)`
`Rightarrow f(0)=underset(x to 0)lim e^(x)((e^(tan x-x)-1))/(tan x-x)+underset(x to 0)lim (log(secx+tanx)-x)/(x^(3)((tan x-x)/(x^(3))))`
`Rightarrow f(0)=1xxe^(0)+3 underset(x to 0)lim ("In"(sec x+tanx )-x)/(x^(3)) [therefore underset(x to 0)lim (tan x-x)/(x^(3))=(1)/(3)]`
`Rightarrow f(0)=1+3 underset(x to 0)lim (sec x-1)/(3x^(2))" "["By L' Hospital's rule"]`
`Rightarrow f(0)=1+3 underset(x to 0)lim (1-cos x)/(3 cos x.x^(2))`
`Rightarrow f(0)=1+3xx(1)/(3)xx(1)/(2)=(3)/(2)`