If `f(x)=log(sec^(2)x)^(cot2)` for `xne0` for x=0 is continuous at x=0, then K is

Options

A. `e^(-1)`
B. `1`
C. `e`
D. 0

Answer

B

Explanation

Given `f(x) ={{:("log"(sec^(2)x)^(cot^(2)x),"for"xne0),(k,"for" x=0):}`
is continuous at `x = 0`
`f(0) = underset(xrarr0)("lim")log(sec^(2)x)^(cot^(2)x)`
`=underset(xrarr0)("lim")cot^(2)xlog(sec^(2)x)`
`[ :' log m^(n) = n log m]`
`= underset(xrarr0)("lim")'1/(tan^(2)x) log(1+tan^(2)x)`
`[ :' sec^(2)x = 1 +tan^(2)x "and" cotx = 1/(tanx)]`
`= underset(xrarr0)("lim")(log(1+tan^(2)x))/(tan^(2)x)`
`= underset(tanx rarr0)("lim")(log(1+tan^(2)x))/(tan^(2)x) = 1`
`[:' "As" x rarr 0, tan x rarr 0]`