If `f(x)=log(sec^(2)x)^(cot^2x)` for `xne0` and `k` 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(xto0)(lim)log(sec^(2)x)^(cot^(2)x)`
`=underset(xto0)(lim)cot^(2)xlog(sec^(2)x)` [`becauselogm^(n)=nlogm`]
`=underset(xto0)(lim)(1)/(tan^(2)x)log(1+tan^(2)x)`
`[because sec^(2)x=1+tan^(2)x and cotx=(1)/(tanx)]`
`=underset(xto0)(lim)(log(1+tan^(2)x))/(tan^(2)x)`
`=underset(tanx to 0)(lim)(log(1+tan^(2)x))/(tan^(2)x)=1" "[because" as "xto0,tanxto0]`