Let `g(x)=((x-1)^(n))/(logcos^(m)(x-1)),0ltxlt2` m and n integers, `m ne0, n gt0` and. If `underset(xrarr1+)(lim)g(x)=-1`, then

Options

A. `n=1, m=1`
B. `n=1, m=-1`
C. `n=2, m=2`
D. `n gt 2, m=n`

Answer

C

Explanation

We have
`f(x)=|x-1|={{:(,x-1,x ge 1),(,1-x,x lt 1):}`
`therefore ` p=Left hand derivative of f(x) at x=1
`Rightarrow p=underset(x to 1^(-))lim (f(x)-f(1))/(x-1) =underset(x to 1^(-))lim (1-x-0)/(x-1)=-1`
Now,
`underset(x to 1^(+))lim g(x)=p`
`Rightarrow underset(h to 0)lim g(1+h)=-1`
`Rightarrow underset(h to 0)lim (h^(n))/(m log cos h)=1`
`Rightarrow (1)/(m) underset(h to 0)lim (h^(n-2))/(((tan h)/(h)))=1`
`Rightarrow n=2 and "in that case "=(n)/(m)=1`
`Rightarrow m=n=2`