If `f(x)={(1-cos(1-cos x/2))/(2^m x^n)1x=0,x!=0` is continous at `x=0` then the value of `m+n` is a. 2 b. 3 c. -3 d. 7

Options

A. 2
B. 3
C. `-3`
D. 7

Answer

C

Explanation

`f(x)={{:((1-cos(1-cos.(x))/(2))/(2^(m)x^(n))",",xne0" is continuous at "x=0),(1",",x-0):}`
Hence, `underset(xrarr0)(lim)([1-cos(1-cos.(x)/(2))])/((1-cos.(x)/(2))^(2))(1-cos.(x)/(2))^(2)/(2^(m).x^(n))=1`
`rArr" "underset(xrarr0)(lim)((1-cos.(x)/(2))^(2))/(2^(m)x^(n))=2" "["Using "underset(thetararr0)(lim)(1-costheta)/(theta^(2))=(1)/(2)]`
`rArr" "underset(xrarr0)(lim)(4sin^(4).(x)/(4))/(2^(m)x^(n))=2`
`rArr" "underset(xrarr0)(lim)(4x^(4))/(4^(4)2^(m)x^(n))=2`
For limit to exist n = 4.
`therefore" "(2)/(2^(8+m))=1`
`therefore" "2^(8+m)=2`
`therefore" "m=-7`
`rArr" "m+n=-3`