If `f(x)=(lim)_(xvecoo)((x^2+a x+1)+x^(2n)(2x^2+x+b))/(1+x^(2n))a n d(lim)_(xvec+-1)f(x)` exist, then The value of `b` is `-1` b. 1 c. `0` d.`2`

Options

A. `-1`
B. 1
C. 0
D. 2

Answer

B

Explanation

`f(x)=underset(nrarroo)(lim)((x^(2)+ax+1)+x^(2n)(2x^(2)+x+b))/(1+x^(2n))`
`={{:(x^(2)+ax+1",",|x|lt1),(2x^(2)+x+b",",|x|gt1),((-a+b+3)/(2)",",x=-1),((a+b+5)/(2)",",x=1):}`
`underset(xrarr-1)(lim)f(x)" exists if"`
`underset(xrarr -1)(lim)f(x)=underset(xrarr-1^(+))(lim)f(x)`
`rArr" "underset(xrarr-1^(-))(lim)(2x^(2)+x+b)=underset(xrarr-1^(+))(lim)(x^(2)+ax+1)`
`rArr" "2-1+b=1-a+1`
`rArr" "a+b=1" (i)"`
`underset(xrarr1)(lim)f(x)" exists if"`
`underset(xrarr1^(-))(lim)f(x)=underset(xrarr1^(+))(lim)f(x)`
`rArr" "underset(xrarr1^(-))(lim)(x^(2)+ax+1)=underset(xrarr1^(+))(lim)(2x^(2)+x+b)`
`rArr 1+a+1=2+1+b`
`rArr a-b=1" (ii)"`
`"Solving Eqs. (i) and (ii), we get a = 1 and b=0."`