If f(x)`=x^(2)+a`, for `xge0`
`=2sqrt(x^(2)+1)+b,"for"xltandf((1)/(2))=2`, is continuous at x = 0 , find a and b .

Explanation

`f(x)=x^(2)+a, "for" x ge0`
`therefore((1)/(2))=((1)/(2))^(2)+a=1/4+a`
But `f((1)/(2))=2 therefore1/4+a=2 thereforea=7/4`
`underset(x to 0^(+))limf(x)=underset(x to 0)lim(x^(2)+a)`
`=0+a=a=7/4`
`underset(x to 0^(-))limf(x)=underset(x to 0)lim"("2sqrt(x^(2)+1)+b`
`=2 sqrt(0+1)+b=2 +b`
Now, f is continous at `x=0`
`thereforeunderset(x to 0^(-))lim f(x) =underset(x to 0^(+))limf(x)`
`therefore2+b=7/4therefore b=-1/4`
Hence, `a=7/4 therefore b=-1/4.`