If `f(x)={{:(x^(2)+alpha," for "xge0),(2sqrt(x^(2)+1)+beta," for "xlt0):}` is continuous at x=0 and `f((1)/(2))=2`, then `alpha^(2)+beta^(2)` is

Options

A. 3
B. `(8)/(25)`
C. `(25)/(8)`
D. `(1)/(3)`

Answer

C

Explanation

We have,
`f(x)={{:(x^(2)+alpha",",xge0),(2sqrt(x^(2)+1)+beta",",x lt 0):}`
Is continuous at x=0,
`therefore underset(xto0^(-))(lim)f(x)=underset(xto0^(+))(lim)f(x)=f(0)`
`implies underset(x to 0^(-))(lim)" "x^(2)+alpha=underset(x to 0^(+))(lim) 2sqrt(x^(2)+1)+beta=alpha" "[becausef(0)=alpha]`
`impliesalpha=2+beta` . . . (i)
`implies alpha-beta=2`
Given, `f((1)/(2))=2`
`f((1)/(2))=((1)/(2))^(2)+alpha" "[because(1)/(2)gt0]`
`implies2=(1)/(4)+alpha`
`implies alpha=(7)/(4)`
On putting the value of `alpha` in Eq. (i), we get
`beta=(7)/(4)-2=-(1)/(4)`
Hence, `alpha^(2)+beta^(2)=((7)/(4))^(2)+((-1)/(4))^(2)`
`=(49+1)/(16)`
`=(50)/(16)=(25)/(8)`