1.

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

Answer» Correct Answer - C
We have,
`f(x)={(x^(2)+alpha",",xge0),(2sqrt(x^(2)+1)+beta",",xlt0):}`
is continuous at x=0.
`:.underset(xto0^(-))(lim)f(x)=underset(xto0^(+))(lim)f(x)=f(0)`
`rArrunderset(xto0^(-))(lim)x^(2)+alpha=underset(xto0^(+))(lim)2sqrt(x^(2)+1)+beta=alpha" "[becausef(0)=alpha]`
`rArr" "alpha=2+beta`
`rArralpha-beta=2` . . . (i)
Given, `f((1)/(2))=2`
`f((1)/(2))=((1)/(2))^(2)+alpha" "[because(1)/(2)lt0]`
`rArr2=(1)/(4)+alpha`
`rArralpha=(7)/(4)`
On putting the value of 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)`


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