`min[(x_1-x^2)^2+(3+sqrt(1-x1 2)-sqrt(4x_2))],AAx_1,x_2 in R ,` is `4sqrt(5)+1` (b) `3-2sqrt(2)` `sqrt(5)+1` (d) `sqrt(5)-1`

Correct Answer - B
(2) Let `y_(1)=3+sqrt(1-x_(1)^(2))andy_(2)=sqrt(4x_(2))`
`orx_(1)^(2)+(y_(1)-3)^(2)=1andy_(2)^(2)=4x_(2)`
Thus, `(x_(1),y_(1))` lies on the circle `x^(2)+(y-3)^(2)=1`.
Also, `(x_(2),y_(2))` lies on the upper half of the parabola `y^(2)=4x`.
Thus, the given expression is square of the shortest distance between the curves `x^(2)+(y-3)^(2)=1andy^(2)=4x`.
Now, the shortest distance always occurs along the common normal to the curves and the normal to the circle passes through the center of the circle.
Normal to the parabola `y^(2)=4x" is "y-mx-2m-m^(3)`. It passes through (0,3).
Therefore, `m^(3)+2m+3=0`, which has only one root, m=-1.
Hence, the required minimum distance is `sqrt(1+1)-1=sqrt(2)-1`.