CUET and UG exam JEE Main Mock Test Series 2025 Mathematics Limit and Continuity Continuity of a function
If \( f(x) = { x }^{ 2 } + \alpha \) for \( x \ge 0 \)
and \(f(x) = 2 \sqrt { { x }^{ 2 } + 1 } + \beta \quad \) for \( x < 0 \) is continuous at \( x = 0 \) and \( f\left(\cfrac{1}{2}\right)=2 \) then \( { \alpha }^{ 2 } + { \beta }^{ 2 } \) is
1
\( 3 \)
2
\( \cfrac { 8 }{ 25 } \)
3
\( \cfrac { 25 }{ 8 } \)
4
\( \cfrac { 1 }{ 3 } \)