JEE Math Practice Question
If $$f(x) = \left\{ {\begin{matrix} {{1 \over {|x|}}} & {;\,|x|\, \ge 1} \\ {a{x^2} + b} & {;\,|x|\, < 1} \\ \end{matrix} } \right.$$ is differentiable at every point of the domain, then the values of a and b are respectively :
- A.$${1 \over 2},{1 \over 2}$$
- B.$${1 \over 2}, - {3 \over 2}$$
- C.$${5 \over 2}, - {3 \over 2}$$
- D.$$ - {1 \over 2},{3 \over 2}$$
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