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Q.Let f(x)={x2x≤1 ax+bx>1f(x) = \begin{cases} x^2 & x \le 1 \ ax + b & x > 1 \end{cases}. If ff is twice differentiable everywhere, then

a
a=2,b=−1a=2, b=-1
b
a=1,b=0a=1, b=0
c
a=2,b=1a=2, b=1
d
a=1,b=1a=1, b=1

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

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