For f(x)={x2x≤1ax+bx>1f(x) = \begin{cases} x^2 & x \le 1 \\ ax + b & x > 1 \end{cases}f(x)={x2ax+bx≤1x>1, find a,ba, ba,b such that fff is differentiable at x=1x=1x=1.
a=2,b=−1a=2, b=-1a=2,b=−1
a=1,b=0a=1, b=0a=1,b=0
a=2,b=1a=2, b=1a=2,b=1
a=0,b=1a=0, b=1a=0,b=1