Find the values of aaa and bbb such that f(x)={ax+3x<2bx=23x−1x>2f(x) = \begin{cases} ax + 3 & x < 2 \\ b & x = 2 \\ 3x - 1 & x > 2 \end{cases}f(x)=⎩⎨⎧ax+3b3x−1x<2x=2x>2 is continuous at x=2x = 2x=2.
a=2,b=5a=2, b=5a=2,b=5
a=1,b=5a=1, b=5a=1,b=5
a=2,b=7a=2, b=7a=2,b=7
a=1,b=3a=1, b=3a=1,b=3