The relationship between a and b making f(x) continuous at x=3, were $f(x) = \begin{cases} ax + 1, & x \le 3 \\ bx + 3, & x > 3 \end{cases}$ is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $3a - 3b = 2$
$\lim\limits_{x→3^-}(ax+1)=f(3)=3a+1$
$\lim\limits_{x→3^+}(bx+3)=3b+3$
$⇒3a+1=3b+3$
so $3a-3b=2$