CUET Preparation Today
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 : |
3a - 3b = 2 a - b = 4 a + b + 4 = 0 a + b = 0 |
3a - 3b = 2 |
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 $3a-3b=2$ |
