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The relationship between a and b making f(x) continuous at x=3, were f(x) = ax + 1, x≤ 3 =bx +3, x > 3 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 $3a-3b=2$ |
