KCET 2026 Mathematics Solved Paper

Concept:For a function to be continuous at $x=1$, the left-hand limit, the function value, and the right-hand limit must all be equal.Explanation:Continuity at $x=1$ requires:$$\lim_{x \to 1^{-}} f(x) = f(1) = \lim_{x \to 1^{+}} f(x)$$First, find $f(1)$.For $x=1$, the definition gives $f(x) = 3x - 1$.So,$$f(1) = 3(1) - 1 = 2$$Now compute the left-hand limit.For $x<1$, we have $f(x) = ax + 7$.Thus,$$\lim_{x \to 1^{-}} f(x) = a(1) + 7 = a + 7$$Setting this equal to $f(1)$:$$a + 7 = 2$$$$a = -5$$Next, compute the right-hand limit.For $x>1$, we have $f(x) = \frac{b}{x+3}$.So,$$\lim_{x \to 1^{+}} f(x) = \frac{b}{1+3} = \frac{b}{4}$$Setting this equal to $f(1)$:$$\frac{b}{4} = 2$$$$b = 8$$Therefore, the required values are $a = -5$ and $b = 8$.Answer:$a = -5$, $b = 8$Hence, the correct option is Option D.