UPSC NDA I 2023 Mathematics Solved Paper

Concept:For a piecewise function to be continuous at a boundary point, the left-hand limit must equal the right-hand limit and the function value at that point.Explanation:Given $f(x) = \begin{cases} x+6, & x\le 1 \\ px+q, & 1 < x<2 \\ 5x, & x\ge 2 \end{cases}$ is continuous.At $x=1$: Left-hand limit = $1+6 = 7$.Right-hand limit = $p(1)+q = p+q$.For continuity: $p+q = 7$.   (Equation 1)At $x=2$: Left-hand limit = $p(2)+q = 2p+q$.Right-hand limit = $5(2) = 10$.For continuity: $2p+q = 10$.   (Equation 2)Subtract Equation 1 from Equation 2:$(2p+q) - (p+q) = 10 - 7$$p = 3$.Substitute $p=3$ into Equation 1:$3+q = 7 \implies q = 4$.Answer:$4$ (Option C).