UPSC NDA I 2023 Mathematics Solved Paper

Concept:The function $f(x) = |x-2| + |3-x| + |4-x|$ is a sum of absolute value terms.Its minimum occurs at points where the derivative changes sign, which can be found by analyzing the function in intervals determined by the critical points $x=2,3,4$.Explanation:For $x<2$: each absolute value opens negatively, so $f(x) = (2-x)+(3-x)+(4-x) = 9-3x$.For $2 \le x < 3$: $f(x) = (x-2)+(3-x)+(4-x) = 5-x$.For $3 \le x < 4$: $f(x) = (x-2)+(x-3)+(4-x) = x-1$.For $x \ge 4$: $f(x) = (x-2)+(x-3)+(x-4) = 3x-9$.Let us find the value of $f(x)$ at each interval boundary and check minima:At $x=2$ (first two intervals give same value): from first interval $9-3(2)=3$, from second $5-2=3$.At $x=3$: from second $5-3=2$, from third $3-1=2$.At $x=4$: from third $4-1=3$, from fourth $3(4)-9=3$.Thus the minimum value among $3,3,2,2,3,3$ is $2$.Alternate method: graph each linear piece; the lowest point is at $x=3$ where $f(3)=2$.Answer:The minimum value of the function is $2$.