UPSC NDA I 2023 Mathematics Solved Paper

Concept:The function uses modulus to create a piecewise linear expression, and the minimum occurs where the sum of distances is smallest.Explanation:We have $f(x)=|x-2|+|3-x|+|4-x|$ for real $x$.The critical points are $x=2, 3, 4$.For $x<2$: $f(x)=2-x+3-x+4-x=9-3x$.This is decreasing for $x<2$, so the smallest in this region is near $x=2$, giving $f(2^-)=3$.For $2\le x<3$: $f(x)=x-2+3-x+4-x=5-x$.It is decreasing from $x=2$ to $x=3$, so the minimum in this interval is at $x=3^-$, giving $f(3^-)=2$.For $3\le x<4$: $f(x)=x-2+x-3+4-x=x-1$.It is increasing from $x=3$ to $x=4$, so the minimum is at $x=3$, giving $f(3)=2$.For $x\ge4$: $f(x)=x-2+x-3+x-4=3x-9$.It is increasing for $x\ge4$, so the smallest is at $x=4$, giving $f(4)=3$.Thus the overall minimum value of $f(x)$ is $2$, and it occurs at $x=3$.Answer:Option B: 3.