NIMCET 2009 — Mathematics PYQ

The problem defines a non-standard distance metric often referred to as the Chebyshev distance or $L_{\infty }$ metric. We need to find the locus of points satisfying the condition:

1. Understanding the Equation

The equation $\max (|x|,|y|)=1$ implies that either $|x|=1$ (while $|y|\le 1$) or $|y|=1$ (while $|x|\le 1$).

2. Breaking Down the Boundaries

Let's analyze the four possible boundary conditions:

• If $|x|=1$: This gives two vertical lines $x=1$ and $x=-1$. For these points to be part of the locus, we must have $|y|\le 1$, meaning $y$ ranges from $-1$ to $1$.

• If $|y|=1$: This gives two horizontal lines $y=1$ and $y=-1$. For these points to be part of the locus, we must have $|x|\le 1$, meaning $x$ ranges from $-1$ to $1$.

3. Identifying the Shape

These four segments form the boundary of a square with vertices at:

4. Calculating the Area

• The length of each side of this square is the distance from $x=-1$ to $x=1$ (or $y=-1$ to $y=1$).

• Side length ($s$) = $1-(-1)=2$ units.

• Area of the square = $s^2=2^2=4$ sq units.

Conclusion

The locus is a square with side length 2 and total area 4 square units.

Correct Option:

(d) a square of area 4 sq units