If the image of an object having length x1 formed by a convex mirror has…

A convex mirror is a diverging mirror: it curves outward, so the reflecting surface spreads the reflected rays apart instead of bringing them together. As a direct consequence, for any position of a real object placed in front of a convex mirror, the image formed is virtual, erect, and always diminished — smaller than the object itself. This is a fixed property of convex mirrors that does not depend on how far the object is from the mirror, unlike a concave mirror, where the image size changes with the object's position. Here, the object has length x1 and its convex-mirror image has length x2. Applying the rule above, the image must always be smaller than the object, so x2 is always less than x1 — that is, x1 > x2, no matter where the object is placed. This can be cross-checked using the mirror magnification relation: for a convex mirror the image distance always works out smaller in magnitude than the object distance for every real object position, so the magnification's magnitude is always less than 1. Since image length equals magnification magnitude times object length, x2 always comes out smaller than x1, confirming x1 > x2. Note that the statement "x1 ≠ x2" is also technically consistent with this result, but it is a weaker, less specific claim than what is asked — the question calls for the definite comparison between the lengths, which is the stronger and complete statement x1 > x2.