Q.Write the difference between real valued function and real function.

A handful of standard functions appear again and again, and JEE Main tests their exact behaviour — their values, domains, ranges, graphs, and the equations built from them. Know each one precisely; the traps live in the fine print (especially at negatives and at integer boundaries).

1 — Greatest-integer (floor) [x]. [x] is the greatest integer ≤ x. So [2.7] = 2, [5] = 5, and — the classic trap — [−2.3] = −3 (the greatest integer not exceeding −2.3, which is −3, not −2). Domain R, range Z. Key facts: [x] ≤ x < [x] + 1; [x + n] = [x] + n for any integer n; and [x] + [−x] = 0 if x is an integer, −1 otherwise. Its graph is a staircase (constant on each [n, n+1), jumping up by 1 at each integer).

2 — Fractional-part {x}. Defined by {x} = x − [x], so it is the "leftover" after the floor. Always 0 ≤ {x} < 1. Example: {2.7} = 0.7, and {−2.3} = −2.3 − (−3) = 0.7. Domain R, range [0, 1). It is periodic with period 1 ({x + 1} = {x}); its graph is a sawtooth. Note {x} = 0 exactly when x ∈ Z.

3 — Modulus |x|. |x| = x if x ≥ 0, −x if x < 0. Domain R, range [0, ∞); graph is a V with vertex at the origin. It is an even function. Useful identity: x = |x|·sgn(x), and x/|x| = sgn(x) for x ≠ 0.

4 — Signum sgn(x). sgn(x) = −1 for x < 0, 0 for x = 0, +1 for x > 0 (equivalently |x|/x for x ≠ 0, and 0 at 0). Domain R, range {−1, 0, 1}; an odd function whose graph is two horizontal rays with a point at the origin.

5 — Domains built from these. When a special function sits inside a root or a denominator, solve its own inequality:

• √([x] − 2) needs [x] − 2 ≥ 0 ⇒ [x] ≥ 2 ⇒ x ≥ 2.
• 1/([x] − 3) is undefined where [x] = 3, i.e. on [3, 4) — exclude that whole interval.
• 1/{x} (and 1/√{x}) is undefined where {x} = 0, i.e. at every integer ⇒ domain R − Z.
• √(x − [x]) = √{x} is defined for all real x ({x} ≥ 0 always).

6 — Equations with [x], {x}, |x|.

• [x] = k (integer k) means x ∈ [k, k+1) — an interval, not a single value.
• Treat [x] as an integer unknown: [x]² − 5[x] + 6 = 0 ⇒ [x] = 2 or 3 ⇒ x ∈ [2, 4).
• {x} = c (with 0 ≤ c < 1) has solutions x = n + c for every integer n. …