Q.Let $f$ be a real function. Show that $h(x) = f(x) + f(-x)$ is always an even function and $g(x) = f(x) - f(-x)$ is always an odd function. Also express $e^x + \sin x$ as the sum of an even function and an odd function.

A function is a relation with one rule broken to none: every input must land on exactly one output, and the whole classification game begins by checking that first. A function f: A → B is a relation from A to B in which each a ∈ A is paired with precisely one f(a) ∈ B — no input left unmapped and no input sent to two places. This is the test hiding inside "is it a function?" traps: a mapping diagram fails if some element of A has no arrow or two arrows; the curve x = y² fails as a function of x because a single positive x gives two y values, whereas y = x² passes; |y| = x and x² + y² = 1 both fail the one-output rule; and a piecewise rule that assigns two different values at a shared boundary point is not a function there. Said graphically, a curve is a function of x exactly when no vertical line meets it more than once. Everything below — one-one, onto, invertible, even/odd — presumes you have already cleared this gate; A is the domain, B the codomain, and the range {f(a) : a ∈ A} is the subset of B actually hit.

One-one (injective) means distinct inputs never collide, and you have three interchangeable weapons to prove it — pick the cheapest for the function in front of you. The definition is a₁ ≠ a₂ ⇒ f(a₁) ≠ f(a₂), which in practice you use in its contrapositive form: assume f(x₁) = f(x₂) and force x₁ = x₂. For (x−1)/(x+1) or a general (ax+b)/(cx+d) the cross-multiplication collapses to x₁ = x₂ in one line; for x/(1+|x|) you split on the sign of x and check the pieces plus the crossover; for x³ + x the algebra x₁³ − x₂³ + x₁ − x₂ = 0 factors to (x₁ − x₂)(x₁² + x₁x₂ + x₂² + 1) = 0 with the second bracket always positive, so x₁ = x₂. When the algebra is ugly the graphical test is faster: f is one-one exactly when no horizontal line meets its graph more than once. If instead two distinct inputs share an output — 1/x² satisfies f(x) = f(−x), x² satisfies f(x) = f(−x), eˣ + e⁻ˣ is symmetric — the function is many-one, and the collision often shows up as f(x₁) = f(x₂) forcing a relation like x₁ + x₂ = constant rather than x₁ = x₂.

The most powerful injectivity tool is monotonicity, and its calculus face — a derivative of fixed sign — turns most Hard one-one problems into a single inequality. A strictly increasing or strictly decreasing function is automatically one-one, because a genuine order on inputs forces a genuine order on outputs, ruling out repeats. When f is differentiable this becomes: if f′(x) > 0 for all x in the domain (or f′(x) < 0 throughout) then f is one-one. So x³ + kx + c is one-one on ℝ precisely when f′(x) = 3x² + k ≥ 0 everywhere, i.e. k ≥ 0; x + sin x is one-one on ℝ because f′ = 1 + cos x ≥ 0 with only isolated zeros; and for 2x³ − 9x² + 12x + 1 you read off the intervals of injectivity directly from where f′ = 6(x−1)(x−2) keeps one sign. The converse warning matters for the trap questions: monotonic ⇒ one-one, but one-one does not ⇒ monotonic — a function can be injective while jumping around discontinuously — so "every one-one function is increasing" is a false statement planted to catch you. If f′ changes sign the function turns back on itself and is many-one on any interval spanning the turning point.

Onto (surjective) is not a property of the formula alone — it is a verdict about the codomain — so you decide it by comparing the range with the stated B. f: A → B is onto when range = codomain, meaning every y ∈ B is f(x) for some x ∈ A; if the range is a proper subset of B the function is into. Operationally you either compute the range and match it against B, or solve f(x) = y for x in the domain and check a solution exists for every y in B. Thus f: ℝ → ℝ, f(x) = x² is into because the range [0, ∞) misses the negatives; f(x) = eˣ is into on ℝ → ℝ since its range is (0, ∞); x/(1+|x|): ℝ → [−1, 1] is into because the range is the open interval (−1, 1); and (x²+1)/(x²+2) has range only a subinterval of ℝ, so it is into. By contrast x³: ℝ → ℝ is onto because every real is a cube. The reliable fix is structural: restricting the codomain to equal the range always makes a function onto, which is why "which codomain makes f onto?" is answered simply by naming the range.

A bijection is the two tests passed together, and it is the exact hinge on which everything invertible turns. f is bijective when it is both one-one and onto — every output is hit exactly once. This is the workhorse single-correct exercise: given an explicit f with a stated domain and codomain, place it in one of four boxes — one-one onto (bijection), one-one into, many-one onto, many-one into — by running an injectivity test and a surjectivity test independently. So 2x − 3: ℝ → ℝ is a bijection while x²: ℝ → ℝ is many-one into; f(n) = 2n on ℤ → ℤ is one-one into while ⌊n/2⌋ is many-one onto; x²: ℝ⁺ → ℝ⁺ becomes a bijection once the domain is cut to the positives, showing how a domain restriction changes the class; and x/(1+|x|): ℝ → (−1, 1) is a bijection because it is both injective and, onto that open interval, surjective. The domain and codomain are decisive, never the algebraic expression on its own — which is why the flagship multiple-correct item lists the same formula twice with different domains and asks you to judge each.

Parameter problems are just the bijection tests read backwards: solve for the values of a, b, k, λ that switch a property on. For injectivity you impose a monotonicity or discriminant condition; for surjectivity you match range to codomain; for a bijection both at once. So x³ − kx is one-one on ℝ iff f′ = 3x² − k ≥ 0 everywhere, i.e. k ≤ 0; ax + b is a bijection on ℝ iff a ≠ 0; a quadratic like x² − 2ax restricted to [a, ∞) is made onto its codomain by choosing a so the minimum value (attained at the vertex x = a) equals the codomain's lower endpoint; and for (x² + ax + 1)/(x² + x + 1) one-one you force the range condition that makes the function monotone. These frequently arrive as integer answers — "the number of integer values of a for which f is a bijection" — or as multiple-correct selections of the λ that keep λx + sin x one-one (f′ = λ + cos x single-signed needs |λ| ≥ 1). Sitting alongside is the integer "how many among this listed set" counter: classify each listed function individually and count those that are one-one, onto, bijective, even, odd or self-inverse — which is not the nⁿ or nPm formula for counting all maps between two sets, a separate concept entirely.

Invertibility has a one-line criterion — f is invertible if and only if it is bijective — and once you know that, building f⁻¹ is a mechanical solve. A bijection pairs A and B one-to-one and onto, so the pairing can be run in reverse: f⁻¹: B → A satisfies f⁻¹(f(x)) = x and f(f⁻¹(y)) = y, with domain(f⁻¹) = range(f) = B and range(f⁻¹) = A. To construct it, set y = f(x), solve for x in terms of y, then swap symbols to write f⁻¹(x). For (2x+3)/(x−1) the algebra returns another rational function; for x/(1+|x|): ℝ → (−1, 1) the inverse is f⁻¹(y) = y/(1−|y|); for 1 + √(x−2) on [2, ∞) the inverse is (x−1)² + 2 with domain the range of f; for log(x + √(x²+1)) the inverse is (eˣ − e⁻ˣ)/2; and eˣ − e⁻ˣ inverts through a quadratic in eˣ. Geometrically the graph of f⁻¹ is the reflection of the graph of f in the line y = x, and (f⁻¹)⁻¹ = f. A many-one function like x² has no inverse on ℝ, but restricting it to a monotone branch and cutting the codomain to the range revives one.

Restricting a parabola to a single arm is the standard way to force invertibility, and the only real skill is picking the correct sign of the square root. A quadratic is many-one on ℝ because its two arms mirror each other about the vertex, so you restrict to one monotone branch — [k, ∞) or (−∞, k] where k is the vertex abscissa — and adjust the codomain to the range so the map is bijective. For f(x) = x² − 2x + 3 = (x−1)² + 2 on [1, ∞) → [2, ∞) the inverse is f⁻¹(x) = 1 + √(x−2), the plus sign chosen because the outputs of f⁻¹ must land in [1, ∞); the same parabola on (−∞, 1] gives the other branch f⁻¹(x) = 1 − √(x−2). The domain of the inverse is always the range of the restricted quadratic, an interval [minimum, ∞), and the restricted domain of f is what resolves the ± ambiguity — get the interval right and the sign follows automatically.

Compositions invert in reverse order — (f∘g)⁻¹ = g⁻¹∘f⁻¹ — because you must undo the last operation first, and reversing the wrong way is the built-in trap. If you apply g then f, then to undo you must strip off f before g, so the inverse of the composite reverses the order of the factors: (f∘g)⁻¹ = g⁻¹∘f⁻¹, never f⁻¹∘g⁻¹. Combined with (f⁻¹)⁻¹ = f this lets you invert composites, evaluate (g∘f)⁻¹ at a point, recover an unknown factor from a stated inverse relation, or invert a chain of three bijections by peeling them off outermost-first. The same idea answers "find h with f∘h = identity": h is exactly f⁻¹. In the correct-statements sets the whole test is often just whether a candidate remembered to reverse the order rather than writing f⁻¹∘g⁻¹. …