MathematicsRelations and FunctionsJEE

Relations and Functions · Part 6 of 20

One-One, Many-One, Onto and Into Functions

Two questions about any function: does each output come from only one input (one-one), and is every element of the codomain reached (onto)? Algebraic tests, the horizontal line test, monotonicity and the role of the codomain.

Builds on: Part 5 · Algebra of Functions and Composite Functions.

One-One, Many-One, Onto and Into Functions

Video coming soon

Injective

One-one vs many-one

f is one-one if $f (x_{1}) = f (x_{2}) \Rightarrow x_{1} = x_{2}$ . 3x + 5 is one-one; x² + 1 is many-one because f(1) = f(−1) = 2.

Two arrow diagrams. — A shared output makes f many-one.

Parabola cut twice by a horizontal line. — Vertical lines: function? Horizontal: one-one?

Horizontal line test

x² + 1 fails

f is one-one exactly when no horizontal line meets its graph twice. y = 2 meets y = x² + 1 at x = ±1. A strictly monotonic function always passes.

Surjective

Onto vs into

Onto: range = codomain. Into: something in the codomain is missed. sin x : R → R is into (range [−1, 1]); with codomain [−1, 1] it is onto, but still many-one.

Graph rising towards y = 1. — Range marked on the y-axis.

Find the codomain

x²/(x² + 1) onto A

x² = y/(1 − y) ≥ 0 gives 0 ≤ y < 1, so A = [0, 1). The graph approaches 1 but never reaches it.

Classify

(x² − 4)/(x² + 1)

Even, so many-one. f = 1 − 5/(x² + 1) has range [−4, 1) ≠ R, so it is into.

Graph with horizontal line and range. — y = −3/2 is hit at x = ±1.

Worked examples

From the video

1. f : R → R, f(x) = sin x. One-one? Onto? Codomain that makes it onto?

Many-one and into; codomain [−1, 1].

2. f : R → A, f(x) = x²/(x² + 1) is surjective. Find A.

[0, 1).

3. Classify f : R → R, f(x) = (x² − 4)/(x² + 1).

Many-one and into.

4. Is sin x one-one on N?

Yes: for m ≠ n, sin m = sin n needs m ± n to be a non-zero multiple of π, impossible since π is irrational.

JEE-style question

Your turn

f : R → [0, ∞), f(x) = x². Then f is:

(a)one-one and onto

(b)many-one and onto

(c)one-one and into

(d)many-one and into

Show the answer and the traps

f(2) = f(−2), so many-one. The range [0, ∞) equals the codomain, so onto. Option (b).

(a) and (c) miss that 2 and −2 share an output.

(d) treats the codomain as R. Here it is [0, ∞): always read the codomain.

Watch out

Common mistakes

Saying a function is onto without its codomainx² is into on R → R but onto on R → [0, ∞).

Confusing the two line testsVertical: is it a function? Horizontal: is it one-one?

Including the asymptote value in the rangex²/(x² + 1) never equals 1.

Practice

Try these

1. Is f : R → R, f(x) = x³ one-one? Onto?

Both (strictly increasing; every real is a cube).

2. Show f : N → N, f(x) = 2x is one-one but not onto.

2x₁ = 2x₂ ⇒ x₁ = x₂; odd numbers have no pre-image.

3. Classify f : R → R, f(x) = x² + 2x + 3.

Many-one (f(−2) = f(0) = 3) and into (range [2, ∞)).

4. Find the codomain that makes f(x) = 2/(1 + x²), x ∈ R, onto.

(0, 2].