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parity (odd/even functions)
4 min read
Updated January 24, 2025
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Core Concept

Parity describes the symmetry properties of a function. A function can be even, odd, or neither.

Prerequisite: For a function to have parity, its domain must be symmetric about the origin (if xx is in the domain, then −x-x must also be in the domain).

Definitions

Even Function (偶函数)

A function ff is even if: f(−x)=f(x)for all x in the domainf(-x) = f(x) \quad \text{for all } x \text{ in the domain}

Graphical property: The graph is symmetric about the y-axis.

Odd Function (奇函数)

A function ff is odd if: f(−x)=−f(x)for all x in the domainf(-x) = -f(x) \quad \text{for all } x \text{ in the domain}

Graphical property: The graph is symmetric about the origin.

Special Property of Odd Functions

If ff is odd and 00 is in the domain, then f(0)=0f(0) = 0.

Proof: f(0)=f(−0)=−f(0)f(0) = f(-0) = -f(0), so 2f(0)=02f(0) = 0, thus f(0)=0f(0) = 0.

Common Functions and Their Parity

FunctionParityVerification
y=xny = x^n (nn even)Even(−x)n=xn(-x)^n = x^n
y=xny = x^n (nn odd)Odd(−x)n=−xn(-x)^n = -x^n
y=∣x∣y = \lvert x \rvertEven∣−x∣=∣x∣\lvert -x \rvert = \lvert x \rvert
y=sin⁡xy = \sin xOddsin⁡(−x)=−sin⁡x\sin(-x) = -\sin x
y=cos⁡xy = \cos xEvencos⁡(−x)=cos⁡x\cos(-x) = \cos x
y=tan⁡xy = \tan xOddtan⁡(−x)=−tan⁡x\tan(-x) = -\tan x
y=axy = a^xNeithera−x≠axa^{-x} \neq a^x and a−x≠−axa^{-x} \neq -a^x
y=log⁡axy = \log_a xNeitherDomain not symmetric

Methods to Determine Parity

Step-by-Step Process

  1. Check domain symmetry: Is −x-x in domain whenever xx is?
  2. Calculate f(−x)f(-x): Substitute −x-x into the function
  3. Compare with f(x)f(x) and −f(x)-f(x):
    • If f(−x)=f(x)f(-x) = f(x) → Even
    • If f(−x)=−f(x)f(-x) = -f(x) → Odd
    • Otherwise → Neither

Example 1: Even Function

Determine the parity of f(x)=x4−2x2+1f(x) = x^4 - 2x^2 + 1.

Step 1: Domain is R\mathbb{R}, which is symmetric about origin. ✓

Step 2: f(−x)=(−x)4−2(−x)2+1=x4−2x2+1=f(x)f(-x) = (-x)^4 - 2(-x)^2 + 1 = x^4 - 2x^2 + 1 = f(x)

Conclusion: ff is even.

Example 2: Odd Function

Determine the parity of f(x)=x3x2+1f(x) = \dfrac{x^3}{x^2 + 1}.

Step 1: Domain is R\mathbb{R}, symmetric about origin. ✓

Step 2: f(−x)=(−x)3(−x)2+1=−x3x2+1=−x3x2+1=−f(x)f(-x) = \dfrac{(-x)^3}{(-x)^2 + 1} = \dfrac{-x^3}{x^2 + 1} = -\dfrac{x^3}{x^2 + 1} = -f(x)

Conclusion: ff is odd.

Example 3: Neither

Determine the parity of f(x)=x+1f(x) = x + 1.

Step 1: Domain is R\mathbb{R}, symmetric. ✓

Step 2: f(−x)=−x+1f(-x) = -x + 1 f(x)=x+1f(x) = x + 1 −f(x)=−x−1-f(x) = -x - 1

Since f(−x)≠f(x)f(-x) \neq f(x) and f(−x)≠−f(x)f(-x) \neq -f(x):

Conclusion: ff is neither even nor odd.

CSCA Practice Problems

💡 Note: The following practice problems are designed based on the CSCA exam syllabus.

Example 1: Basic (Difficulty ★★☆☆☆)

Determine the parity of f(x)=x3−xf(x) = x^3 - x.

Solution: f(−x)=(−x)3−(−x)=−x3+x=−(x3−x)=−f(x)f(-x) = (-x)^3 - (-x) = -x^3 + x = -(x^3 - x) = -f(x)

Answer: Odd function


Example 2: Intermediate (Difficulty ★★★☆☆)

If f(x)f(x) is an odd function and f(2)=3f(2) = 3, find f(−2)+f(0)f(-2) + f(0).

Solution:

Since ff is odd:

  • f(−2)=−f(2)=−3f(-2) = -f(2) = -3
  • f(0)=0f(0) = 0 (property of odd functions)

f(−2)+f(0)=−3+0=−3f(-2) + f(0) = -3 + 0 = -3

Answer: −3-3


Example 3: Advanced (Difficulty ★★★★☆)

If f(x)=ax3+bx2+cx+df(x) = ax^3 + bx^2 + cx + d is an odd function, find the values of bb and dd.

Solution:

For odd function: f(−x)=−f(x)f(-x) = -f(x)

f(−x)=a(−x)3+b(−x)2+c(−x)+d=−ax3+bx2−cx+df(-x) = a(-x)^3 + b(-x)^2 + c(-x) + d = -ax^3 + bx^2 - cx + d

−f(x)=−ax3−bx2−cx−d-f(x) = -ax^3 - bx^2 - cx - d

Comparing: −ax3+bx2−cx+d=−ax3−bx2−cx−d-ax^3 + bx^2 - cx + d = -ax^3 - bx^2 - cx - d

This requires: bx2=−bx2bx^2 = -bx^2 and d=−dd = -d

Therefore: b=0b = 0 and d=0d = 0

Answer: b=0b = 0, d=0d = 0


Example 4: Advanced (Difficulty ★★★★☆)

If f(x)=ln⁡(x2+1+x)f(x) = \ln\left(\sqrt{x^2+1} + x\right), prove that f(x)f(x) is odd.

Solution:

f(−x)=ln⁡((−x)2+1+(−x))=ln⁡(x2+1−x)f(-x) = \ln\left(\sqrt{(-x)^2+1} + (-x)\right) = \ln\left(\sqrt{x^2+1} - x\right)

Now compute f(x)+f(−x)f(x) + f(-x): f(x)+f(−x)=ln⁡(x2+1+x)+ln⁡(x2+1−x)f(x) + f(-x) = \ln\left(\sqrt{x^2+1} + x\right) + \ln\left(\sqrt{x^2+1} - x\right)

=ln⁡[(x2+1+x)(x2+1−x)]= \ln\left[\left(\sqrt{x^2+1} + x\right)\left(\sqrt{x^2+1} - x\right)\right]

=ln⁡[(x2+1)−x2]=ln⁡(1)=0= \ln\left[(x^2+1) - x^2\right] = \ln(1) = 0

Therefore: f(−x)=−f(x)f(-x) = -f(x)

Conclusion: f(x)f(x) is odd.

Parity and Operations

Sum of Functions

ffggf+gf + g
EvenEvenEven
OddOddOdd
EvenOddNeither (generally)

Product of Functions

ffggf⋅gf \cdot g
EvenEvenEven
OddOddEven
EvenOddOdd

Common Mistakes

❌ Mistake 1: Ignoring Domain Symmetry

Wrong: f(x)=xf(x) = \sqrt{x} is even because x≥0\sqrt{x} \geq 0 ✗

Correct: Domain [0,+∞)[0, +\infty) is not symmetric about origin, so parity is undefined. ✓

❌ Mistake 2: Forgetting f(0)=0f(0) = 0 for Odd Functions

If ff is odd and defined at x=0x = 0, then f(0)=0f(0) = 0.

❌ Mistake 3: Only Checking One Value

Wrong: f(1)=f(−1)f(1) = f(-1), so ff is even. ✗

Correct: Must verify f(−x)=f(x)f(-x) = f(x) for ALL xx in domain. ✓

Study Tips

  1. ✅ Check domain first: Symmetric about origin is required
  2. ✅ Use algebraic verification: Don't rely on graphs alone
  3. ✅ Remember special property: Odd functions pass through origin
  4. ✅ Know product rules: odd × odd = even

💡 Exam Tip: For polynomials, odd functions have only odd-degree terms, even functions have only even-degree terms!

How the real exam tests it

Questions from real CSCA papers that test this term. Work each one out before you open the answer.

  1. CSCA Math December 2025 Past Paper
    Which of the following functions is odd?
    • A.y=cos⁡xy=\cos x
    • B.y=x+x3y=x+x^3
    • C.y=x4y=x^4
    • D.y=xsin⁡xy=x\sin x
    Show answer & solution

    Answer: B

    Answer: B For f(x)=x+x3f(x)=x+x^3, f(−x)=−f(x)f(-x)=-f(x), so it is odd. Choose B.
    See it in the full CSCA Math December 2025 Past Paper
  2. CSCA Math June 2026 Past Paper
    Which of the following statements about the odd function y=f(x)y = f(x) is incorrect?
    • A.The domain of the function is symmetric about the origin
    • B.In the domain of the function, f(−x)=f(x)f(-x) = f(x)
    • C.The graph of the function is symmetric about the origin
    • D.In the domain of the function, f(−x)=−f(x)f(-x) = -f(x)
    Show answer & solution

    Answer: B

    For an odd function, both the domain and the graph are symmetric about the origin, so (A) and (C) are true; by definition f(−x)=−f(x)f(-x) = -f(x), so (D) is true. Statement (B) is the property of an even function and is therefore the incorrect statement. Choose (B).
    Take the full CSCA Math June 2026 Past Paper — timed & graded

Practice Exercises

Practice to reinforce your understanding