Sets, Relations and Functions — MHT-CET Maths

Formulas (4)

• Sets and Cartesian Products: Counting Subsets and Double Counting · Counting with sets
  $$n(A \times B) = n(A)\,n(B) \qquad \#\text{subsets} = 2^m \qquad n(A \cup B) = n(A) + n(B) - n(A \cap B)$$
• One-One and Onto: Test Injectivity by f(x₁) = f(x₂), Surjectivity by Solving for x · One-one and onto
  $$f(x_1) = f(x_2) \Rightarrow x_1 = x_2 \ (\text{one-one}) \qquad \forall y\ \exists x: f(x) = y \ (\text{onto}) \qquad f' > 0 \Rightarrow \text{one-one}$$
• Equations in [x]: Solve for the Integer, Then Widen to the Interval · Greatest integer
  $$[x] = k \iff k \le x < k + 1$$
• Identities Like f(x + 1) − f(x) = 8x + 3: Compare Coefficients · Comparing coefficients
  $$f(x) = bx^2 + cx + d \Rightarrow f(x+1) - f(x) = 2bx + (b + c)$$

Watch out for (4)

• Subtracting only the empty set
  → Sets and Cartesian Products: Counting Subsets and Double Counting
• Calling a linear-fractional function onto ℝ
  → One-One and Onto: Test Injectivity by f(x₁) = f(x₂), Surjectivity by Solving for x
• Closing the right end
  → Equations in [x]: Solve for the Integer, Then Widen to the Interval
• Substituting one value of x
  → Identities Like f(x + 1) − f(x) = 8x + 3: Compare Coefficients

Formulas (3)

• Domain: Write One Condition Per Piece, Then Intersect · Standard conditions
  $$\sqrt{u}: u \ge 0 \quad \frac{1}{\sqrt{u}}: u > 0 \quad \log u: u > 0 \quad \sin^{-1}u: -1 \le u \le 1 \quad \frac{1}{u}: u \ne 0$$
• Domain Through a Quadratic Inequality: sin⁻¹ of a Rational Function With |x| · Clearing a positive denominator
  $$\frac{p(x)}{q(x)} \le 1,\ q(x) > 0 \iff p(x) \le q(x)$$
• Range of a Rational Function: Set y = f(x), Clear, and Demand a Real x · Discriminant method
  $$y = f(x) \ \Rightarrow\ a(y)x^2 + b(y)x + c(y) = 0 \ \Rightarrow\ b(y)^2 - 4a(y)c(y) \ge 0$$

Watch out for (3)

• Closing the bracket at a root in the denominator
  → Domain: Write One Condition Per Piece, Then Intersect
• Solving x² − x − 4 ≥ 0 as x ≥ the smaller root
  → Domain Through a Quadratic Inequality: sin⁻¹ of a Rational Function With |x|
• Guessing the bracket from the shape
  → Range of a Rational Function: Set y = f(x), Clear, and Demand a Real x

Formulas (4)

• Evaluate a Composite at a Point: Innermost First · Composition
  $$(f \circ g)(x) = f(g(x)) \qquad \text{evaluate inside} \to \text{outside}$$
• Composite as a Formula: Substitute in Two Steps, Then Simplify · Two-step composition
  $$(g \circ g \circ f)(x) = g\big(g(u)\big)\Big|_{u = f(x)}$$
• Recover f From f(g(x)): Match the Shape, or Evaluate at the Right x · Shape matching
  $$f(g(x)) = \Phi\big(g(x)\big) \ \Rightarrow\ f = \Phi$$
• Functional Identities: f(2x/(1 + x²)) = 2f(x) and the Cubic Twin · The two identities
  $$f(x) = \log\frac{1 - x}{1 + x}:\quad f\!\left(\frac{2x}{1 + x^2}\right) = 2f(x),\qquad f\!\left(\frac{3x + x^3}{1 + 3x^2}\right) = 3f(x)$$

Watch out for (4)

• Reading f(g(g(f(x)))) as (f∘g)² or as f²g²
  → Evaluate a Composite at a Point: Innermost First
• Matching α² = 1 alone
  → Composite as a Formula: Substitute in Two Steps, Then Simplify
• Solving for f(x) when only f(2) is asked
  → Recover f From f(g(x)): Match the Shape, or Evaluate at the Right x
• Reading the sign of the ratio backwards
  → Functional Identities: f(2x/(1 + x²)) = 2f(x) and the Cubic Twin

Formulas (4)

• Inverse of (ax + b)/(cx + d): Solve for x, and the Swap-and-Negate Shortcut · Linear-fractional inverse
  $$f(x) = \frac{ax + b}{cx + d} \ \Rightarrow\ f^{-1}(x) = \frac{dx - b}{-cx + a}$$
• Self-Inverse: f(f(x)) = x Fixes the Parameter · Involution test
  $$f(f(x)) = x \ \iff\ f = f^{-1} \qquad \frac{ax + b}{cx + d} \text{ self-inverse} \iff a + d = 0$$
• Inverse With a Square Root: Choose the Branch From the Domain · Order of inverses
  $$(f \circ g)^{-1} = g^{-1} \circ f^{-1}$$
• Solving f(x) = f⁻¹(x): For an Increasing f, Solve f(x) = x · Fixed points
  $$f \text{ increasing}:\quad f(x) = f^{-1}(x) \iff f(x) = x$$

Watch out for (4)

• Taking the reciprocal
  → Inverse of (ax + b)/(cx + d): Solve for x, and the Swap-and-Negate Shortcut
• Solving f(x) = x instead of f(f(x)) = x
  → Self-Inverse: f(f(x)) = x Fixes the Parameter
• Keeping the '−' branch
  → Inverse With a Square Root: Choose the Branch From the Domain
• Including the complex roots
  → Solving f(x) = f⁻¹(x): For an Increasing f, Solve f(x) = x