Q: How do I solve modulus equations that involve inequalities, like |2x - 1| < 3?

To solve an inequality like |2x - 1| 3, you split into two separate inequalities: 2x - 1 3, leading to x 2. Always check your solutions by substituting a value from each interval back into the original inequality to ensure they work. For more complex equations with two modulus expressions, sketch the graphs or consider critical values where each modulus expression changes sign, then solve linear equations in each region.

Question 1

What is the difference between the domain and range of a function?

Accepted Answer

The domain of a function is the set of all possible input values (usually x) for which the function is defined. For example, f(x) = √x has domain x ≥ 0 because you cannot take the square root of a negative number. The range is the set of all possible output values (f(x)) that the function can produce. For f(x) = √x, the range is f(x) ≥ 0, since square roots are non-negative. In exam questions, you may need to find the domain by considering restrictions like denominators (cannot be zero) or even roots (cannot be negative), and the range by analysing the function's behaviour, often via a graph. Remember, when finding the inverse function, the domain and range swap: the domain of f⁻¹ is the range of f, and vice versa.

Question 2

How do I remember the correct order for graph transformations?

Accepted Answer

Think of transformations as having a 'do to the inside first' rule. For a function y = f(x), transformations inside the function argument (like f(x + 2)) affect the x-coordinates horizontally. If you have both horizontal and vertical transformations, apply those closest to the x first. For example, to get from f(x) to 2f(3x - 1), you'd first deal with the inner part: f(3x) is a horizontal stretch by factor 1/3, then f(3x - 1) is a translation by vector (1/3, 0) (since you need to write as f(3(x - 1/3))). After horizontal changes, apply vertical ones: multiply by 2 gives a vertical stretch by factor 2. A common mnemonic is 'Change inside, opposite; change outside, straightforward'. Practice by listing steps clearly in the correct order and always check a key point, like the origin, to verify your final graph.

Question 3

When do I use the factor theorem and what is the remainder theorem for?

Accepted Answer

The factor theorem states that for a polynomial p(x), if p(a) = 0, then (x - a) is a factor. Use this to find one factor of a cubic or higher-degree polynomial, typically by trying small integer values like ±1, ±2. Once you have a factor, you can divide the polynomial (using long division or synthetic division) to reduce it to a quadratic, which you can then factorise further or solve. The remainder theorem tells you that when you divide p(x) by (x - a), the remainder is p(a). This is useful for quickly finding the remainder without full division, or for forming equations when given remainder conditions. Together, they are powerful tools for factorising polynomials and sketching their graphs, as you can systematically find all roots.

Question 4

How do I solve modulus equations that involve inequalities, like |2x - 1| < 3?

Accepted Answer

To solve an inequality like |2x - 1| < 3, you transform it into a compound inequality without the modulus: -3 < 2x - 1 < 3. Then solve the two parts simultaneously: add 1 to all sides: -2 < 2x < 4, then divide by 2: -1 < x < 2. For a greater-than inequality such as |2x - 1| > 3, you split into two separate inequalities: 2x - 1 < -3 or 2x - 1 > 3, leading to x < -1 or x > 2. Always check your solutions by substituting a value from each interval back into the original inequality to ensure they work. For more complex equations with two modulus expressions, sketch the graphs or consider critical values where each modulus expression changes sign, then solve linear equations in each region.

Question 5

What are the most common mistakes when doing partial fractions?

Accepted Answer

The most frequent error is using the wrong form for the denominator. For a linear factor (ax+b), the numerator is a constant (A); for a repeated linear factor (ax+b)^2, you need two fractions: A/(ax+b) + B/(ax+b)^2; and for an irreducible quadratic factor (ax^2+bx+c), the numerator is linear: (Ax+B)/(ax^2+bx+c). Another mistake is forgetting to check if the fraction is improper: if the degree of the numerator is greater than or equal to the degree of the denominator, perform polynomial long division first to obtain a quotient polynomial plus a proper fraction, then decompose the proper fraction. Also, students often make algebraic slips when multiplying both sides by the original denominator or when equating coefficients. Always compare your final decomposition with the original by giving a common denominator to verify.

Question 6

How do I complete the square for a quadratic expression like x^2 - 6x + 5?

Accepted Answer

To complete the square for x^2 - 6x + 5, take the coefficient of x (which is -6), halve it to get -3, and square it to get 9. Then rewrite the expression as (x^2 - 6x + 9) - 9 + 5 = (x - 3)^2 - 4. So the completed square form is (x - 3)^2 - 4. This tells you the vertex (minimum point) of the parabola y = x^2 - 6x + 5 is at (3, -4). If the coefficient of x^2 is not 1, first factor it out from the x terms. For example, 2x^2 + 8x + 3 becomes 2(x^2 + 4x) + 3, then complete the square inside the brackets: 2[(x+2)^2 - 4] + 3 = 2(x+2)^2 - 8 + 3 = 2(x+2)^2 - 5. Completing the square is vital for finding the turning point and for solving equations where factorisation is not possible.

B: Algebra and functions

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Related Topics in AQA vocational Mathematics

A: Proof

C: Coordinate geometry in the ( x , y ) plane

D: Sequences and series

E: Trigonometry

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