OxfordAQA International AS and A-level Mathematics: Transformations and rational functions (9660) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover the second half of section P2.1 (Algebra and functions) in the OxfordAQA International AS and A-level Mathematics specification, Version 5.2: combined transformations, the Factor and Remainder Theorems, simplifying rational expressions, algebraic division and partial fractions. The specification places this content in International A-level (A2) Unit P2 (Pure Maths); it is not examined at International AS.

Related pages: the OxfordAQA Mathematics hub, the printable checklist, the study guide and the revision notes. Try the free 10-minute diagnostics as well.

Questions

1. Describe a sequence of transformations that maps the graph of y = e^x onto the graph of y = 3e^(2x + 1). [3]

2. The graph of y = sec x is transformed to give the graph of y = 2 sec 3x.

(a) Write down the equation of the vertical asymptote in the interval 0 < x < π/3. [2] (b) Write down the coordinates of the turning points for 0 ≤ x ≤ π/3. [2]

3. The curve y = f(x) has a maximum point at (4, −1) and passes through (0, −9). Find the coordinates of the images of both points on the curve y = 3 − 2f(x/2 + 1), and state whether the image of the maximum point is a maximum or a minimum. [5]

4. h(x) = 4x³ + ax² − 11x + b. The expression (2x + 1) is a factor of h(x), and the remainder when h(x) is divided by (x − 2) is −10.

(a) Find a and b. [4] (b) Solve h(x) = 0, giving exact answers. [3]

5. (a) Find the remainder when 5x³ − 4x² + x − 7 is divided by (2x − 1). [2] (b) Simplify (6x² + x − 2)/(9x² − 4). [3]

6. p(x) = 2x³ + x² − 25x + 12.

(a) Show that (x − 3) is a factor of p(x). [2] (b) Hence simplify p(x)/(x² + 3x − 4). [3]

7. (a) Express (6x + 7)/(2x + 1) in the form A + B/(2x + 1). [2] (b) Express (8x³ − 2x² + 11x + 8)/(4x + 1) in the form Ax² + Bx + C + D/(4x + 1). [4]

8. The fraction (2x² + 12x − 2)/((x − 1)(x + 3)) is identical to A + (Bx + C)/((x − 1)(x + 3)).

(a) Find A, B and C. [4] (b) Hence express the original fraction as a constant plus two partial fractions. [4]

9. Express (12x² − 13x − 25)/((x + 2)(2x − 3)²) in partial fractions. [6]

10. Express (2x³ + 5x² − 16x)/((x − 1)²(x + 2)) as a constant plus partial fractions. [8]

11. The curve C has equation y = (2x + 7)/(x + 2), x ≠ −2.

(a) Show that y = 2 + 3/(x + 2). [2] (b) Describe a sequence of three transformations that maps y = 1/x onto C. [3] (c) Write down the equations of the asymptotes of C. [1] (d) Find the coordinates of the points where C meets the axes. [2]

Answers

1. y = 3e^(2x + 1) = 3f(2(x + ½)) where f(x) = e^x. Stretch, scale factor ½, parallel to the x-axis [1]; translation of (−½, 0) [1]; stretch, scale factor 3, parallel to the y-axis [1]. (Translating by (−1, 0) first and then stretching by ½ is equally valid.) [3] Examiner insight: Whichever order you choose, say which transformation comes first and give its scale factor or vector with direction; a bare list of words is hard to credit.

2. (a) sec 3x is undefined when cos 3x = 0, so 3x = π/2 [1]. The asymptote is x = π/6 [1]. (b) At x = 0, y = 2 sec 0 = 2 [1]. At x = π/3, y = 2 sec π = −2 [1]. The turning points are (0, 2) and (π/3, −2). [4] Examiner insight: The factor 3 inside shortens the period, so turning points sit at multiples of π/3, not π. Check the endpoints of the interval before you write coordinates.

3. Solve x/2 + 1 = 4: x = 6 [1]. y = 3 − 2(−1) = 5 [1]. Solve x/2 + 1 = 0: x = −2 [1]. y = 3 − 2(−9) = 21 [1]. The images are (6, 5) and (−2, 21). Because of the −2 the graph is reflected in the x-axis, so the maximum becomes a minimum [1]. [5] Examiner insight: Work with the inside first (solve for the old x-value), then the outside. Saying “minimum” with a reason about the reflection shows you understand the sign.

4. (a) h(−½) = 0: −½ + a/4 + 11/2 + b = 0, so a + 4b = −20 [1]. h(2) = −10: 32 + 4a − 22 + b = −10, so 4a + b = −20 [1]. Solve simultaneously [1]: a = −4, b = −4 [1]. (b) h(x) = (2x + 1)(2x² − 3x − 4) [1]. So x = −½ [1], and from the quadratic formula x = (3 ± √41)/4 [1]. [7] Examiner insight: Write the two conditions as separate equations before solving. Leave the quadratic roots in surd form when the question asks for exact answers.

5. (a) Remainder = f(½) [1] = 5/8 − 1 + ½ − 7 = −55/8 [1]. (b) 6x² + x − 2 = (3x + 2)(2x − 1) [1]; 9x² − 4 = (3x − 2)(3x + 2) [1]; the result is (2x − 1)/(3x − 2) [1]. [5] Examiner insight: For (2x − 1), use ½, not −½ or 2. In (b), cancel the bracket, not the 2x terms.

6. (a) p(3) = 54 + 9 − 75 + 12 [1] = 0, so (x − 3) is a factor by the Factor Theorem [1]. (b) p(x) = (x − 3)(2x² + 7x − 4) [1] = (x − 3)(2x − 1)(x + 4) [1]. The denominator is (x + 4)(x − 1), so the fraction simplifies to (x − 3)(2x − 1)/(x − 1) [1]. [5] Examiner insight: “Show that” asks for the evaluation and a stated conclusion. Reaching the factorised form without showing the division leaves your factorisation unsupported.

7. (a) 6x + 7 = 3(2x + 1) + 4 [1], so the fraction is 3 + 4/(2x + 1) [1]. (b) 8x³ ÷ 4x gives 2x², and 2x²(4x + 1) = 8x³ + 2x² leaves −4x² + 11x + 8 [1]. Then −x(4x + 1) = −4x² − x leaves 12x + 8 [1]. Then 3(4x + 1) = 12x + 3 leaves 5 [1]. The result is 2x² − x + 3 + 5/(4x + 1) [1]. [6] Examiner insight: Set the working out in columns and bring down every term, including zero coefficients. A remainder of 5 must appear over the divisor.

8. (a) 2x² + 12x − 2 ≡ A(x − 1)(x + 3) + Bx + C. Matching x²: A = 2 [1]. A(x² + 2x − 3) = 2x² + 4x − 6 [1]. Matching x: 12 = 4 + B, B = 8 [1]. Matching constants: −2 = −6 + C, C = 4 [1]. (b) 8x + 4 ≡ P(x + 3) + R(x − 1) [1]. x = 1: 12 = 4P, P = 3 [1]. x = −3: −20 = −4R, R = 5 [1]. The result is 2 + 3/(x − 1) + 5/(x + 3) [1]. [8] Examiner insight: Part (b) uses the Bx + C found in part (a), so expand your identity to check it before you start splitting.

9. Write the form A/(x + 2) + B/(2x − 3) + C/(2x − 3)² [1]. Then 12x² − 13x − 25 ≡ A(2x − 3)² + B(x + 2)(2x − 3) + C(x + 2) [1]. x = −2: 49 = 49A, A = 1 [1]. x = 3/2: −35/2 = 7C/2, C = −5 [1]. Matching x²: 12 = 4A + 2B, so B = 4 [1]. The result is 1/(x + 2) + 4/(2x − 3) − 5/(2x − 3)² [1]. [6] Examiner insight: The repeated factor needs both (2x − 3) and (2x − 3)² terms. Check with a spare value such as x = 0: both sides of the identity equal −25.

10. The denominator expands to x³ − 3x + 2 [1]. The numerator and denominator have the same degree, so the quotient is 2 [1], leaving 5x² − 10x − 4 [1]. Write 5x² − 10x − 4 ≡ A(x − 1)(x + 2) + B(x + 2) + C(x − 1)², giving the form A/(x − 1) + B/(x − 1)² + C/(x + 2) [1]. x = 1: −9 = 3B, B = −3 [1]. x = −2: 36 = 9C, C = 4 [1]. Matching x²: 5 = A + C, A = 1 [1]. The result is 2 + 1/(x − 1) − 3/(x − 1)² + 4/(x + 2) [1]. [8] Examiner insight: Compare degrees before writing any unknowns. Forgetting to divide first leaves you with no way to match the x² coefficient.

11. (a) 2x + 7 = 2(x + 2) + 3 [1], so (2x + 7)/(x + 2) = 2 + 3/(x + 2) [1]. (b) Translation of (−2, 0) gives 1/(x + 2) [1]. Stretch, scale factor 3, parallel to the y-axis gives 3/(x + 2) [1]. Translation of (0, 2) gives 2 + 3/(x + 2) [1]. (c) x = −2 and y = 2 [1]. (d) y = 0: 2x + 7 = 0, so (−7/2, 0) [1]. x = 0: y = 7/2, so (0, 7/2) [1]. [8] Examiner insight: Part (a) is the link, because the transformations cannot be read from the single fraction. State the equation after each step.

Where marks are usually lost

• Reading f(x + a) as a shift to the right, or f(2x) as a stretch of factor 2.
• Giving the images of points without solving the inside bracket for x.
• Using −b/a the wrong way round in the Remainder Theorem.
• Cancelling a term in a numerator against a term in a denominator.
• Stopping at the quotient when the question needs the remainder fraction.
• Starting partial fractions on a fraction with numerator degree equal to or above the denominator’s.
• Missing the squared term for a repeated factor.
• Giving a decimal for an answer that should be exact, such as (3 ± √41)/4.

Next steps

• Re-read the revision notes and the study guide.
• Go back to the OxfordAQA Mathematics hub and tick items off the checklist.
• Take all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, published by OxfordAQA. For International AS exams from May/June 2018 and International A-level exams from May/June 2019. Section P2.1: Algebra and functions.