Algebra

Factorise x² + 5x + 6.

• A(x + 1)(x + 6)
• B(x + 2)(x + 3)✓
• C(x + 6)(x − 1)
• D(x − 2)(x − 3)

Answer: Find two numbers that multiply to 6 and add to 5: 2 and 3. So x² + 5x + 6 = (x + 2)(x + 3).

Solve x² = 49.

• Ax = ±49
• Bx = 7 only
• Cx = 7 or x = −7✓
• Dx = 24.5

Answer: Square rooting gives two solutions: x = 7 or x = −7 (since (−7)² = 49).

Expand and simplify (x + 3)(x − 2).

• Ax² − 6
• Bx² + x − 6✓
• Cx² + 5x − 6
• Dx² − x − 6

Answer: (x + 3)(x − 2) = x² − 2x + 3x − 6 = x² + x − 6.

Solve the simultaneous equations x + y = 10 and x − y = 4.

• Ax = 5, y = 5
• Bx = 6, y = 4
• Cx = 7, y = 3✓
• Dx = 3, y = 7

Answer: Add the equations: 2x = 14, so x = 7. Then y = 10 − 7 = 3.

Make x the subject of y = 3x − 5.

• Ax = (y + 5)/3✓
• Bx = (y − 5)/3
• Cx = (y + 5)·3
• Dx = 3y − 5

Answer: Add 5: y + 5 = 3x. Divide by 3: x = (y + 5)/3.

What is the gradient of the line y = 4x − 7?

• A4✓
• B−7
• C−4
• D7

Answer: In y = mx + c, the gradient m is the coefficient of x, which is 4.

Solve the inequality 2x + 1 > 9.

• Ax < 4
• Bx > 5
• Cx > 8
• Dx > 4✓

Answer: Subtract 1: 2x > 8. Divide by 2: x > 4.

The nth term of a sequence is 3n + 2. What is the 10th term?

• A23
• B30
• C32✓
• D35

Answer: Substitute n = 10: 3(10) + 2 = 32.

Solve 3(x − 2) = 2x + 4.

• Ax = 10✓
• Bx = 2
• Cx = −10
• Dx = 6

Answer: 3x − 6 = 2x + 4 → x = 10.

Simplify (3x²y)(2xy³).

• A6x³y³
• B5x³y⁴
• C6x²y³
• D6x³y⁴✓

Answer: Multiply coefficients (3×2=6) and add indices: x²·x = x³, y·y³ = y⁴, giving 6x³y⁴.

Solve the quadratic x² − 5x + 6 = 0.

• Ax = 5 or 6
• Bx = −2 or −3
• Cx = 1 or 6
• Dx = 2 or 3✓

Answer: Factorise to (x − 2)(x − 3) = 0, so x = 2 or x = 3.

Write x² + 6x + 11 in the form (x + a)² + b.

• A(x + 3)² − 2
• B(x + 3)² + 2✓
• C(x + 3)² + 20
• D(x + 6)² − 25

Answer: Halve the coefficient of x: (x + 3)² = x² + 6x + 9, which is 9 too big, so subtract 9 from the 11. That leaves (x + 3)² + 2.

The curve y = x² − 8x + 3 has one turning point. Work out its coordinates.

• A(4, 13)
• B(−4, −13)
• C(4, −13)✓
• D(8, −13)

Answer: Completing the square gives y = (x − 4)² − 16 + 3 = (x − 4)² − 13. The squared bracket is smallest when x = 4, giving y = −13, so the minimum is at (4, −13).

Use the quadratic formula to solve 3x² − 5x − 4 = 0. Give each solution to 2 decimal places.

• Ax = 4.51 or x = −1.18
• Bx = 0.59 or x = −2.26
• Cx = 2.26 or x = 0.59
• Dx = 2.26 or x = −0.59✓

Answer: With a = 3, b = −5, c = −4 the discriminant is (−5)² − 4×3×(−4) = 25 + 48 = 73, and √73 = 8.544. So x = (5 ± 8.544) ÷ 6, giving 2.26 and −0.59.

Which statement about the roots of 2x² + 3x + 5 = 0 is correct?

• AThe discriminant is 49, so there are two real roots.
• BThe discriminant is −31, so there are no real roots.✓
• CThe discriminant is 0, so there is one repeated root.
• DThe discriminant is −31, so there are two real roots.

Answer: b² − 4ac = 3² − 4×2×5 = 9 − 40 = −31. A negative discriminant means the square root is not a real number, so the curve never meets the x-axis and the equation has no real roots.

The equation x² + kx + 16 = 0 has two equal roots, where k is a positive number. Work out the value of k.

• Ak = 64
• Bk = 16
• Ck = 8✓
• Dk = 4

Answer: Equal roots means the discriminant is zero: k² − 4×1×16 = 0, so k² = 64 and k = ±8. Taking the positive value gives k = 8.

Solve the simultaneous equations y = x + 2 and x² + y² = 10.

• Ax = 1, y = 3 and x = −3, y = −1✓
• Bx = 3, y = 5 and x = −1, y = 1
• Cx = 1, y = −1 and x = −3, y = 3
• Dx = √3, y = √3 + 2 and x = −√3, y = 2 − √3

Answer: Substituting gives x² + (x + 2)² = 10, so 2x² + 4x − 6 = 0 and x² + 2x − 3 = 0. Factorising gives (x + 3)(x − 1) = 0, so x = 1 with y = 3, or x = −3 with y = −1.

Solve the simultaneous equations y = 2x − 1 and y = x² − 4.

• Ax = 3, y = 5 and x = −1, y = −3✓
• Bx = −3, y = −7 and x = 1, y = 1
• Cx = 3, y = 5 and x = 5, y = 9
• Dx = 3, y = 5 only

Answer: Setting the expressions equal gives x² − 4 = 2x − 1, so x² − 2x − 3 = 0 and (x − 3)(x + 1) = 0. Then x = 3 gives y = 5, and x = −1 gives y = −3; both intersections are needed.

Simplify fully (x² − 9) ÷ (x² + 7x + 12).

• A(x − 3)/(x − 4)
• B−9/(7x + 12)
• C(x + 3)/(x + 4)
• D(x − 3)/(x + 4)✓

Answer: The numerator is a difference of two squares, (x − 3)(x + 3), and the denominator factorises to (x + 3)(x + 4). The common factor (x + 3) cancels, leaving (x − 3)/(x + 4).

Write 3/(x + 1) + 2/(x − 2) as a single fraction in its simplest form.

• A5/(2x − 1)
• B(5x − 4)/(x² − x − 2)✓
• C(5x + 8)/(x² − x − 2)
• D(5x − 4)/(x² + x − 2)

Answer: The common denominator is (x + 1)(x − 2) = x² − x − 2. The numerator becomes 3(x − 2) + 2(x + 1) = 3x − 6 + 2x + 2 = 5x − 4.

Solve (x + 4)/3 − (x − 2)/5 = 2.

• Ax = 8
• Bx = 4
• Cx = 2✓
• Dx = −12

Answer: Multiplying every term by 15 gives 5(x + 4) − 3(x − 2) = 30, so 5x + 20 − 3x + 6 = 30. That simplifies to 2x + 26 = 30, so 2x = 4 and x = 2.

The graph of y = f(x) has a minimum point at (2, −5). Work out the coordinates of the minimum point on the graph of y = f(x + 3).

• A(−1, −5)✓
• B(5, −5)
• C(2, −2)
• D(2, −8)

Answer: Adding a number inside the bracket translates the graph horizontally in the negative direction, so y = f(x + 3) is a translation 3 units to the left. The y-coordinate is unchanged, giving (−1, −5).

The graph of y = f(x) passes through the point (3, 7). Which point must lie on the graph of y = −f(x)?

• A(−3, 7)
• B(7, 3)
• C(−3, −7)
• D(3, −7)✓

Answer: The minus sign is applied after f, so every output is negated while the input stays the same. This reflects the graph in the x-axis, sending (3, 7) to (3, −7).

The graph of y = f(x) has a maximum point at (4, 2). Work out the coordinates of the maximum point on the graph of y = 3f(x) − 1.

• A(4, 6)
• B(12, 6)
• C(4, 4)
• D(4, 5)✓

Answer: Multiplying f(x) by 3 stretches the graph vertically by scale factor 3, so the y-value 2 becomes 6, and subtracting 1 translates it down one unit to 5. The x-coordinate is unaffected, giving (4, 5).

f(x) = (3x − 1)/4. Work out an expression for the inverse function f⁻¹(x).

• Af⁻¹(x) = (4x − 1)/3
• Bf⁻¹(x) = (4x + 1)/3✓
• Cf⁻¹(x) = (3x + 1)/4
• Df⁻¹(x) = 4/(3x − 1)

Answer: Write y = (3x − 1)/4 and rearrange for x: 4y = 3x − 1, so 3x = 4y + 1 and x = (4y + 1)/3. Replacing y with x gives f⁻¹(x) = (4x + 1)/3.

f(x) = 2x + 5 and g(x) = x². Work out the value of fg(4).

• A208
• B169
• C37✓
• D21

Answer: In fg(4) the function g is applied first: g(4) = 4² = 16. Then f(16) = 2×16 + 5 = 37.

n is an integer. Which working proves that the sum of any three consecutive integers is always a multiple of 3?

• An + (n + 1) + (n + 2) = 3n + 3 = 3(n + 1), a multiple of 3✓
• Bn + (n + 1) + (n + 2) = 3n + 2, which is 2 more than a multiple of 3
• Cn + 2n + 3n = 6n, a multiple of 3
• D4 + 5 + 6 = 15 and 7 + 8 + 9 = 24, so the rule holds for every integer

Answer: Three consecutive integers are n, n + 1 and n + 2. Their sum is 3n + 3, which factorises to 3(n + 1) — an integer multiplied by 3, so it is a multiple of 3 for every value of n. Checking numerical examples shows the rule works in those cases but proves nothing in general.

The equation x³ − 6x + 2 = 0 can be rearranged to give the iterative formula xₙ₊₁ = (xₙ³ + 2)/6. Starting with x₀ = 0, work out x₃ correct to 4 decimal places.

• Ax₃ = 0.0391
• Bx₃ = 0.3333
• Cx₃ = 0.3395
• Dx₃ = 0.3399✓

Answer: x₁ = (0³ + 2)/6 = 0.333333, then x₂ = (0.333333³ + 2)/6 = 0.339506, then x₃ = (0.339506³ + 2)/6 = 0.339855, which is 0.3399 to 4 decimal places. Stopping at x₁ or x₂ gives an answer one or two iterations short.

Work out an expression for the nth term of the quadratic sequence 3, 8, 15, 24, 35, …

• An² + 2
• B2n² + n
• Cn² + 2n✓
• Dn² + 5n − 3

Answer: The first differences are 5, 7, 9, 11 and the second difference is 2, so the sequence starts with n². Subtracting n² (1, 4, 9, 16, 25) from the sequence leaves 2, 4, 6, 8, 10, which is 2n, giving n² + 2n.