Algebra & Functions — Gradora Demo Drill Pack

M1 Substitute x = 2: f(2) = 8 − 12 + 4 = 0. AO1 — applying the factor theorem Common mistake: Substituting x = −2 instead of x = +2.

A1 Since f(2) = 0, by the Factor Theorem, (x − 2) is a factor of f(x). ✓ AO2 — stating the conclusion referencing the theorem

M1 Halve the coefficient of x: ½(−6) = −3. Form (x − 3)². AO1 — completing the square procedure Common mistake: Forgetting to subtract the squared term: writing (x−3)² + 11 without subtracting 9.

A1 Subtract 3² = 9: x² − 6x + 11 = (x − 3)² − 9 + 11 = (x − 3)² + 2. AO1 — correct completed-square form

A1 The minimum value is 2, occurring at x = 3 (since the squared term ≥ 0). AO2 — reading the minimum from completed-square form

M1 b² − 4ac = (−5)² − 4(3)(4) = 25 − 48 = −23. AO1 — applying discriminant formula Common mistake: Using b = 5 rather than b = −5 (the negative sign does not affect b², but errors in ac are common).

A1 Since −23 < 0, the equation has no real roots (two complex conjugate roots). AO2 — interpreting the sign of the discriminant

M1 Substitute y = x + 1 into y = x² − 1: x + 1 = x² − 1. AO1 — setting up the substitution

M1 Rearrange: x² − x − 2 = 0. Factorise: (x − 2)(x + 1) = 0, so x = 2 or x = −1. AO1 — solving the resulting quadratic Common mistake: Forgetting to find y values after finding x.

A1 When x = 2: y = 3. When x = −1: y = 0. AO1 — pairing solutions correctly

A1 Solution pairs: (2, 3) and (−1, 0). AO2 — presenting both pairs clearly

M1 Binomial: (1+2x)⁵ = ∑ ⁵C_r (2x)^r. First three terms use r = 0, 1, 2. AO1 — identifying the correct binomial form

A1 r=0: 1. r=1: 5(2x) = 10x. r=2: 10(4x²) = 40x². AO1 — computing each coefficient Common mistake: Forgetting to raise the whole of 2x to the power r, e.g. writing 5⋅2x instead of 5⋅(2x)¹.

A1 First three terms: 1 + 10x + 40x². AO1 — final answer

M1 By the Remainder Theorem, remainder = f(3) = 27 + 18 − 15 + 1. AO1 — applying the remainder theorem

A1 f(3) = 27 + 18 − 15 + 1 = 31. The remainder is 31. AO1 — correct arithmetic

M1 Verify: f(−2) = −16 + 4 + 10 + 2 = 0. ✓. Divide 2x³ + x² − 5x + 2 by (x + 2). AO1 — setting up the division

M2 Quotient by inspection/long division: 2x² − 3x + 1. So f(x) = (x + 2)(2x² − 3x + 1). AO1 — polynomial division Common mistake: A common error: sign errors in the quotient. Check by expanding to verify.

A1 Factorise 2x² − 3x + 1 = (2x − 1)(x − 1). AO1 — factorising the quadratic factor

A1 Full factorisation: f(x) = (x + 2)(2x − 1)(x − 1). AO2 — presenting the complete factorised form

M1 Discriminant: b² − 4ac = (−3)² − 4(1)(k) = 9 − 4k. AO1 — forming the discriminant

M1 For no real roots: 9 − 4k < 0 ⇒ k > 9/4. ✓ AO2 — setting up and solving the inequality

A1 For exactly one real root (repeated root): discriminant = 0, so 9 − 4k = 0 ⇒ k = 9/4. AO1 — solving the equality condition

A1 At k = 9/4, the equation has one repeated root at x = 3/2. AO2 — interpreting the result

M1 Factorise: 2x² + 5x − 3 = (2x − 1)(x + 3) = 0, so x = 1/2 or x = −3. AO1 — factorising the quadratic

A1 Solutions to first equation: x = 1/2 or x = −3. AO1 — correct roots

M1 Let u = y − 1. Then 2u² + 5u − 3 = 0 has solutions u = 1/2 or u = −3. AO2 — recognising the substitution structure Common mistake: Not substituting back to find y.

A1 y − 1 = 1/2 ⇒ y = 3/2.   y − 1 = −3 ⇒ y = −2. AO1 — back-substitution

A1 Solutions: y = 3/2 and y = −2. AO2 — final paired answer

M1 Set kx − 1 = x² + 2x + 3 ⇒ x² + (2 − k)x + 4 = 0. AO1 — forming the quadratic

M1 Discriminant > 0 for two distinct intersections: (2 − k)² − 16 > 0. AO2 — applying the discriminant condition

M2 (2−k)² > 16 ⇒ |2−k| > 4 ⇒ 2−k > 4 or 2−k < −4 ⇒ k < −2 or k > 6. AO2 — solving the inequality with two cases Common mistake: Only finding one of the two intervals from |2−k| > 4.

A1 Range: k < −2 or k > 6. AO1 — correct final answer

M1 f(1) = 0: 1 + a + b − 6 = 0 ⇒ a + b = 5.   [Equation 1] AO1 — applying factor theorem at x = 1

M1 f(−2) = 0: −8 + 4a − 2b − 6 = 0 ⇒ 4a − 2b = 14 ⇒ 2a − b = 7.   [Equation 2] AO1 — applying factor theorem at x = −2 Common mistake: Sign errors when substituting x = −2 into the cubic.

M1 Solve simultaneous: Eq1 + Eq2: 3a = 12 ⇒ a = 4. AO1 — solving the system

A1 Substitute a = 4 into Eq1: b = 1. AO1 — finding b

A1 a = 4, b = 1. Verify: f(x) = x³ + 4x² + x − 6 = (x−1)(x+2)(x+3). ✓ AO2 — verification with all three factors

A1 Check: (x−1)(x+2)(x+3) = (x−1)(x²+5x+6) = x³+4x²+x−6. ✓ AO2 — expanding to verify

M1 Write out terms: ⁴C₀(3)⁴ + ⁴C₁(3)³(2x) + ⁴C₂(3)²(2x)² + ⁴C₃(3)(2x)³ + ⁴C₄(2x)⁴. AO1 — setting up the binomial expansion

M1 Compute each term: 81;   4⋅27⋅2x = 216x;   6⋅9⋅4x² = 216x²;   4⋅3⋅8x³ = 96x³;   16x⁴. AO1 — computing the coefficients Common mistake: Forgetting to raise the whole 2x to the required power, e.g. computing 6(9)(2x²) = 108x² instead of 216x².

A2 (3 + 2x)⁴ = 81 + 216x + 216x² + 96x³ + 16x⁴. AO1 — all five terms correct

A1 Verify leading term: (2x)⁴ = 16x⁴. ✓ Verify constant: 3⁴ = 81. ✓ AO2 — sanity-checking leading and constant terms

M1 From 3x − 2y = 1: x = (1 + 2y)/3. AO1 — making x the subject

M2 Substitute: ⅓(1+2y)²/3 ⋅3 + ⅓(1+2y)⋅y ⋅3... Simplify: (1+2y)²/9 + y(1+2y)/3 = 4. Multiply through by 9: (1+2y)² + 3y(1+2y) = 36. AO2 — forming the substituted equation

M1 Expand: 1 + 4y + 4y² + 3y + 6y² = 36 ⇒ 10y² + 7y − 35 = 0. AO1 — expanding and collecting Common mistake: Sign or arithmetic errors in expansion.

A1 y = (−7 ± √(49 + 1400))/20 = (−7 ± √1449)/20. √1449 = 3√161. AO1 — quadratic formula application

A1 y = (−7 ± 3√161)/20, with corresponding x values from x = (1+2y)/3. AO2 — presenting exact form answers

M1 General term: ⁶C_r ⋅ x^6−r ⋅ (3/x)^r = ⁶C_r ⋅ 3^r ⋅ x^6−r ⋅ x^−r = ⁶C_r ⋅ 3^r ⋅ x^6−2r. AO1 — setting up the general term

M1 Independent of x when power = 0: 6 − 2r = 0 ⇒ r = 3. AO2 — finding the required value of r

A1 T₄ = ⁶C₃ ⋅ 3³ = 20 ⋅ 27 = 540. AO1 — computing the term

A1 The term independent of x is 540. AO2 — stating the answer clearly

M1 Factor 3: 3(x² + 4x) − 7. AO1 — factoring out the leading coefficient Common mistake: Forgetting to factor out 3 and treating the expression as x² + 12x − 7.

M1 Complete the square inside: x² + 4x = (x + 2)² − 4. AO1 — completing the square

A1 3[(x+2)² − 4] − 7 = 3(x+2)² − 12 − 7 = 3(x+2)² − 19. AO1 — combining constants

A1 Minimum value: −19 at x = −2. AO2 — reading off minimum and its location

A1 Since a = 3 > 0, the parabola opens upward, confirming this is a minimum. AO2 — justifying minimum rather than maximum

M1 Remainder = f(3) = 1: 2(27) − 9(9) + 3(3) + k = 1 ⇒ 54 − 81 + 9 + k = 1 ⇒ −18 + k = 1. AO1 — applying remainder theorem

A1 k = 19. AO1 — correct value of k

M1 Test (2x−1): f(1/2) = 2(1/8) − 9(1/4) + 3(1/2) + 19 = 1/4 − 9/4 + 3/2 + 19. AO1 — substituting x = 1/2 Common mistake: Substituting x = −1/2 instead of x = 1/2 for the factor (2x−1).

A1 1/4 − 9/4 + 6/4 + 76/4 = 74/4 = 37/2 ≠ 0. AO1 — evaluating correctly

A1 Since f(1/2) ≠ 0, (2x − 1) is NOT a factor of f(x). AO2 — clear conclusion with reasoning

M1 f(−1) = 6: −2 + a + 5 + b = 6 ⇒ a + b = 3.   [Eq 1] AO1 — applying given condition f(−1)=6

M1 f(1/2) = 0 (factor theorem): 2(1/8) + a(1/4) − 5/2 + b = 0 ⇒ 1/4 + a/4 − 5/2 + b = 0. Multiply by 4: 1 + a − 10 + 4b = 0 ⇒ a + 4b = 9.   [Eq 2] AO1 — applying factor theorem for (2x−1) Common mistake: Not recognising that (2x−1) factor ⇒ f(1/2) = 0.

M1 Eq2 − Eq1: 3b = 6 ⇒ b = 2. Then a = 3 − 2 = 1. AO1 — solving simultaneous equations

A1 a = 1, b = 2. AO1 — correct values

M1 So f(x) = 2x³ + x² − 5x + 2. Since (2x−1) is a factor, divide: 2x³ + x² − 5x + 2 = (2x−1)(x² + px − 2). AO2 — setting up polynomial division

A1 By inspection or long division: quotient is x² + x − 2. AO1 — correct quotient

A1 Factorise the quotient: x² + x − 2 = (x + 2)(x − 1). Full factorisation: f(x) = (2x−1)(x + 2)(x − 1). AO2 — presenting the factorised form

A1 Verify: (2x−1)(x+2)(x−1) expands to 2x³ + x² − 5x + 2 = f(x). ✓ Roots: x = 1/2, −2, 1. AO2 — verification step

M1 Discriminant: k² − 4(k+3) > 0 ⇒ k² − 4k − 12 > 0. AO1 — setting up discriminant inequality

M1 Factorise: (k − 6)(k + 2) > 0. AO1 — factorising the quadratic

A2 Sign table or sketch: (k−6)(k+2) > 0 when k < −2 or k > 6. AO2 — solving the quadratic inequality Common mistake: Writing the answer as −2 < k < 6 (confusing > 0 with < 0 for a positive quadratic). Sketch a parabola to check.

A1 Range: k < −2 or k > 6. AO2 — correct interval notation

M1 (a) (1−2x)⁶: terms for r=0,1,2,3: 1; 6(−2x); 15(4x²); 20(−8x³). AO1 — identifying binomial terms Common mistake: Forgetting the negative sign from (−2x): writing +12x instead of −12x.

A1 (a) (1−2x)⁶ ≈ 1 − 12x + 60x² − 160x³ (up to x³). AO1 — correct first four terms

M1 (b) Set 1−2x = 0.98 ⇒ x = 0.01. Substitute. AO2 — identifying the correct substitution

A1 (0.98)⁶ ≈ 1 − 12(0.01) + 60(0.0001) − 160(0.000001) = 1 − 0.12 + 0.006 − 0.00016. AO1 — correct substitution

A1 (0.98)⁶ ≈ 0.8859 (4 s.f.). (Exact: 0.88584… ✓) AO1 — correct 4 s.f. answer

A1 The approximation is valid because x = 0.01 is small, so higher-power terms contribute negligibly. AO2 — justifying the validity of the approximation

M1 Express: x = 3 − 2y. AO1 — making x the subject

M1 Substitute: (3−2y)² − (3−2y)y + y² = 7. Expand: 9−12y+4y² −3y+2y²+y² = 7. AO1 — substituting and expanding Common mistake: Errors in expanding (3−2y)² or (3−2y)y.

M1 Collect: 7y² − 15y + 9 = 7 ⇒ 7y² − 15y + 2 = 0. AO1 — simplifying the quadratic

A1 Factorise: (7y − 1)(y − 2) = 0 ⇒ y = 1/7 or y = 2. AO1 — solving the quadratic

A1 y = 1/7: x = 3 − 2/7 = 19/7.   y = 2: x = 3 − 4 = −1. AO1 — back-substituting for x

A1 Solution pairs: (19/7, 1/7) and (−1, 2). AO2 — presenting both pairs clearly

A1 Verify (−1, 2): (−1) + 2(2) = 3 ✓; 1+2+4 = 7 ✓. AO2 — verification of both equations