Printable · GCSE Higher · ages 14-16
Algebra worksheet — GCSE Higher
Fifteen questions across the algebra statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Algebra worksheet — GCSE Higher
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- 1.A cyclist's speed-time graph is described as follows: speed increases steadily from 0 m/s to 8 m/s over the first 4 seconds, stays constant at 8 m/s for the next 6 seconds, then decreases steadily to 0 m/s over the final 2 seconds. Work out the total distance travelled.
- 2.A student is asked whether 3(x − 4) = 3x − 4 is an identity. Which statement gives the correct verdict and reason?
- 3.By completing the square, find the turning point of the curve y = x² + 6x + 2.y = x² + 6x + 2
- 4.For the graph of y = 1/x, where x cannot be zero, which of these statements is correct?
- 5.Work out the value of (x − 4)/2 + 3 when x = 10
- 6.Make x the subject of the formula y = 4(x − 6).
- 7.Solve the inequality 5 − x ≤ 2.
- 8.The height, h metres, of a ball t seconds after a stopwatch is started follows h = (t − 1)(5 − t) for 1 ≤ t ≤ 5, where h = 0 means the ball is at ground level. Work out the two times at which the ball is at ground level.
- 9.The graph of y = x² − 4x is translated by the vector (3, 0). Work out the equation of the image, giving your answer in the form y = x² + bx + c.y = x² − 4xy = x²
- 10.f(x) = x² − 1 and g(x) = 3x. Work out fg(4).y = x² − 1
- 11.A sequence has the position-to-term rule n² + 2, where n is the position number. Work out the 6th term.
- 12.The graph of y = f(x) has x-intercepts at x = −1 and x = 4. Which statement correctly describes the x-intercepts of y = f(2 − x)?
- 13.The nth term of a sequence is 2n² + 1. Work out the 4th term of the sequence.
- 14.ABCD is a parallelogram. A has coordinates (−3, 1), B has coordinates (2, 1) and C has coordinates (4, 4). Work out the coordinates of D.
- 15.The curve y = x² − 1 and the line y = 3x − 3 meet at two points. Work out the x-coordinates of those two points.y = 3x − 3y = x² − 1
Answer key
- (c) 72 m — The first section is a triangle with base 4 s and height 8 m/s, area = 1/2 × 4 × 8 = 16 m. The middle section is a rectangle with base 6 s and height 8 m/s, area = 6 × 8 = 48 m. The final section is a triangle with base 2 s and height 8 m/s, area = 1/2 × 2 × 8 = 8 m. Total distance = 16 + 48 + 8 = 72 m. Treating both sloping sections as full rectangles instead of triangles — forgetting the factor of a half — gives 4 × 8 + 6 × 8 + 2 × 8 = 96 m, which is wrong because a sloping speed-time section is a triangle, not a rectangle. Leaving out the final, decreasing section entirely gives 16 + 48 = 64 m, which is wrong because the cyclist is still moving and covering distance while slowing down. Leaving out the middle, constant-speed section gives 16 + 8 = 24 m, which is wrong because the cyclist covers the most distance of all during the 6 seconds of constant speed.
- (b) It is not even an ordinary equation with a solution: expanding the left-hand side gives 3x − 12, and 3x − 12 = 3x − 4 would require −12 = −4, which is never true. — Expanding the left-hand side, 3(x − 4) = 3x − 12. Setting this equal to the right-hand side, 3x − 12 = 3x − 4, gives −12 = −4 once the 3x terms are removed from both sides — a statement that is never true, so no value of x satisfies the equation at all, and it is certainly not an identity. The option about substituting a specific value misunderstands algebraic expansion, which holds for every x, not one chosen value. The option matching the first term wrongly assumes that is enough to prove equivalence. The option about multiplying the 4 by 3 on both sides is nonsensical, since there is only one bracket to expand, on the left-hand side.
- (b) x = −3, y = −7 — x² + 6x + 2 = (x + 3)² − 3² + 2 = (x + 3)² − 7. Substituting x = −3: (−3)² = 9, 6 × (−3) = −18, so 9 − 18 + 2 = −7, confirming the minimum value −7 at x = −3: turning point (−3, −7). Using 6 instead of half of 6 inside the bracket gives (x + 6)² − 34, turning point (−6, −34) — wrong, because only half the coefficient of x belongs inside the bracket. Reading the bracket's sign directly as the turning point's x-coordinate gives (3, −7) — wrong, because (x + 3)² is zero at x = −3, not x = 3. Computing 9 − 2 = 7 instead of 2 − 9 = −7 flips the sign of the constant, giving (−3, 7) — wrong, since the completed square's constant must be evaluated as 2 minus 9, not 9 minus 2. Always check a turning point by substituting its x-value back into the original equation.
- (d) The graph never crosses either axis — Since x ≠ 0, there is no point on the graph where x = 0, so it cannot cross the y-axis; likewise 1/x is never equal to 0 for any x, so it cannot cross the x-axis either — the graph never touches either axis. A candidate who forgets the restriction x ≠ 0 might think the graph behaves like other graphs and passes through the origin, (0, 0). A candidate who correctly rules out the x-axis but forgets that x = 0 is also excluded might say the graph crosses the y-axis but never the x-axis. A candidate who only pictures the branch where x and y are both positive might say the graph has only one branch, in quadrant 1, forgetting the second branch where x and y are both negative.
- (c) 6 — Method: substitute the value, work out the top of the fraction first, then the division, and add the 3 last. Working: the top gives 10 − 4 = 6, dividing by 2 gives 6 ÷ 2 = 3, and adding 3 gives 3 + 3 = 6. Answer: 6. The distractors: 11 comes from dividing only the 4 by 2 instead of the whole of the top, giving 10 − 2 + 3; 4.5 comes from dividing the + 3 by 2 as well, giving (10 − 4 + 3) ÷ 2; 0 comes from subtracting the wrong way round on the top, giving (4 − 10) ÷ 2 = −3 and then −3 + 3.
- (c) x = (y + 24)/4 — Method: expand the bracket first, then undo the operations done to x in reverse order. Working: y = 4(x − 6) = 4x − 24, so y + 24 = 4x, so x = (y + 24)/4. Answer: x = (y + 24)/4. x = (y + 6)/4 comes from expanding the bracket incorrectly, treating 4(x − 6) as 4x − 6 instead of 4x − 24. x = 4y + 96 comes from multiplying by 4 instead of dividing by 4 to undo the multiplication, giving 4(y + 24) = 4y + 96. x = y/4 − 6 comes from dividing by 4 first without expanding the bracket, then subtracting 6 as if the bracket's operation still applied afterwards.
- (c) x ≥ 3 — Method: collect the number terms first; the x term is negative, so the final step multiplies both sides by −1, and that is the one step that turns the inequality sign round. Working: subtracting 5 from both sides of 5 − x ≤ 2 gives −x ≤ −3; multiplying both sides by −1 turns −x into x and −3 into 3, and because the multiplier is negative the ≤ becomes ≥, so x ≥ 3. Answer: x ≥ 3. The distractors: x ≤ 3 comes from multiplying by −1 without turning the sign round, the commonest slip on this type; x ≤ −3 comes from reading −x ≤ −3 as though the minus sign could simply be rubbed off the left-hand side; x ≥ −3 comes from turning the sign round correctly but leaving the right-hand side at −3 instead of multiplying it by −1 as well.
- (c) t = 1 second and t = 5 seconds — The ball is at ground level when h = 0, which happens when either bracket is zero. t − 1 = 0 gives t = 1, and 5 − t = 0 gives t = 5. The option with t = −1 makes a sign error solving the first bracket. The option with t = −5 makes a sign error solving the second bracket, treating 5 − t = 0 as if it gave a negative solution. The option with t = 4 seconds is an arithmetic slip in solving 5 − t = 0.
- (d) y = x² − 10x + 21 — A translation by the vector (3, 0) moves the graph 3 units in the positive x-direction, which means replacing every x in the equation with (x − 3). Substitute into x² − 4x: (x − 3)² − 4(x − 3). Expand (x − 3)² to x² − 6x + 9, and expand −4(x − 3) to −4x + 12. Collecting like terms, x² − 6x + 9 − 4x + 12 = x² − 10x + 21, so the image is y = x² − 10x + 21. Substituting (x + 3) instead of (x − 3) — translating in the wrong direction — gives y = x² + 2x − 3. Adding 3 straight onto the original equation, treating the translation as vertical instead of horizontal, gives y = x² − 4x + 3. Expanding (x − 3)² as x² − 3x + 9, using −3x instead of −6x for the middle term, and then combining with −4(x − 3) gives y = x² − 7x + 21.
- (b) 143 — fg(4) means f(g(4)): work out g(4) first, then substitute the result into f. g(4) = 3 × 4 = 12, then f(12) = 12² − 1 = 144 − 1 = 143. Working out gf(4) instead swaps the order: f(4) = 4² − 1 = 15, then g(15) = 3 × 15 = 45 — that is the wrong composition. Treating f(x) as x − 1 (forgetting to square the input) gives f(12) = 12 − 1 = 11. Applying g twice instead of applying g then f gives g(g(4)) = g(12) = 3 × 12 = 36, which mixes up which function should be applied second.
- (a) 38 — Method: substitute the position number into the rule and follow the order of operations, so the squaring is carried out before the 2 is added. Working: n = 6 gives 6² + 2; 6² means 6 × 6 = 36, and then 2 is added to 36. Answer: 38. The distractors: 14 comes from multiplying the position by 2 instead of squaring it, 6 × 2 + 2; 36 comes from squaring correctly and then forgetting to add the 2; 64 comes from adding the 2 first and squaring afterwards, (6 + 2)².
- (d) Reflect in the y-axis, +2 in x; roots 3, −2 — f(2 − x) is zero exactly when 2 − x equals one of f's roots: 2 − x = −1 or 2 − x = 4. Solving each correctly (x = 2 − (−1) = 3, and x = 2 − 4 = −2) gives the new roots x = 3 and x = −2. This is the same as reflecting y = f(x) in the y-axis to get f(−x), then translating 2 units in the positive x-direction to get f(−(x − 2)) = f(2 − x). Solving 2 − x = k as x = k − 2 instead of x = 2 − k is a sign slip when rearranging, and gives x = −3 and x = 2. Translating +2 in x with no reflection at all uses f(x − 2), whose roots are the original roots plus 2: x = 1 and x = 6 — this misses the reflection completely. Assuming 'no overall change' wrongly treats a reflection-and-translation pair as always cancelling out, when here the roots genuinely move, from x = −1 and x = 4 to x = 3 and x = −2.
- (a) 33 — Substitute n=4 into 2n²+1: 2×4²+1=2×16+1=33. A candidate who computes n² as 2×n instead of n×n would compute 2×(2×4)+1=2×8+1=17. A candidate who correctly finds 2×16 but forgets to add the constant 1 would stop at 32. A candidate who squares the whole term 2n, rather than squaring n before multiplying by 2, would compute (2×4)²+1=64+1=65.
- (d) (−1, 4) — In parallelogram ABCD the side DC is parallel and equal to the side AB, so D = C − AB. The vector from A to B is (2 − (−3), 1 − 1) = (5, 0), so D = (4 − 5, 4 − 0) = (−1, 4). A candidate who adds this vector to C instead of subtracting it gets (4 + 5, 4 + 0) = (9, 4). A candidate who subtracts A's coordinates from C's rather than the vector AB, and drops the minus sign on −3 while doing so, works out (4 − 3, 4 − 1) and gets (1, 3). A candidate who makes only the y-part of that slip, working out 4 − 1 instead of 4 − 0, gets (−1, 3).
- (d) x = 1 and x = 2 — Method: where a line meets a curve the two expressions for y are equal, so set them equal and solve the quadratic that results. Working: x² − 1 = 3x − 3 collects to x² − 3x + 2 = 0; factorising gives (x − 1)(x − 2) = 0, so x = 1 or x = 2, and each value gives the same y on both graphs. Answer: x = 1 and x = 2. The distractors: x = −1 and x = −2 come from factorising as (x + 1)(x + 2) and so reversing the sign of both roots; x = −1 and x = 4 come from moving the −3 across the equals sign without changing its sign, which gives x² − 3x − 4 = 0; x = 1 and x = −1 come from setting each expression equal to zero separately instead of equal to each other.
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