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.
Answer key: Algebra worksheet — GCSE Higher
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- (d) x₂ closer: f(x₂) = 0.0225, nearer to 0 — f(2.6) = 2.6² − 7 = 6.76 − 7 = −0.24, and f(2.65) = 2.65² − 7 = 7.0225 − 7 = 0.0225. The closer a value of x is to the root, the closer f(x) is to zero — regardless of sign. Since |0.0225| = 0.0225 is much smaller than |−0.24| = 0.24, x₂ = 2.65 is closer to the root. 'x₁ closer: −0.24 is the smaller value' comes from comparing the SIGNED values of f(x) rather than their distances from zero — −0.24 is indeed less than 0.0225 as a number, but that does not mean x₁ is closer to the root. 'x₁ closer: f(x₁) negative ⇒ nearer root' invents a rule that a negative f(x) means x is closer to the root; the sign of f(x) only tells you which side of the root x is on, not how close it is. 'x₂ is the exact root, since f(x₂) ≈ 0' misreads f(x₂) = 0.0225 as zero; the true root is √7 ≈ 2.6458, so f(2.65) is close to zero but not equal to it, and x₂ is an approximation, not the exact root.
- (a) An identity, true for every value of x — Expanding the bracket on the left gives 3x + 12, which matches the right-hand side exactly, so the statement is true for every value of x — this makes it an identity. A candidate who reasons that any statement with an equals sign must be an equation picks that option, missing that an equation is only true for particular value(s) of x, not all of them. A candidate who confuses an identity with a formula, because both relate two expressions, picks the formula option — but a formula connects two different quantities, such as area and side length, not two equivalent forms of the same expression. A candidate who assumes it can be solved for a single value of x, as with a normal equation, picks that option, not realising there is no single solution here.
- (c) −4 — f(2) = 7, and y = f(x) + a passing through (2, 3) means f(2) + a = 3, so 7 + a = 3, giving a = 3 − 7 = −4. Writing a = 4 comes from subtracting the wrong way round, 7 − 3 instead of 3 − 7. Writing a = 10 comes from adding the two y-values instead of subtracting one from the other. Writing a = 3 comes from taking the new y-coordinate as the value of a directly, without accounting for the original y-value of 7 at all.
- (d) £195 — Substituting x = 3 into y = 45x + 60 gives y = 45 × 3 + 60 = 135 + 60 = 195. A candidate who forgets to add the call-out fee would get only 45 × 3 = £135. A candidate who adds the hours to the fee and the rate instead of multiplying would get 45 + 60 + 3 = £108. A candidate who multiplies both the hourly rate and the call-out fee by the number of hours would get 45 × 3 + 60 × 3 = £315.
- (b) 12n — Distributing the minus sign across the second bracket gives n² + 6n + 9 − n² + 6n − 9, and the n² terms and the +9/−9 cancel, leaving 6n + 6n = 12n. Writing 18 comes from only negating the first term of the second bracket, n², and treating the −6n and +9 as unchanged, which gives n² + 6n + 9 − n² − 6n + 9 = 18. Writing 2n² + 18 comes from adding the two brackets instead of subtracting them, (n² + 6n + 9) + (n² − 6n + 9) = 2n² + 18. Writing 6n comes from correctly negating the bracket but then only counting one of the two 6n terms, missing that they add rather than cancel.
- (a) x = 6 — Method: a minus sign in front of a bracket changes the sign of every term inside it, so expand the bracket first and then simplify. Working: expanding gives 4x − 2x + 6 = 18, which simplifies to 2x + 6 = 18; subtracting 6 from both sides gives 2x = 12, and dividing both sides by 2 gives x = 6. Answer: x = 6. The distractors: x = 12 comes from leaving the −6 unchanged when the bracket is removed, giving 4x − 2x − 6 = 18 and so 2x = 24; x = 24 comes from reaching 2x = 12 correctly and then multiplying by 2 instead of dividing by 2; x = 3 comes from dividing by 2 too early, at 2x + 6 = 18, and dividing only the 2x and the 18 while leaving the 6 untouched, which gives x + 6 = 9.
- (a) t = 0 or t = 4 — The ball is at ground level exactly when h = 0. From the table, h = 0 at t = 0 and at t = 4, so those are the two times. Distractor origins: t = 1 or t = 3 picks the times with equal (but non-zero) height instead of ground level; t = 2 picks the time of maximum height instead of ground level; t = 0 finds only the starting time and misses the second one.
- (d) −2 — Method: run the machine backwards, undoing the operations in the opposite order and swapping each one for its inverse. Working: the machine added 7 last, so take 7 off the output: 1 − 7 = −6; before that the machine had multiplied by 3, so divide: −6 ÷ 3, and a negative divided by a positive stays negative. Answer: −2, which checks because 3 × (−2) + 7 = −6 + 7 = 1. The distractors: 2 comes from dividing 6 by 3 and losing the minus sign; −6 comes from taking the 7 off and stopping there, never undoing the multiplication; −18 comes from multiplying −6 by 3 instead of dividing by 3.
- (d) g is one-to-one: f(3) = f(−3), removed by x ≥ 0 — A function has an inverse only if it is one-to-one: every output must come from exactly one input. f(3) = 9 and f(−3) = 9, so two different inputs give the same output, and there is no way to send 9 back to a single input — f is not one-to-one over all real x. Restricting the domain to x ≥ 0 removes one of the two inputs behind every such pair, so g is one-to-one and does have an inverse. 'g's outputs are positive; f's could be negative' is wrong because f(x) = x² also only gives outputs of 0 or more — the outputs of f and g are identical sets; it is the INPUTS that differ, not the outputs. 'Restricting any domain always creates an inverse' is wrong because a restriction only helps if it actually removes the repeated outputs: restricting f(x) = x² to x ≥ −3 still leaves f(1) = f(−1) = 1, so that restricted function is still not one-to-one and still has no inverse. 'Squares can never be reversed, under any conditions' is wrong because √9 = 3 does reverse 3² = 9 once you know the input was non-negative — a square root just cannot tell you WHICH of two inputs you started from unless the domain has already ruled one of them out.
- (a) y = 2ˣ — An exponential graph y = 2ˣ passes through (0, 1) since 2⁰ = 1, rises more and more steeply for positive x, and has the x-axis as an asymptote as x becomes very negative, since 2ˣ gets closer to 0 without ever reaching it. y = x² + 1 also passes through (0, 1) and also rises steeply for positive x, but as x becomes very negative it rises to infinity too, rather than settling towards the x-axis — mistaking any curve that gets steeper for an exponential misses this. y = x³ + 1 passes through (0, 1) and rises for positive x, but as x becomes very negative it falls towards negative infinity rather than approaching the x-axis from above. y = 1 − x² also passes through (0, 1), but it falls for large positive x rather than rising — a candidate who checks only the y-intercept, without reading the described shape of the curve, could pick this.
- (d) 6th term — The terms are aₙ = 3 × (√3)ⁿ⁻¹: a₁ = 3, a₂ = 3√3 ≈ 5.196, a₃ = 9, a₄ = 9√3 ≈ 15.588, a₅ = 27, a₆ = 27√3 ≈ 46.765. Since a₅ = 27 is still below 30 and a₆ ≈ 46.8 is above 30, the 6th term is the first to exceed 30. 4th term comes from treating the common ratio as 3 instead of √3: 3, 9, 27, 81, … — under that wrong ratio, 81, the 4th term, is the first over 30. 5th term comes from misjudging a₅ = 27 as already greater than 30. 7th term comes from tracking only the whole-number terms (3, 9, 27, 81, …) and reporting the ORIGINAL position of 81 in the full sequence — 81 is indeed a₇, but a₆ ≈ 46.8 already exceeds 30 first.
- (c) −6 — Method: substitute the value, work out the bracket first, then multiply what the bracket comes to by 2. Working: the bracket gives (−5) + 2 = −3, and multiplying by 2 gives 2 × (−3) = −6. Answer: −6. The distractors: −3 comes from working out the bracket correctly but stopping there and never multiplying by 2; −1 comes from multiplying the 2 inside the bracket by the 2 outside before adding, giving −5 + 4; −8 comes from multiplying only the x by 2 and leaving the 2 inside the bracket unchanged, giving −10 + 2.
- (a) 11 — 2n − 7 = 15, so adding 7 to both sides gives 2n = 22, and dividing by 2 gives n = 11. A candidate who makes a sign error and forms the equation 2n + 7 = 15 gets 2n = 8 and n = 4. A candidate who correctly adds 7 to get 2n = 22 but forgets to divide by 2 gets n = 22. A candidate who ignores the −7 altogether and solves 2n = 15 gets n = 7.5.
- (b) An identity, since both sides are equal for every x. — Expanding the left-hand side: 5(2x−3)=10x−15, which is exactly the same as the right-hand side for every value of x, so it is an identity, not an equation that is only true for one particular x. A candidate who treats every equals-sign statement as an equation, without checking whether it holds for all values of x, would choose the equation option. A candidate who mistakes it for a formula is assuming it relates two different letters or quantities, but only x appears — there is no second variable such as area or cost — so it is not a formula. A candidate who mistakes it for an inequality is assuming the two sides are only equal for particular values of x, but expanding shows they are identical for every value, not just some.
- (a) x = 6 or x = −4 — We need two numbers that multiply to −24 and add to −2: these are −6 and 4, since −6 × 4 = −24 and −6 + 4 = −2. So x² − 2x − 24 = (x − 6)(x + 4) = 0, giving x = 6 or x = −4. A candidate who swaps the signs, using 6 and −4 the wrong way round in the brackets, gets x = −6 or x = 4. A candidate who picks the wrong factor pair, 8 and −3 (which multiply to −24 but add to +5, not −2), gets (x + 8)(x − 3) = 0 and answers x = −8 or x = 3. A candidate who makes both factors negative gets x = −6 or x = −4, which would require the constant term to be +24, not −24.
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