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Edexcel International A Level Maths · P1 · Pure Mathematics 1
28 free, high-quality past-paper-style questions on Indices, surds and quadratics for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
Give exact answers to this question.
Find the value of 1252/3.
Find the value of (4⁄9)−3/2.
Simplify (x−3)−2/3, where x > 0.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 1251/3 = 5 | M1 | |
| 52 = 25 | A1 | |
| Part (b) | ||
| reciprocal and root used, e.g. (9⁄4)3/2 or (2⁄3)−3 or (3⁄2)3 | M1 | |
| = 27⁄8 | A1 | |
| Part (c) | ||
| x2 | B1 | |
Common error: A negative index does not make the answer negative. It turns the base upside down: (4/9)−3/2 = (9/4)3/2.
In this question, show all your working and give each answer in its simplest form.
Simplify √98 − √32 + √8, giving your answer in the form k√2, where k is an integer.
Expand and simplify (3 + 2√5)(3 − √5).
Show that (√6 + √2)2 = 8 + 4√3.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| at least two of √98 = 7√2, √32 = 4√2, √8 = 2√2 | M1 | |
| 5√2 | A1 | |
| Part (b) | ||
| 9 − 3√5 + 6√5 − 2 × 5 (four terms, at least three correct) | M1 | |
| −1 + 3√5 | A1 | |
| Part (c) | ||
| 6 + 2√12 + 2 (or 6 + √12 + √12 + 2) | M1 | |
| √12 = 2√3, so 8 + 4√3 with no errors seen | A1 | |
Exam tip: In a “show that” part the answer is given, so each step must be visible: write √12 and then 2√3, not just the final line.
The expression x2 − 8x + 21 is to be written in completed square form.
Write x2 − 8x + 21 in the form (x + a)2 + b, where a and b are integers.
Hence state the minimum value of x2 − 8x + 21 and the value of x at which it occurs.
Hence explain why the equation x2 − 8x + 21 = 0 has no real roots.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| (x − 4)2 seen | M1 | |
| (x − 4)2 + 5 | A1 | |
| Part (b) | ||
| minimum value 5 ft their b | B1 | |
| when x = 4 ft their −a | B1 | |
| Part (c) | ||
| (x − 4)2 ≥ 0 for all real x, so x2 − 8x + 21 ≥ 5 | M1 | |
| the expression is never zero, so the equation has no real roots | A1 | |
Exam tip: “Hence” means use the completed square. A correct discriminant argument (64 − 84 = −20 < 0) is true but does not answer a “hence” part.
The curve C has equation y = 10 + 3x − x2.
Factorise 10 + 3x − x2 completely.
Write down the coordinates of the points where C meets the coordinate axes.
Find the equation of the line of symmetry of C and the coordinates of its maximum point.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| attempt at two brackets whose constant terms multiply to give ±10 | M1 | |
| (5 − x)(2 + x) or −(x − 5)(x + 2) | A1 | |
| Part (b) | ||
| (−2, 0) and (5, 0) ft their factors | B1 | |
| (0, 10) | B1 | |
| Part (c) | ||
| x = 3⁄2 (midway between the roots, or by completing the square) | M1 | |
| y = 10 + 9⁄2 − 9⁄4: maximum point (3⁄2, 49⁄4) | A1 | |
Common error: Losing the sign of the x2 term when factorising. Expand your brackets to check: (5 − x)(2 + x) = 10 + 3x − x2.
Figure 1 shows part of the curve with equation y = x2 + bx + c, where b and c are constants. The curve meets the x-axis at the points A and B and the y-axis at the point C, as marked. The point V is the minimum point of the curve.
Use the points A and B to find the value of b and the value of c.
Write x2 + bx + c in the form (x + p)2 + q, where p and q are integers.
Hence state the coordinates of V and the equation of the line of symmetry of the curve.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| y = (x + 2)(x − 4) | M1 | |
| = x2 − 2x − 8, so b = −2, c = −8 | A1 | |
| Part (b) | ||
| (x − 1)2 seen | M1 | |
| (x − 1)2 − 9 ft their b, c | A1 | |
| Part (c) | ||
| V(1, −9) ft | B1 | |
| x = 1 ft | B1 | |
Common error: Writing the roots straight into the brackets with the wrong sign. A root at x = −2 gives the factor (x + 2), not (x − 2). Check by putting x = 0: (2)(−4) = −8 matches the point C.
Figure 1 shows the triangle PQR, in which angle PQR = 90°. The lengths of PQ and QR are shown on the figure. Give each answer in exact, simplified form.
Write the length of PQ and the length of QR each in the form k√5 cm, where k is an integer.
Find the area of the triangle.
Find the perimeter of the triangle, giving your answer in the form (a√5 + √b) cm, where a and b are integers.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| PQ = √(4 × 5) = 2√5 cm | B1 | |
| QR = √(9 × 5) = 3√5 cm | B1 | |
| Part (b) | ||
| ½ × 2√5 × 3√5 (or ½ × √900) | M1 | |
| = ½ × 6 × 5 = 15 cm2 | A1 | |
| Part (c) | ||
| PR2 = 20 + 45 = 65, so PR = √65 | M1 | |
| perimeter = (5√5 + √65) cm | A1 | |
Common error: Adding surds that do not share a root: √20 + √45 is not √65. Simplify each to a multiple of √5 first, then add the multiples: 2√5 + 3√5 = 5√5.
Figure 1 shows the curve with equation y = kxn, x ≥ 0, where k and n are constants. The curve passes through the marked points A and B. Answer this question without using logarithms.
Use the point A to find the value of k.
Use the point B to show that n = 3⁄2.
Find the value of y when x = 16.
Find the value of x for which y = 1⁄4.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 1n = 1, so k = 2 | B1 | |
| Part (b) | ||
| 2 × 9n = 54, so 9n = 27 | M1 | |
| 32n = 33, so 2n = 3 and n = 3⁄2 | A1 | |
| Part (c) | ||
| 2 × 163/2 = 2 × 64 = 128 | B1 | |
| Part (d) | ||
| x3/2 = 1⁄8 | M1 | |
| x = (1⁄8)2/3 = (1⁄2)2 = 1⁄4 | A1 | |
Exam tip: To undo a power of 3⁄2, raise both sides to the power 2⁄3: take the cube root first, then square. Working with the root first keeps the numbers small.
The three quadratic expressions f, g and h are given by
f(x) = x2 + x − 6, g(x) = x2 − 6x + 9, h(x) = 2x2 + 3x + 4.
Figure 1 shows sketches of the curves y = f(x), y = g(x) and y = h(x), in a different order and not to the same scale.
Calculate the value of the discriminant of each of f(x), g(x) and h(x).
Hence match each expression to Graph 1, Graph 2 or Graph 3, giving a reason for each.
Find the coordinates of the points where the curve in Graph 2 meets the x-axis.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| f: 12 − 4 × 1 × (−6) = 25 | B1 | |
| g: (−6)2 − 4 × 1 × 9 = 0 | B1 | |
| h: 32 − 4 × 2 × 4 = −23 | B1 | |
| Part (b) | ||
| any two correct with reasons, e.g. h: discriminant < 0, no real roots, so the curve does not meet the x-axis ft | B1 | |
| all three: f → Graph 2 (two distinct roots), g → Graph 3 (equal roots, touches), h → Graph 1 ft | B1 | |
| Part (c) | ||
| (x + 3)(x − 2) = 0 | M1 | |
| (−3, 0) and (2, 0) | A1 | |
Exam tip: The sign of b2 − 4ac tells you how a curve meets the x-axis: positive, it crosses twice; zero, it touches; negative, it misses. You do not need to solve the equation to decide.
Given that y = (4x3 − 5√x) ÷ (2x2), where x > 0,
express y in the form Axp + Bxq, where A, B, p and q are constants.
Find the exact value of y when x = 4.
Solve the equation x−2/3 = 4⁄9, where x > 0.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 4x3 ÷ 2x2 = 2x | B1 | |
| √x written as x1/2 and powers subtracted: x1/2 − 2 | M1 | |
| y = 2x − 5⁄2x−3/2 | A1 | |
| Part (b) | ||
| 2 × 4 − 5⁄2 × 4−3/2 with 4−3/2 = 1⁄8 | M1 | |
| = 8 − 5⁄16 = 123⁄16 | A1 | |
| Part (c) | ||
| x2/3 = 9⁄4 or x1/3 = 3⁄2 | M1 | |
| x = 27⁄8 | A1 | |
Exam tip: Split the fraction into one term per numerator term before applying the index laws, and keep each coefficient separate from its power of x.
Answer this question without using logarithms.
Write √2 and 8x + 2 each as a power of 2.
Hence solve 42x − 1 = 8x + 2 ÷ √2.
Solve 9y × 27y − 1 = 38.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| √2 = 21/2 | B1 | |
| 8x + 2 = 23x + 6 | B1 | |
| Part (b) | ||
| 42x − 1 = 24x − 2 | B1 | |
| indices equated: 4x − 2 = 3x + 6 − ½ | M1 | |
| x = 15⁄2 | A1 | |
| Part (c) | ||
| 32y × 33y − 3 = 38 so 5y − 3 = 8 | M1 | |
| y = 11⁄5 | A1 | |
Exam tip: Choose the smallest base that every number is a power of (here 2, then 3), and write the whole of each side as a single power before equating indices.
Give each answer in its simplest form, showing every step of your working.
Write 12 ÷ √3 in the form k√3, where k is an integer.
Write (5 − √2) ÷ (3 + 2√2) in the form a + b√2, where a and b are integers.
Show that 1 ÷ (√11 − 3) − 1 ÷ (√11 + 3) = 3.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| multiply numerator and denominator by √3: 12√3 ÷ 3 | M1 | |
| 4√3 | A1 | |
| Part (b) | ||
| numerator and denominator multiplied by 3 − 2√2 | M1 | |
| denominator 9 − 8 = 1 | B1 | |
| numerator 15 − 10√2 − 3√2 + 4 = 19 − 13√2 | A1 | |
| Part (c) | ||
| combine over the common denominator (√11 − 3)(√11 + 3), or rationalise each fraction | M1 | |
| denominator 11 − 9 = 2 | B1 | |
| numerator (√11 + 3) − (√11 − 3) = 6, and 6 ÷ 2 = 3 | A1 | |
Common error: Squaring 2√2 as 4 instead of 8. (2√2)2 = 4 × 2 = 8, so (3 + 2√2)(3 − 2√2) = 9 − 8 = 1.
The equation (k − 1)x2 + 4x + (k + 2) = 0, where k is a constant and k ≠ 1, has equal roots.
Show that k2 + k − 6 = 0.
Hence find the possible values of k.
For each value of k, find the repeated root of the equation.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| discriminant with a = k − 1, b = 4, c = k + 2: 16 − 4(k − 1)(k + 2) | M1 | |
| set equal to 0 and expand (k − 1)(k + 2) = k2 + k − 2 | M1 | |
| 16 − 4k2 − 4k + 8 = 0 ⇒ k2 + k − 6 = 0, no errors seen | A1 | |
| Part (b) | ||
| (k + 3)(k − 2) = 0 | M1 | |
| k = −3 or k = 2 | A1 | |
| Part (c) | ||
| x = −4 ÷ [2(k − 1)], or substitute k and factorise | M1 | |
| k = 2: x2 + 4x + 4 = 0, x = −2 | A1 | |
| k = −3: −4x2 + 4x − 1 = 0, x = ½ | A1 | |
Exam tip: When a or c contains k, put brackets round it in 4ac before expanding. Equal roots of ax2 + bx + c = 0 are at x = −b/2a.
Each part of this question asks for a different method of solving a quadratic equation.
By factorising, solve 10x2 − 9x − 9 = 0.
By completing the square, solve x2 + 5x − 3 = 0, giving your answers in exact form.
Solve 4x2 − 3x − 5 = 0, giving your answers to 3 significant figures.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| (5x + 3)(2x − 3) = 0 | M1 | |
| x = −3⁄5 or x = 3⁄2 | A1 | |
| Part (b) | ||
| (x + 5⁄2)2 − 25⁄4 − 3 = 0 | M1 | |
| (x + 5⁄2)2 = 37⁄4 | A1 | |
| x = (−5 ± √37) ÷ 2 | A1 | |
| Part (c) | ||
| formula with a = 4, b = −3, c = −5: (3 ± √(9 + 80)) ÷ 8 | M1 | |
| (3 ± √89) ÷ 8 | A1 | |
| x = 1.55 or x = −0.804 | A1 | |
Common error: Writing −b as −3 when b = −3. With a negative b, −b is positive: here it is +3.
The curve with equation y = 3x2 + bx + c, where b and c are constants, passes through the points (0, −4) and (2, 6).
Find the value of b and the value of c.
Calculate the value of the discriminant of 3x2 + bx + c, and hence state the number of points where the curve meets the x-axis.
Find the x-coordinates of the points where the curve meets the x-axis.
Write down the equation of the line of symmetry of the curve.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| c = −4 | B1 | |
| substitute (2, 6): 12 + 2b − 4 = 6 | M1 | |
| b = −1 | A1 | |
| Part (b) | ||
| (−1)2 − 4 × 3 × (−4) = 49 ft their b, c | M1 | |
| 49 > 0, so two points | A1 | |
| Part (c) | ||
| (3x − 4)(x + 1) = 0 | M1 | |
| x = 4⁄3 and x = −1 | A1 | |
| Part (d) | ||
| x = 1⁄6 ft | B1 | |
Model sketch
Exam tip: A discriminant that is a perfect square (here 49 = 72) tells you the quadratic factorises with rational roots.
A jet of water leaves a nozzle N, which is 0.8 m above horizontal ground. The path of the water is modelled by
h = a − b(x − c)2, h ≥ 0,
where h metres is the height of the water above the ground and x metres is its horizontal distance from the nozzle, and a, b and c are positive constants. Figure 1 shows the path, which starts at N and has its highest point at M.
Use Figure 1 to find the values of a, b and c.
Find the horizontal distance from the nozzle at which the water reaches the ground. Give your answer in the form p + √q, where p and q are integers, and also correct to 3 significant figures.
A fence of height 1.5 m stands on the ground, perpendicular to the plane of the path, 5.5 m horizontally from the nozzle. Determine whether the water passes over the fence.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| highest point (3, 2.6): c = 3 and a = 2.6 | B1 | |
| substitute (0, 0.8): 0.8 = 2.6 − 9b | M1 | |
| b = 0.2 | A1 | |
| Part (b) | ||
| 2.6 − 0.2(x − 3)2 = 0, so (x − 3)2 = 13 | M1 | |
| x = 3 ± √13 | M1 | |
| 3 − √13 < 0 is rejected: x = 3 + √13 = 6.61 m | A1 | |
| Part (c) | ||
| h = 2.6 − 0.2 × 2.52 = 1.35 | M1 | |
| 1.35 < 1.5, so the water does not pass over the fence | A1 | |
Exam tip: In the form a − b(x − c)2, the highest point is (c, a). Read those from the graph first; one more point on the curve then gives the last constant.
Figure 1 shows the trapezium ABCD, in which AB is parallel to DC and angle DAB = angle ADC = 90°. The lengths of AB, DC and AD are shown on the figure. The area of the trapezium is (6 + 4√3) cm2. Give all answers in exact form.
Show that h = 1 + √3.
Show that BC2 = 8 + 2√3.
Express AB ÷ DC in the form p + q√3, where p and q are integers.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ½(6 + 2√3)h = 6 + 4√3, so h = (6 + 4√3) ÷ (3 + √3) | M1 | |
| multiply numerator and denominator by 3 − √3; denominator 9 − 3 = 6 | M1 | |
| numerator 18 − 6√3 + 12√3 − 12 = 6 + 6√3, so h = 1 + √3, no errors seen | A1 | |
| Part (b) | ||
| horizontal difference AB − DC = 2, so BC2 = 22 + (1 + √3)2 | M1 | |
| (1 + √3)2 = 4 + 2√3, so BC2 = 8 + 2√3, no errors seen | A1 | |
| Part (c) | ||
| (4 + √3) ÷ (2 + √3) multiplied top and bottom by 2 − √3 | M1 | |
| denominator 4 − 3 = 1 | B1 | |
| numerator 8 − 4√3 + 2√3 − 3 = 5 − 2√3 | A1 | |
Common error: Multiplying only the denominator by the conjugate. The numerator must be multiplied by the same bracket, otherwise the value of the fraction changes.
Figure 1 shows the curves C1 and C2 with equations
C1: y = 4x and C2: y = 9 × 2x − 8.
The curves meet at the points A and B. Answer this question without using logarithms.
Using the substitution u = 2x, show that the x-coordinates of A and B satisfy u2 − 9u + 8 = 0.
Hence find the coordinates of A and B.
The point E on C1 has y-coordinate 1⁄8. Find the x-coordinate of E.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 4x = (22)x = (2x)2 = u2 | B1 | |
| u2 = 9u − 8, so u2 − 9u + 8 = 0 | A1 | |
| Part (b) | ||
| (u − 1)(u − 8) = 0 | M1 | |
| u = 1 or u = 8 | A1 | |
| 2x = 1 gives x = 0; 2x = 8 gives x = 3 | A1 | |
| A(0, 1) and B(3, 64) | A1 | |
| Part (c) | ||
| 22x = 2−3 | M1 | |
| x = −3⁄2 | A1 | |
Exam tip: Look for a power that is the square of another: 4x = (2x)2. The equation is then a quadratic in 2x; solve for u and convert each value back to x.
Figure 1 shows the curve with equation y = f(x), where f(x) = p − (x − q)2 and p and q are constants. The curve has a maximum point at M and meets the y-axis at P.
Use Figure 1 to write down the values of p and q, and show that f(x) = 1 + 4x − x2.
Solve f(x) = 0, giving your answers in simplest surd form.
State the set of values of k for which the equation f(x) = k has two distinct real roots.
The line with equation y = c − 2x, where c is a constant, is a tangent to the curve. Find the value of c and the coordinates of the point of contact.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| p = 5, q = 2 | B1 | |
| 5 − (x2 − 4x + 4) = 1 + 4x − x2 | A1 | |
| Part (b) | ||
| (x − 2)2 = 5 | M1 | |
| x = 2 ± √5 | A1 | |
| Part (c) | ||
| k < 5 ft their p | B1 | |
| Part (d) | ||
| 1 + 4x − x2 = c − 2x, so x2 − 6x + (c − 1) = 0 | M1 | |
| equal roots: 36 − 4(c − 1) = 0 | M1 | |
| c = 10 | A1 | |
| x2 − 6x + 9 = 0, x = 3: point (3, 4) | A1 | |
Exam tip: A line is a tangent when the equation formed by setting curve = line has equal roots. Put everything on one side first, then use b2 − 4ac = 0 with the brackets kept around c − 1.
Figure 1 shows the curve with equation y = 2x2 − 5x − 4, drawn on a grid.
Use Figure 1 to write down estimates, to 1 decimal place, of the two roots of the equation 2x2 − 5x − 4 = 0.
Use the quadratic formula to solve 2x2 − 5x − 4 = 0, giving your answers to 3 significant figures.
Write 2x2 − 5x − 4 in the form a(x + b)2 + c, where a, b and c are constants.
Hence find the set of values of the constant k for which the equation 2x2 − 5x − 4 = k has no real roots.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| both roots, allow 3.1 to 3.2 and −0.7 to −0.6 | B1 | |
| Part (b) | ||
| a = 2, b = −5, c = −4: (5 ± √(25 + 32)) ÷ 4 | M1 | |
| (5 ± √57) ÷ 4 | A1 | |
| x = 3.14 or x = −0.637 | A1 | |
| Part (c) | ||
| 2(x2 − 5⁄2x) − 4 | M1 | |
| 2[(x − 5⁄4)2 − 25⁄16] − 4 | M1 | |
| 2(x − 5⁄4)2 − 57⁄8 | A1 | |
| Part (d) | ||
| k < −57⁄8 ft | B1 | |
Exam tip: A graph gives estimates; the formula gives accurate values. Use the graph to check that the accurate roots are sensible: 3.14 and −0.637 sit where the curve crosses the x-axis.
Figure 1 shows a cuboid. Its edges have lengths 2x cm, x3/2 cm and x1/2 cm, as marked, where x > 0. A and G are opposite vertices of the cuboid.
Show that the volume of the cuboid is 2x3 cm3.
Given that the volume of the cuboid is 54 cm3, find the value of x.
For this value of x, find the total surface area of the cuboid, giving your answer in the form (a + b√3) cm2, where a and b are integers.
Find the exact length AG.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 2x × x3/2 × x1/2, adding indices 1 + 3⁄2 + 1⁄2 | M1 | |
| = 2x3, no errors seen | A1 | |
| Part (b) | ||
| x3 = 27, x = 3 | B1 | |
| Part (c) | ||
| edges 6, 3√3 and √3 | M1 | |
| face areas 6 × 3√3 = 18√3, 6 × √3 = 6√3, 3√3 × √3 = 9 | A1 | |
| 2(18√3 + 6√3 + 9) = (18 + 48√3) cm2 | A1 | |
| Part (d) | ||
| AG2 = 62 + (3√3)2 + (√3)2 = 36 + 27 + 3 | M1 | |
| AG = √66 cm | A1 | |
Common error: Multiplying the indices instead of adding them. x × x3/2 × x1/2 = x1 + 3/2 + 1/2 = x3, not x3/4.
f(x) = 5 − 8x − 2x2, x ∈ ℝ.
Write f(x) in the form a + b(x + c)2, where a, b and c are integers.
State the maximum value of f(x) and the equation of the line of symmetry of the curve y = f(x).
Using your answer to part (a), solve f(x) = 0, giving your answers in simplest surd form.
State the value of the constant q for which the equation f(x) = q has exactly one solution.
The equation f(x) = 2x + t, where t is a constant, has no real roots. Find the set of possible values of t.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| −2(x2 + 4x) + 5 | M1 | |
| −2[(x + 2)2 − 4] + 5 | M1 | |
| 13 − 2(x + 2)2 | A1 | |
| Part (b) | ||
| maximum value 13 ft | B1 | |
| x = −2 ft | B1 | |
| Part (c) | ||
| (x + 2)2 = 13⁄2 | M1 | |
| x + 2 = ±√(13⁄2) | M1 | |
| x = −2 ± ½√26 | A1 | |
| Part (d) | ||
| q = 13 ft | B1 | |
| Part (e) | ||
| 2x2 + 10x + (t − 5) = 0 | M1 | |
| 102 − 4 × 2 × (t − 5) < 0 | M1 | |
| t > 35⁄2 | A1 | |
Model sketch
Exam tip: With a negative x2 coefficient, take out −2 (not 2) from the x2 and x terms, and remember the −2 multiplies the −4 you subtract inside the bracket.
Each equation below can be solved as a quadratic equation after a suitable substitution. Find all the real solutions of each.
Using the substitution u = x1/3, solve x2/3 + 2 = 3x1/3.
Solve 32x + 1 − 28 × 3x + 9 = 0.
Solve (x2 + x)2 − 8(x2 + x) + 12 = 0.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| u2 − 3u + 2 = 0 | M1 | |
| u = 1 or u = 2 | A1 | |
| x = u3 | M1 | |
| x = 1 or x = 8 | A1 | |
| Part (b) | ||
| 32x + 1 = 3(3x)2, giving 3u2 − 28u + 9 = 0 with u = 3x | B1 | |
| (3u − 1)(u − 9) = 0 | M1 | |
| u = 1⁄3 or u = 9 | A1 | |
| x = −1 or x = 2 | A1 | |
| Part (c) | ||
| with w = x2 + x: w2 − 8w + 12 = (w − 2)(w − 6) = 0 | M1 | |
| w = 2 or w = 6 | A1 | |
| solve x2 + x − 2 = 0 or x2 + x − 6 = 0 | M1 | |
| x = −3, −2, 1, 2 | A1 | |
Exam tip: After solving for the new letter, every value must be turned back into x. Check each one in the original equation; in (c) there are four solutions, not two.
A rectangle has length (3 + √2) cm and area (8 + 5√2) cm2. Give all answers in exact form.
Show that the width of the rectangle is (2 + √2) cm.
Find the perimeter of the rectangle in the form (a + b√2) cm.
Show that the square of the length of a diagonal of the rectangle is (17 + 10√2) cm2.
Verify that 2 + √2 is a root of x2 − 4x + 2 = 0 and, by completing the square, find the other root.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| width = (8 + 5√2) ÷ (3 + √2) | M1 | |
| multiply numerator and denominator by 3 − √2, denominator 9 − 2 = 7 | M1 | |
| numerator 24 − 8√2 + 15√2 − 10 = 14 + 7√2, so width = 2 + √2, no errors seen | A1 | |
| Part (b) | ||
| (10 + 4√2) cm | B1 | |
| Part (c) | ||
| (3 + √2)2 + (2 + √2)2 with at least one square expanded | M1 | |
| 11 + 6√2 and 6 + 4√2 | A1 | |
| sum 17 + 10√2, no errors seen | A1 | |
| Part (d) | ||
| (2 + √2)2 − 4(2 + √2) + 2 = 6 + 4√2 − 8 − 4√2 + 2 = 0 | B1 | |
| (x − 2)2 − 4 + 2 = 0 ⇒ (x − 2)2 = 2 | M1 | |
| other root 2 − √2 | A1 | |
Exam tip: Dividing by a surd expression is the same skill as rationalising: multiply top and bottom by the conjugate, and check the answer by multiplying it back out.
f(x) = x2 + 2px + 3p + 4, where p is a constant.
Write f(x) in the form (x + a)2 + b, where a and b are expressions in p.
Hence write down, in terms of p, the coordinates of the vertex of the curve y = f(x).
Given that the vertex lies on the x-axis, find the possible values of p.
Given instead that p = 2, show that the equation f(x) = 0 has no real roots.
Given instead that x = −2 is a root of f(x) = 0, find the value of p and the other root.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| (x + p)2 seen | M1 | |
| (x + p)2 − p2 + 3p + 4 | A1 | |
| Part (b) | ||
| x-coordinate −p ft | B1 | |
| y-coordinate −p2 + 3p + 4 ft | B1 | |
| Part (c) | ||
| −p2 + 3p + 4 = 0, i.e. p2 − 3p − 4 = 0 | M1 | |
| (p − 4)(p + 1) = 0 | M1 | |
| p = 4 or p = −1 | A1 | |
| Part (d) | ||
| f(x) = (x + 2)2 + 6, or discriminant 42 − 4 × 1 × 10 = −24 | M1 | |
| (x + 2)2 + 6 ≥ 6 > 0, or −24 < 0, so no real roots | A1 | |
| Part (e) | ||
| 4 − 4p + 3p + 4 = 0 | M1 | |
| p = 8 | A1 | |
| x2 + 16x + 28 = (x + 2)(x + 14), other root −14 | A1 | |
Exam tip: The vertex lies on the x-axis exactly when the equation has equal roots, so part (c) can be checked with the discriminant: 4p2 − 4(3p + 4) = 0 gives the same equation.
Figure 1 shows the triangle ABC, in which angle ABC = 90°, AB = x cm, BC = (x + 2) cm and AC = 6 cm. The point D lies on AC so that BD is perpendicular to AC, and BD = d cm. Give all answers in exact form.
Show that x2 + 2x − 16 = 0.
By completing the square, find the exact value of x.
Without substituting your value of x, show that the area of the triangle is 8 cm2.
Show that AB ÷ BC = (9 − √17) ÷ 8.
Find the value of d.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| x2 + (x + 2)2 = 62 | M1 | |
| 2x2 + 4x + 4 = 36, so x2 + 2x − 16 = 0, no errors seen | A1 | |
| Part (b) | ||
| (x + 1)2 − 1 − 16 = 0 | M1 | |
| x = −1 ± √17 | A1 | |
| x > 0 for a length, so x = √17 − 1 | A1 | |
| Part (c) | ||
| area = ½x(x + 2) = ½(x2 + 2x) | M1 | |
| x2 + 2x = 16 from part (a), so area = 8 cm2 | A1 | |
| Part (d) | ||
| (√17 − 1) ÷ (√17 + 1) | M1 | |
| multiply top and bottom by √17 − 1; denominator 17 − 1 = 16 | M1 | |
| numerator 17 − 2√17 + 1 = 18 − 2√17, and (18 − 2√17) ÷ 16 = (9 − √17) ÷ 8, no errors seen | A1 | |
| Part (e) | ||
| area = ½ × AC × BD: ½ × 6 × d = 8 | M1 | |
| d = 8⁄3 | A1 | |
Exam tip: Part (c) is quicker without the surd: the area needs x2 + 2x, and part (a) already tells you that equals 16. Look for an earlier result before substituting a messy value.
Figure 1 shows the curve C with equation y = x2 − 2x + 4 and the straight line L with equation y = mx, where m is a constant. The line L passes through the origin O and touches C at the point T.
Show that the x-coordinates of any points where the line y = mx meets C satisfy x2 − (m + 2)x + 4 = 0.
Find the two values of m for which the line y = mx is a tangent to C.
Using Figure 1, state which of these values is the gradient of L, giving a reason, and find the coordinates of T.
Find the set of values of m for which the line y = mx meets C at two distinct points.
By completing the square, find the coordinates of the minimum point of C.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| x2 − 2x + 4 = mx | M1 | |
| x2 − 2x − mx + 4 = 0, so x2 − (m + 2)x + 4 = 0 | A1 | |
| Part (b) | ||
| (m + 2)2 − 4 × 1 × 4 = 0 | M1 | |
| m + 2 = ±4 | M1 | |
| m = 2 or m = −6 | A1 | |
| Part (c) | ||
| L has a positive gradient (T is in the first quadrant), so m = 2 | B1 | |
| x2 − 4x + 4 = 0, (x − 2)2 = 0, x = 2 | M1 | |
| T(2, 4) | A1 | |
| Part (d) | ||
| (m + 2)2 > 16, critical values 2 and −6 | M1 | |
| m < −6 or m > 2 | A1 | |
| Part (e) | ||
| (x − 1)2 + 3 | M1 | |
| (1, 3) | A1 | |
Common error: Choosing the inside region for a two-distinct-points condition. (m + 2)2 > 16 holds outside the critical values, so the answer is m < −6 or m > 2, not −6 < m < 2.
Figure 1 shows a square of side 6 cm. A right-angled isosceles triangle, with equal sides of length x cm, is cut from each corner of the square, leaving a regular octagon. Give all answers in exact, simplified form.
Show that x = 6 ÷ (2 + √2).
Hence show that x = 6 − 3√2.
Find the length of one side of the octagon in the form (a√2 + b) cm, where a and b are integers.
Show that the area of the octagon is 72(√2 − 1) cm2.
Find the perimeter of the octagon.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| each cut edge (hypotenuse) = √(x2 + x2) = x√2 | M1 | |
| regular octagon: x√2 = 6 − 2x, so x(2 + √2) = 6 and x = 6 ÷ (2 + √2) | A1 | |
| Part (b) | ||
| multiply numerator and denominator by 2 − √2 | M1 | |
| denominator 4 − 2 = 2 | B1 | |
| 6(2 − √2) ÷ 2 = 6 − 3√2, no errors seen | A1 | |
| Part (c) | ||
| x√2 = (6 − 3√2)√2 | M1 | |
| = (6√2 − 6) cm | A1 | |
| Part (d) | ||
| area = 36 − 4 × ½x2 = 36 − 2x2 | M1 | |
| x2 = 36 − 36√2 + 18 = 54 − 36√2 | A1 | |
| 36 − 108 + 72√2 = 72√2 − 72 = 72(√2 − 1), no errors seen | A1 | |
| Part (e) | ||
| 8(6√2 − 6) = (48√2 − 48) cm ft | B1 | |
Exam tip: Write down what 'regular' gives you as an equation before doing any algebra: here, a cut edge equals the part of the square's side that is left, x√2 = 6 − 2x.
Figure 1 shows the curve with equation y = f(x), where
f(x) = x − 5√x + 4, x ≥ 0.
The curve meets the y-axis at P and the x-axis at A and B, and has a minimum point at M.
Using the substitution u = √x, find the x-coordinates of A and B.
By completing the square in u, find the exact coordinates of M.
Using Figure 1 and your answer to part (a), write down the set of values of x for which f(x) < 0.
The equation x − 5√x + k = 0, where k is a positive constant, has exactly one real solution. Find the value of k and the solution.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| x = u2, giving u2 − 5u + 4 = 0 | B1 | |
| (u − 1)(u − 4) = 0 | M1 | |
| u = 1 or u = 4 | A1 | |
| x = u2: x = 1 and x = 16 | A1 | |
| Part (b) | ||
| (u − 5⁄2)2 − 25⁄4 + 4 = (u − 5⁄2)2 − 9⁄4 | M1 | |
| least value −9⁄4 when u = 5⁄2 | A1 | |
| x = (5⁄2)2: M(25⁄4, −9⁄4) | A1 | |
| Part (c) | ||
| 1 < x < 16 ft | B1 | |
| Part (d) | ||
| u2 − 5u + k = 0 must have equal roots: 25 − 4k = 0 | M1 | |
| k = 25⁄4 | A1 | |
| u = 5⁄2, so x = 25⁄4 | A1 | |
Exam tip: With u = √x, only values of u ≥ 0 give solutions for x. In part (d), for 0 < k < 25⁄4 both roots in u are positive, so there would be two solutions for x: equal roots are the only way to get exactly one.
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