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A-Level · Pure Mathematics

P2 Pure: Algebra and Functions

EDEXCEL 9MA0 · Calculator allowed · about 150 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

This question tests indices and surds. Show all algebraic working; do not use a calculator to evaluate surd expressions.

(a)Simplify fully (8x⁹)^2/3.(2)

(b)Show that (3sqrt(5) - 2)(3sqrt(5) + 2) = 41.(2)

(c)Express 10/(√5 - 2) in the form a + b*√5, where a and b are integers.(3)

(Total for Question 1 is 7 marks)

f(x) = x² - 8x + 19.

(a)Express f(x) in the form (x - p)² + q, stating the values of p and q.(2)

(b)State the minimum value of f(x) and the value of x at which it occurs.(2)

(c)Show that the equation f(x) = 2 has no real roots.(3)

(Total for Question 2 is 7 marks)

Simplify the following algebraic fractions, giving each answer in its simplest form. State any values of x for which the fractions are undefined.

(a)Simplify fully (x² - 9)/(x² + x - 6).(3)

(b)Hence, or otherwise, simplify (x² - 9)/(x² + x - 6) / ((2x - 4)/(x + 3)), giving your answer as a single fraction in its simplest form.(4)

(Total for Question 3 is 7 marks)

Solve the following inequalities.

(a)Solve the inequality 2x² - 5x - 3 ≤ 0.(3)

(b)Solve the inequality |2x - 3| < 3.(3)

(c)Hence find the set of values of x for which both 2x² - 5x - 3 ≤ 0 and |2x - 3| < 3 are satisfied.(2)

(Total for Question 4 is 8 marks)

Functions f and g are defined by f(x) = 2/(x - 1), x is real, x not equal to 1, and g(x) = x² + 4, x is real.

(a)State the range of g.(1)

(b)Find the value of fg(3).(2)

(c)Find f^-1(x), the inverse function of f, stating its domain.(4)

(d)Show that gf(x) ≥ 4 for all x in the domain of f, without finding an explicit expression for gf(x).(2)

(Total for Question 5 is 9 marks)

The function f is defined by f(x) = (x - 3)² - 4, x is real.

(a)State the coordinates of the turning point of the graph of y = f(x), and hence state the range of f.(2)

(b)The function g is defined by g(x) = f(x - 2) + 5, x is real. Describe fully the two geometric transformations that map the graph of y = f(x) onto the graph of y = g(x), and hence state the coordinates of the turning point of y = g(x).(4)

(c)Given that h(x) = |f(x)|, find the set of values of x for which h(x) = f(x).(3)

(Total for Question 6 is 9 marks)

Prove each of the following results using an appropriate method of proof.

(a)Prove that x² - 6x + 11 > 0 for all real values of x.(3)

(b)Prove, by contradiction, that there is no smallest positive rational number.(3)

(Total for Question 7 is 6 marks)

A line l has equation y = 3x - 2. A curve C has equation y = x² + x - 4. The line l intersects the curve C at the points A and B.

(a)Show that the x-coordinates of A and B satisfy the equation x² - 2x - 2 = 0.(2)

(b)Hence find the coordinates of A and B, giving your answers in exact (surd) form.(5)

(c)The line l is translated to give the line l' with equation y = 3x + k. Find the range of values of k for which l' does not intersect C.(4)

(Total for Question 8 is 11 marks)

f(x) = 2x³ + x² - 13x + 6.

(a)Use the factor theorem to show that (x - 2) is a factor of f(x).(2)

(b)Hence factorise f(x) completely.(4)

(c)Solve f(x) = 0, and hence find the set of values of x for which f(x) > 0.(4)

(Total for Question 9 is 10 marks)

h(x) = |2x + 5|.

(a)State the coordinates of the vertex of the graph of y = h(x), and the coordinates of the point where the graph crosses the y-axis.(2)

(b)Solve the equation |2x + 5| = 3x - 1.(4)

(c)Hence, or otherwise, solve the inequality |2x + 5| ≤ 3x - 1.(3)

(Total for Question 10 is 9 marks)

(2x² - 3x + 4)/((x-2)²(x+1)) = A/(x-2) + B/(x-2)² + C/(x+1)

(a)Find the values of the constants A, B and C.(6)

(b)Express (9x + 1)/((x-3)(2x+1)) in the form A/(x-3) + B/(2x+1), finding integer values of A and B.(4)

(Total for Question 11 is 10 marks)

The function f is defined by f(x) = (3x + 2)/(x - 3), x is real, x not equal to 3.

(a)Find f^-1(x) algebraically, and show that f^-1(x) = f(x) for all x in the domain of f.(5)

(b)Hence write down the value of ff(4).(1)

(c)Solve the equation f(x) = x, giving your answers in exact (surd) form.(4)

(Total for Question 12 is 10 marks)

Mark scheme · P2 Pure: Algebra and Functions

Question 1

(a) M1 8^2/3 evaluated as 4, or x⁹ raised to power 2/3 giving x⁶
(a) A1 4x⁶ cao
(a) Answer: 4x⁶
(b) M1 uses difference of two squares to write (3sqrt(5))² - 2²
(b) A1 45 - 4 = 41 cso, answer printed
(b) Answer: 41 (shown)
(c) M1 multiplies numerator and denominator by (√5 + 2)
(c) M1 correct denominator 5 - 4 = 1 and expands numerator 10(√5 + 2)
(c) A1 20 + 10sqrt(5) oe, i.e. a = 20, b = 10 cao
(c) Answer: 20 + 10sqrt(5) (a = 20, b = 10)

Question 2

(a) M1 attempt to complete the square, (x - 4)² + k
(a) A1 (x - 4)² + 3, i.e. p = 4, q = 3 cao
(a) Answer: (x - 4)² + 3
(b) B1ft minimum value = 3 (ft from part a)
(b) B1ft occurs at x = 4 (ft from part a)
(b) Answer: minimum value 3 at x = 4
(c) M1 rearranges to x² - 8x + 17 = 0
(c) M1 attempts discriminant b² - 4ac with correct values
(c) A1 discriminant = -4 < 0 so no real roots, cso
(c) Answer: discriminant = -4, no real roots (shown)

Question 3

(a) M1 factorises numerator as (x-3)(x+3)
(a) M1 factorises denominator as (x+3)(x-2)
(a) A1 (x-3)/(x-2) cao, x not equal to -3, 2
(a) Answer: (x - 3)/(x - 2), x not equal to -3 or 2
(b) M1 rewrites division as multiplication by the reciprocal (x+3)/(2x-4)
(b) M1 uses result of part (a) and factorises 2x - 4 as 2(x-2)
(b) A1 correct combined fraction (x-3)(x+3)/[2(x-2)²]
(b) A1 (x² - 9)/(2(x-2)²) cao, fully simplified
(b) Answer: (x² - 9)/(2(x - 2)²)

Question 4

(a) M1 factorises 2x²-5x-3 as (2x+1)(x-3)
(a) A1 critical values x = -1/2 and x = 3
(a) A1 -1/2 ≤ x ≤ 3 cao, correct inequality direction
(a) Answer: -0.5 ≤ x ≤ 3
(b) M1 sets up compound inequality -3 < 2x - 3 < 3 (or two linear cases)
(b) A1 critical values x = 0 and x = 3
(b) A1 0 < x < 3 cao
(b) Answer: 0 < x < 3
(c) B1 identifies the need to intersect the two sets from parts (a) and (b)
(c) A1 0 < x < 3 cao (the tighter set)
(c) Answer: 0 < x < 3

Question 5

(a) B1 g(x) ≥ 4 oe
(a) Answer: g(x) ≥ 4
(b) M1 finds g(3) = 13
(b) A1 fg(3) = 1/6 cao
(b) Answer: 1/6
(c) M1 sets y = 2/(x-1) and attempts to make x the subject, e.g. y(x-1) = 2
(c) dM1 correctly isolates x, x = 1 + 2/y
(c) A1 f^-1(x) = 1 + 2/x oe cao
(c) B1ft domain x not equal to 0 (ft from range of f)
(c) Answer: f^-1(x) = 1 + 2/x, x not equal to 0
(d) B1 recognises that g(t) = t² + 4 ≥ 4 for any real number t, since t² ≥ 0
(d) B1 correct conclusion: since f(x) is real for all x not equal to 1, gf(x) = g(f(x)) ≥ 4 for all such x, cso
(d) Answer: gf(x) ≥ 4 for all x not equal to 1 (shown)

Question 6

(a) B1 turning point (3, -4)
(a) B1ft range f(x) ≥ -4 (ft from turning point)
(a) Answer: turning point (3, -4); range f(x) ≥ -4
(b) B1 translation by vector (2, 0), i.e. 2 units in the positive x-direction
(b) B1 translation by vector (0, 5), i.e. 5 units in the positive y-direction
(b) B1 both transformations correctly described together (order not required to be justified)
(b) A1ft turning point of g is (5, 1) (ft from part a translated correctly)
(b) Answer: translations (2,0) then (0,5); turning point of g is (5, 1)
(c) M1 recognises h(x) = f(x) requires f(x) ≥ 0
(c) M1 solves (x-3)² ≥ 4 correctly, e.g. x - 3 ≤ -2 or x - 3 ≥ 2
(c) A1 x ≤ 1 or x ≥ 5 cao
(c) Answer: x ≤ 1 or x ≥ 5

Question 7

(a) M1 completes the square: (x - 3)² + 2
(a) A1 correct completed square form (x-3)² + 2
(a) A1 valid conclusion: (x-3)² ≥ 0 for all real x, so expression ≥ 2 > 0, cso
(a) Answer: x² - 6x + 11 = (x-3)² + 2 ≥ 2 > 0 (proved)
(b) M1 states the assumption for contradiction: suppose q is the smallest positive rational number
(b) M1 constructs a valid smaller positive rational, e.g. q/2, and justifies it is rational and positive
(b) A1 correct contradiction stated: q/2 < q contradicts q being smallest, so no smallest positive rational exists, cso
(b) Answer: no smallest positive rational number exists (proved by contradiction)

Question 8

(a) M1 sets x² + x - 4 = 3x - 2
(a) A1 correctly rearranges to x² - 2x - 2 = 0, cso, answer printed
(a) Answer: x² - 2x - 2 = 0 (shown)
(b) M1 correct use of the quadratic formula (or completing the square) with a=1, b=-2, c=-2
(b) A1 x = 1 + √3 and x = 1 - √3 oe, simplified
(b) M1 substitutes an x-value into y = 3x - 2 (or into C) to find corresponding y
(b) A1 y = 1 + 3sqrt(3) and y = 1 - 3sqrt(3), both correct
(b) A1 both coordinates correctly stated as ordered pairs: (1+√3, 1+3sqrt(3)) and (1-√3, 1-3sqrt(3))
(b) Answer: A = (1 + √3, 1 + 3sqrt(3)), B = (1 - √3, 1 - 3sqrt(3))
(c) M1 sets x² + x - 4 = 3x + k and rearranges to a 3-term quadratic equal to zero
(c) M1 writes correct discriminant expression, e.g. 4 - 4(1)(-4-k)
(c) M1 sets discriminant < 0 and attempts to solve the resulting inequality in k
(c) A1 k < -5 cao
(c) Answer: k < -5

Question 9

(a) M1 substitutes x = 2 into f(x)
(a) A1 f(2) = 0, so (x-2) is a factor, cso
(a) Answer: f(2) = 0, so (x-2) is a factor (shown)
(b) M1 divides f(x) by (x-2), or compares coefficients, to obtain quadratic factor 2x² + 5x - 3
(b) A1 correct quadratic factor 2x² + 5x - 3
(b) M1 factorises the quadratic factor
(b) A1 f(x) = (x-2)(2x-1)(x+3) cao, fully factorised
(b) Answer: f(x) = (x - 2)(2x - 1)(x + 3)
(c) B1 roots x = -3, x = 1/2, x = 2 identified
(c) M1 considers the sign of f(x) in each of the four regions determined by the roots
(c) A1 -3 < x < 1/2 identified as part of the solution set
(c) A1 x > 2 identified, giving full solution -3 < x < 1/2 or x > 2, oe
(c) Answer: -3 < x < 1/2 or x > 2

Question 10

(a) B1 vertex (-2.5, 0) oe
(a) B1 y-intercept (0, 5)
(a) Answer: vertex (-2.5, 0); y-intercept (0, 5)
(b) M1 squares both sides (or considers two linear cases)
(b) M1 forms correct quadratic 5x² - 26x - 24 = 0 (or equivalent pair of linear equations)
(b) A1 x = 6 and x = -0.8 found
(b) A1 rejects x = -0.8 since 3x-1 < 0 there (invalid as LHS is non-negative); x = 6 only, cao
(b) Answer: x = 6 (only valid solution)
(c) M1 considers the two cases x ≥ -2.5 and x < -2.5 separately
(c) A1 x ≥ 6 obtained from the case x ≥ -2.5
(c) B1 correctly justifies no solutions arise from the case x < -2.5 (consistent with rejected root in part b)
(c) Answer: x ≥ 6

Question 11

(a) M1 sets up correct identity 2x²-3x+4 = A(x-2)(x+1) + B(x+1) + C(x-2)²
(a) M1 substitutes x = 2 into the identity
(a) A1 B = 2
(a) M1 substitutes x = -1 into the identity
(a) A1 C = 1
(a) A1 A = 1, found by comparing coefficients or a further substitution, cao
(a) Answer: A = 1, B = 2, C = 1
(b) M1 sets up correct identity 9x + 1 = A(2x+1) + B(x-3)
(b) M1 substitutes a suitable value, e.g. x = 3 or x = -1/2
(b) A1 A = 4
(b) A1 B = 1
(b) Answer: A = 4, B = 1

Question 12

(a) M1 sets y = (3x+2)/(x-3) and cross-multiplies: y(x-3) = 3x+2
(a) M1 collects x terms on one side: x(y-3) = 3y+2
(a) A1 x = (3y+2)/(y-3)
(a) A1 f^-1(x) = (3x+2)/(x-3)
(a) A1 states this is identical to f(x), so f is self-inverse, cso
(a) Answer: f^-1(x) = (3x+2)/(x-3) = f(x), so f is self-inverse (shown)
(b) B1ft ff(4) = 4, using the self-inverse property (ft from part a)
(b) Answer: 4
(c) M1 sets up equation (3x+2)/(x-3) = x and cross-multiplies
(c) M1 rearranges to a 3-term quadratic equal to zero: x² - 6x - 2 = 0
(c) M1 correct use of the quadratic formula (or completing the square)
(c) A1 x = 3 + √11 or x = 3 - √11, both values, simplified, oe
(c) Answer: x = 3 + √11 or x = 3 - √11

Pure: Algebra and Functions - Worksheets, Questions and Revision