Pure: Algebra and Functions Depth - Worksheets, Questions and Revision

12 original exam-style questions - 3 pages of questions with a full mark scheme - free printable PDF.

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A-Level · Pure Mathematics

P11 Pure: Algebra and Functions Depth

EDEXCEL 9MA0 · Calculator allowed · about 160 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

f(x) = 2x³ - 3x² - 11x + 6.

(a)Show that (x - 3) is a factor of f(x).(2)

(b)Hence factorise f(x) completely.(3)

(c)Solve f(x) = 0.(2)

(Total for Question 1 is 7 marks)

h(x) = x³ - 2x² + kx - 5, where k is a constant. When h(x) is divided by (x - 2), the remainder is 3.

(a)Find the value of k.(3)

(b)Using this value of k, find the remainder when h(x) is divided by (x + 1).(3)

(Total for Question 2 is 6 marks)

g(x) = x³ + ax² + bx - 12, where a and b are constants. (x - 1) and (x + 3) are both factors of g(x).

(a)Show that a + b = 11, form a second equation connecting a and b, and hence find the values of a and b.(5)

(b)Fully factorise g(x).(3)

(c)Hence write down all the roots of g(x) = 0.(2)

(Total for Question 3 is 10 marks)

(5x + 7) / ((x + 2)(x - 1)), x is real, x not equal to -2 or 1.

(a)Express (5x + 7) / ((x + 2)(x - 1)) in partial fractions.(4)

(b)State the two values of x for which the partial fraction expression is not valid.(2)

(c)Find the exact value of (5x + 7) / ((x + 2)(x - 1)) when x = 3, using both the original expression and the partial fraction form, showing they agree.(2)

(Total for Question 4 is 8 marks)

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

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

(b)Explain why the partial fraction expression requires no additional polynomial term.(2)

(c)Find the exact value of (3x² - x + 4) / ((x + 1)(x - 1)²) when x = 2, using the partial fraction form, and verify using the original expression.(3)

(Total for Question 5 is 11 marks)

f(x) = (x³ + x² + x + 3) / (x² + x - 2), x not equal to -2 or 1.

(a)Show that f(x) can be written in the form x + A/(x + 2) + B/(x - 1), stating the values of A and B.(7)

(b)Hence write down the equation of the oblique asymptote of the curve y = f(x).(2)

(c)Find f(3) as an exact fraction, and verify it agrees with the value obtained by direct substitution into the original expression for f(x).(3)

(Total for Question 6 is 12 marks)

f(x) = (5 + x) / ((1 + 2x)(1 - x)), |x| < 1/2.

(a)Express f(x) in the form A/(1 + 2x) + B/(1 - x), stating the values of A and B.(5)

(b)Hence expand f(x) in ascending powers of x, up to and including the term in x², giving each coefficient as an exact number. State the range of values of x for which the expansion is valid.(4)

(Total for Question 7 is 9 marks)

Functions f and g are defined by f(x) = 2x/(x - 3), x is real, x not equal to 3, and g(x) = x + 1, x is real.

(a)Find fg(x), simplifying your answer fully.(3)

(b)Find f^-1(x), stating its domain.(4)

(c)Solve the equation fg(x) = f^-1(x), showing that there is no solution.(3)

(Total for Question 8 is 10 marks)

h(x) = |2x - 5| - 3.

(a)Solve h(x) = 4.(4)

(b)Solve h(x) ≤ 2, giving your answer in set notation.(4)

(c)State the coordinates of the vertex of y = h(x), and the coordinates of the points where the graph crosses the coordinate axes.(3)

(Total for Question 9 is 11 marks)

Solve the equation |2x + 1| = 3x - 5.

(a)By considering the equation (2x + 1) = (3x - 5) and the equation (2x + 1) = -(3x - 5), find two possible values of x.(4)

(b)By checking each value in the original equation |2x + 1| = 3x - 5, determine which value(s) are valid solutions, justifying any rejection.(4)

(Total for Question 10 is 8 marks)

f(x) = x² - 6x + 5, x ≥ 3.

(a)Express f(x) in the form (x - a)² - b, stating the values of a and b.(3)

(b)State the range of f, given the domain x ≥ 3.(2)

(c)Find f^-1(x) and state its domain.(4)

(Total for Question 11 is 9 marks)

(x + 1)/(x - 1) - (x - 1)/(x + 1), x not equal to 1 or -1.

(a)Show that (x + 1)/(x - 1) - (x - 1)/(x + 1) can be written as 4x/(x² - 1).(4)

(b)Hence solve (x + 1)/(x - 1) - (x - 1)/(x + 1) = 3, giving your answer(s) as exact simplified surds.(4)

(Total for Question 12 is 8 marks)

Mark scheme · P11 Pure: Algebra and Functions Depth

Question 1

(a) M1 substitutes x = 3 into f(x), or attempts algebraic division by (x - 3)
(a) A1 f(3) = 2(27) - 3(9) - 11(3) + 6 = 0, so (x - 3) is a factor, cso
(a) Answer: f(3) = 0, so (x - 3) is a factor of f(x) (shown)
(b) M1 divides f(x) by (x - 3) to obtain a quadratic quotient
(b) A1 quotient 2x² + 3x - 2 oe
(b) A1 f(x) = (x - 3)(2x - 1)(x + 2) oe, fully factorised
(b) Answer: f(x) = (x - 3)(2x - 1)(x + 2)
(c) B1 x = 3 and x = -2 stated
(c) B1 x = 1/2 stated, all three roots oe
(c) Answer: x = 3, x = 1/2, x = -2

Question 2

(a) M1 applies the remainder theorem: h(2) = 3
(a) M1 forms 8 - 8 + 2k - 5 = 3, i.e. 2k - 5 = 3
(a) A1 k = 4 cao
(a) Answer: k = 4
(b) M1 applies the remainder theorem with x = -1: remainder = h(-1)
(b) M1 h(-1) = -1 - 2 - 4 - 5 oe
(b) A1 remainder = -12 cao
(b) Answer: Remainder = -12

Question 3

(a) M1 uses g(1) = 0 to form a + b = 11 oe (1 + a + b - 12 = 0)
(a) M1 uses g(-3) = 0 to form 3a - b = 13 oe (-27 + 9a - 3b - 12 = 0)
(a) M1 solves the simultaneous equations
(a) A1 a = 6
(a) A1 b = 5
(a) Answer: a = 6, b = 5
(b) M1 divides g(x) = x³ + 6x² + 5x - 12 by (x - 1)(x + 3) = x² + 2x - 3
(b) A1 third factor (x + 4) obtained
(b) A1 g(x) = (x - 1)(x + 3)(x + 4) oe
(b) Answer: g(x) = (x - 1)(x + 3)(x + 4)
(c) B1 x = 1 and x = -3 stated
(c) B1 x = -4 stated, all three roots oe
(c) Answer: x = 1, x = -3, x = -4

Question 4

(a) M1 writes (5x+7)/((x+2)(x-1)) = A/(x+2) + B/(x-1) and forms the identity 5x + 7 = A(x - 1) + B(x + 2)
(a) M1 substitutes x = 1 (or x = -2) to find one constant
(a) A1 A = 1
(a) A1 B = 4 oe
(a) Answer: 1/(x + 2) + 4/(x - 1)
(b) B1 x = -2 stated
(b) B1 x = 1 stated
(b) Answer: x = -2 and x = 1
(c) M1 substitutes x = 3 into both the original expression and the partial fraction form
(c) A1 both give 11/5 (= 2.2) cao
(c) Answer: 11/5 (= 2.2)

Question 5

(a) M1 forms the identity 3x² - x + 4 = A(x-1)² + B(x+1)(x-1) + C(x+1)
(a) M1 substitutes x = 1 to find C
(a) A1 C = 3
(a) M1 substitutes x = -1 to find A
(a) A1 A = 2
(a) A1 B = 1 oe, found by comparing coefficients of x² (or a further substitution)
(a) Answer: A = 2, B = 1, C = 3
(b) B1 identifies that the numerator has degree 2 and the denominator has degree 3
(b) B1 states that, since 2 < 3, the fraction is already proper, so no polynomial division / extra term is needed oe
(b) Answer: The numerator 3x² - x + 4 has degree 2 and the denominator (x+1)(x-1)² has degree 3; since 2 < 3, the fraction is proper and no extra polynomial term is required
(c) M1 substitutes x = 2 into the partial fraction form 2/(x+1) + 1/(x-1) + 3/(x-1)²
(c) A1 value = 14/3 cao
(c) B1 verifies using the original expression: (3(4)-2+4)/((3)(1)) = 14/3
(c) Answer: 14/3

Question 6

(a) M1 attempts algebraic division of x³+x²+x+3 by x²+x-2
(a) A1 quotient x, remainder 3x + 3 obtained oe
(a) M1 writes (3x+3)/((x+2)(x-1)) = A/(x+2) + B/(x-1) and forms the identity 3x+3 = A(x-1) + B(x+2)
(a) M1 substitutes x = 1 (or x = -2) to find one constant
(a) A1 B = 2
(a) A1 A = 1, so f(x) = x + 1/(x+2) + 2/(x-1), cso
(a) Answer: f(x) = x + 1/(x + 2) + 2/(x - 1), so A = 1, B = 2
(b) M1 recognises that 1/(x+2) tends to 0 and 2/(x-1) tends to 0 as x tends to ± infinity
(b) A1 y = x oe
(b) Answer: y = x
(c) M1 substitutes x = 3 into f(x) = x + 1/(x+2) + 2/(x-1)
(c) A1 f(3) = 21/5 cao
(c) B1 verifies using the original expression: (27+9+3+3)/(9+3-2) = 42/10 = 21/5
(c) Answer: 21/5

Question 7

(a) M1 forms the identity 5 + x = A(1 - x) + B(1 + 2x)
(a) M1 substitutes x = 1 (or x = -1/2) to find one constant
(a) A1 B = 2
(a) A1 A = 3
(a) A1 f(x) = 3/(1+2x) + 2/(1-x) oe cao
(a) Answer: 3/(1 + 2x) + 2/(1 - x)
(b) M1 expands 3(1+2x)^-1 = 3(1 - 2x + 4x² - ...) using the binomial series
(b) M1 expands 2(1-x)^-1 = 2(1 + x + x² + ...) using the binomial series
(b) A1 f(x) = 5 - 4x + 14x² + ... oe cao
(b) B1 valid for |x| < 1/2 oe, identified as the more restrictive of the two ranges
(b) Answer: f(x) = 5 - 4x + 14x² + ..., valid for |x| < 1/2

Question 8

(a) M1 attempts f(g(x)) = f(x+1) = 2(x+1)/((x+1)-3)
(a) A1 simplifies the denominator to (x - 2)
(a) A1 fg(x) = 2(x+1)/(x-2), x not equal to 2, oe cao
(a) Answer: fg(x) = 2(x + 1)/(x - 2), x not equal to 2
(b) M1 sets y = 2x/(x-3) and rearranges: y(x-3) = 2x
(b) M1 collects x terms: x(y-2) = 3y oe
(b) A1 f^-1(x) = 3x/(x-2) oe cao
(b) B1ft domain: x is real, x not equal to 2, ft from the denominator of f^-1(x)
(b) Answer: f^-1(x) = 3x/(x - 2), domain: x is real, x not equal to 2
(c) M1 sets 2(x+1)/(x-2) = 3x/(x-2)
(c) M1 multiplies both sides by (x-2), x not equal to 2, to obtain 2(x+1) = 3x, giving x = 2
(c) A1 rejects x = 2 as it is excluded from the domain of both functions, so there is no solution, cso
(c) Answer: No solution (the only candidate, x = 2, is excluded from the domain)

Question 9

(a) M1 forms |2x - 5| = 7
(a) M1 sets up two linear equations 2x-5=7 and 2x-5=-7
(a) A1 x = 6
(a) A1 x = -1 oe
(a) Answer: x = 6 or x = -1
(b) M1 forms |2x - 5| ≤ 5
(b) M1 forms the compound inequality -5 ≤ 2x-5 ≤ 5
(b) A1 0 ≤ x ≤ 5 oe
(b) A1 correct set notation {x : 0 ≤ x ≤ 5} oe
(b) Answer: {x : 0 ≤ x ≤ 5}
(c) B1 vertex (2.5, -3) oe
(c) B1 y-intercept (0, 2)
(c) B1 x-intercepts (1, 0) and (4, 0)
(c) Answer: Vertex (2.5, -3); y-intercept (0, 2); x-intercepts (1, 0) and (4, 0)

Question 10

(a) M1 solves 2x+1 = 3x-5
(a) A1 x = 6
(a) M1 solves 2x+1 = -(3x-5)
(a) A1 x = 0.8 oe (4/5)
(a) Answer: x = 6 or x = 0.8
(b) M1 substitutes x = 6: LHS = |13| = 13, RHS = 3(6)-5 = 13, equal so valid
(b) A1 x = 6 is a valid solution
(b) M1 substitutes x = 0.8: RHS = 3(0.8)-5 = -2.6, and notes RHS < 0 so cannot equal a modulus (which is always ≥ 0)
(b) A1 x = 0.8 is rejected, cso
(b) Answer: x = 6 is the only valid solution; x = 0.8 is rejected since it gives a negative value for 3x - 5, which cannot equal a modulus

Question 11

(a) M1 attempts to complete the square on x² - 6x + 5
(a) A1 a = 3
(a) A1 b = 4 oe, i.e. f(x) = (x-3)² - 4
(a) Answer: f(x) = (x - 3)² - 4, so a = 3, b = 4
(b) M1 recognises that for x ≥ 3, (x-3) ≥ 0 so (x-3)² ≥ 0, and f is increasing on this domain
(b) A1 f(x) ≥ -4 oe
(b) Answer: f(x) ≥ -4
(c) M1 sets y = (x-3)² - 4 and rearranges to (x-3)² = y+4
(c) M1 takes the square root and selects the positive root, consistent with domain x ≥ 3: x-3 = √y+4
(c) A1 f^-1(x) = 3 + √x+4 oe cao
(c) B1ft domain x ≥ -4, ft from the range of f found in part (b)
(c) Answer: f^-1(x) = 3 + √x + 4, domain: x ≥ -4