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Admissions tests / ESAT / Maths 2 / Algebra, functions and sequences

Demanding. 15 questions, 15 marks, about 28 minutes.

ESAT Mathematics 2: Algebra, functions and sequences, set 3

Laws of indices, surds, quadratics and the discriminant, completing the square, simultaneous equations, inequalities, polynomials, the factor theorem, sequences, series and the binomial expansion.

Download the questions (PDF) Download with worked solutions (PDF)

  • Answer all questions. No calculator.
  • Each question has exactly one correct answer.
  • This is a demanding set: several questions need multi-step working, and there is less signposting than on the real exam.
  1. 11 mark

    Given that y^(2/3) = 4, where y > 0, find the value of y^(1/3).

    1. A 4
    2. B 8
    3. C 2
    4. D -2
  2. 21 mark

    Solve the equation x - sqrt(x) - 6 = 0, for x > 0.

    1. A x = 9
    2. B x = 9 or x = 4
    3. C x = 4
    4. D x = 36
  3. 31 mark

    The equation x^2 + (k - 2)x + (k + 1) = 0 has equal roots. Find the possible value(s) of k.

    1. A k = 8 only
    2. B k = 0 or k = 6
    3. C No real value of k
    4. D k = 0 or k = 8
  4. 41 mark

    The line y = x + k intersects the curve y = x^2 - 4x + 1 at two points whose x-coordinates differ by 5. Find the value of k.

    1. A k = -4
    2. B k = 1
    3. C k = 5
    4. D k = 4
  5. 51 mark

    Solve the inequality 6 + x - x^2 > 0.

    1. A x < -2 or x > 3
    2. B -3 < x < 2
    3. C -2 <= x <= 3
    4. D -2 < x < 3
  6. 61 mark

    f(x) = x^3 + ax^2 + bx - 6. Given that (x - 1) is a factor of f(x), and that dividing f(x) by (x + 2) leaves a remainder of -6, find the value of a.

    1. A -9
    2. B -1
    3. C 3
    4. D -3
  7. 71 mark

    When f(x) = x^3 - 4x^2 + 2x + 10 is divided by (x^2 - x - 2), the remainder is of the form ax + b. Find the value of a + b.

    1. A 5
    2. B 3
    3. C 9
    4. D -2
  8. 81 mark

    A sequence satisfies x_(n+1) = 2x_n + c for n >= 1, where c is a constant. Given that x_1 = 3 and x_2 = 7, find x_4.

    1. A 40
    2. B 31
    3. C 15
    4. D 30
  9. 91 mark

    A geometric series has first term 6 and common ratio 1/3. A new series is formed by squaring each term of the original series. Find the sum to infinity of this new series.

    1. A 81
    2. B 54
    3. C 81/2
    4. D 32
  10. 101 mark

    Given that nC2 = 45, find the value of n, where n is a positive integer.

    1. A 9
    2. B 11
    3. C -9
    4. D 10
  11. 111 mark

    Find the coefficient of x^2 in the expansion of (x + 2)(x - 1)(x + 5).

    1. A -10
    2. B 6
    3. C 8
    4. D 3
  12. 121 mark

    An arithmetic series has third term 11 and eighth term 31. Find the sum of the first 20 terms.

    1. A 820
    2. B 900
    3. C 860
    4. D 790
  13. 131 mark

    Find the range of values of c for which the equation x^2 - 6x + c = 0 has two distinct positive roots.

    1. A c < 9
    2. B c < 0
    3. C 0 < c < 9
    4. D c > 9
  14. 141 mark

    Factorise x^3 - 7x + 6 completely.

    1. A (x - 1)(x - 3)(x + 2)
    2. B (x - 1)(x^2 + x + 6)
    3. C (x - 1)^2(x + 6)
    4. D (x - 1)(x - 2)(x + 3)
  15. 151 mark

    Solve |2x - 5| = |x + 4|.

    1. A x = 9 only
    2. B x = 9 or x = 1/3
    3. C x = 9 or x = 3
    4. D x = 1 or x = 3

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

  1. Question 1Answer: C

    1. Since (y^(1/3))^2 = y^(2/3), squaring y^(1/3) gives the same result as y^(2/3), which is 4.
    2. So y^(1/3) is a number whose square is 4, meaning y^(1/3) = 2 or y^(1/3) = -2.
    3. Because y > 0, its cube root y^(1/3) must also be positive, so the negative value is rejected.
    4. Therefore y^(1/3) = 2, and the answer is C.
    • Why not A: Assumes y^(1/3) must equal the given value of y^(2/3) itself, without carrying out any calculation to relate the two different powers of y.
    • Why not B: Correctly finds y = 4^(3/2) = 8 by raising both sides of y^(2/3) = 4 to the power 3/2, but then stops there and reports y itself instead of going on to find y^(1/3).
    • Why not D: Correctly recognises that (y^(1/3))^2 = y^(2/3) = 4, so y^(1/3) = +-2, but reports the negative root instead of rejecting it, even though y > 0 means y^(1/3) must also be positive.

  2. Question 2Answer: A

    1. Let u = sqrt(x), so u >= 0 and x = u^2. The equation becomes u^2 - u - 6 = 0.
    2. Factorise: the two numbers that multiply to -6 and add to -1 are -3 and 2, giving (u - 3)(u + 2) = 0, so u = 3 or u = -2.
    3. Since u = sqrt(x) cannot be negative, reject u = -2, leaving u = 3.
    4. Then x = u^2 = 9, and checking: 9 - sqrt(9) - 6 = 9 - 3 - 6 = 0, so the answer is A.
    • Why not B: Correctly reaches u^2 - u - 6 = 0 and factorises to (u - 3)(u + 2) = 0, giving u = 3 or u = -2, but fails to reject u = -2 (impossible since u = sqrt(x) cannot be negative), squaring both values to give x = 9 or x = 4.
    • Why not C: Sets up the substituted equation with a sign error, using u^2 + u - 6 = 0 instead of u^2 - u - 6 = 0 (misreading the minus sign on sqrt(x)), which factorises as (u + 3)(u - 2) = 0; rejecting the negative root u = -3 leaves u = 2, so x = 4.
    • Why not D: Misreads the equation x - sqrt(x) - 6 = 0 as if the x terms cancel to leave sqrt(x) = 6 directly, without any valid algebraic justification, then squares both sides to get x = 36.

  3. Question 3Answer: D

    1. For equal roots, the discriminant b^2 - 4ac must equal 0, with a = 1, b = (k - 2) and c = (k + 1).
    2. So (k - 2)^2 - 4(k + 1) = 0. Expand: k^2 - 4k + 4 - 4k - 4 = 0, which simplifies to k^2 - 8k = 0.
    3. Factorise: k(k - 8) = 0, so k = 0 or k = 8, and both are valid roots of this equation in their own right.
    4. Checking k = 0 gives x^2 - 2x + 1 = (x - 1)^2 = 0, and k = 8 gives x^2 + 6x + 9 = (x + 3)^2 = 0, both repeated roots, so the answer is D.
    • Why not A: Correctly reduces the discriminant condition to k(k - 8) = 0, but then divides both sides by k to leave k - 8 = 0, treating k = 0 as a factor to cancel rather than as a valid root in its own right, so reports only k = 8.
    • Why not B: Expands (k - 2)^2 as k^2 - 2k + 4 instead of k^2 - 4k + 4, using (a - b)^2 = a^2 - ab + b^2 and forgetting the factor of 2 on the middle term, and so reaches k(k - 6) = 0 instead of k(k - 8) = 0.
    • Why not C: Sets up the discriminant condition with the wrong sign, adding 4(k + 1) to (k - 2)^2 instead of subtracting it, which gives k^2 + 8 = 0, and wrongly concludes that no real value of k makes the roots equal.

  4. Question 4Answer: B

    1. Substitute the line into the curve: x + k = x^2 - 4x + 1, which rearranges to x^2 - 5x + (1 - k) = 0.
    2. If the two roots are p and q, then p + q = 5 and pq = 1 - k, so (p - q)^2 = (p + q)^2 - 4pq = 25 - 4(1 - k) = 21 + 4k.
    3. Since the x-coordinates differ by 5, (p - q)^2 = 5^2 = 25, so 21 + 4k = 25, giving 4k = 4.
    4. So k = 1, and checking: the equation becomes x^2 - 5x = 0, giving x = 0 or x = 5, which do differ by 5, so the answer is B.
    • Why not A: Correctly forms x^2 - 5x + (1 - k) = 0 and the identity (p - q)^2 = 21 + 4k, but sets this equal to the difference itself (5) rather than the square of the difference (25), giving 21 + 4k = 5 and so k = -4.
    • Why not C: Makes a sign error moving the x from the line's equation across to combine with the -4x term, obtaining x^2 - 3x + (1 - k) = 0 instead of x^2 - 5x + (1 - k) = 0, then correctly sets (p - q)^2 = 9 - 4(1 - k) = 25, giving 5 + 4k = 25 and k = 5.
    • Why not D: Expands 4(1 - k) as 4 - k instead of 4 - 4k when forming the discriminant, forgetting to multiply the k term by 4 as well as the 1, so solves 21 + k = 25 instead of 21 + 4k = 25, giving k = 4.

  5. Question 5Answer: D

    1. Rearrange by multiplying both sides by -1, remembering to reverse the inequality: -x^2 + x + 6 > 0 becomes x^2 - x - 6 < 0.
    2. Factorise: the two numbers that multiply to -6 and add to -1 are -3 and 2, giving (x - 3)(x + 2) < 0.
    3. This upward-opening parabola is negative between its roots, x = -2 and x = 3.
    4. So the solution is -2 < x < 3, and the answer is D.
    • Why not A: Multiplies through by -1 to reach x^2 - x - 6 < 0 but forgets to reverse the direction of the inequality when doing so, treating it as x^2 - x - 6 > 0 instead, which gives the region outside the roots, x < -2 or x > 3.
    • Why not B: Attempts to factorise 6 + x - x^2 directly and mistypes the sign of the middle term, effectively factorising 6 - x - x^2 instead, which gives -(x + 3)(x - 2) and leads to the interval -3 < x < 2.
    • Why not C: Correctly finds the critical values x = -2 and x = 3, but loses track of the strict inequality when multiplying through by -1, writing the solution with <= instead of <, as -2 <= x <= 3.

  6. Question 6Answer: C

    1. By the Factor Theorem, f(1) = 0: 1 + a + b - 6 = 0, so a + b = 5.
    2. By the Remainder Theorem, dividing by (x + 2) means f(-2) equals the remainder: -8 + 4a - 2b - 6 = -6, which simplifies to 4a - 2b = 8, so 2a - b = 4.
    3. Adding the two equations a + b = 5 and 2a - b = 4 eliminates b: 3a = 9, so a = 3.
    4. This gives b = 2, and checking f(-2) = -8 + 12 - 4 - 6 = -6 confirms it, so the answer is C.
    • Why not A: Misapplies the Remainder Theorem for the divisor (x + 2), evaluating f(2) instead of f(-2): f(2) = 8 + 4a + 2b - 6 = -6 gives 2a + b = -4, and combining with the correct factor-theorem equation a + b = 5 gives a = -9.
    • Why not B: Drops the sign on the constant term when applying the Factor Theorem, treating f(x) as if it ended in +6 rather than -6: f(1) = 1 + a + b + 6 = 0 gives a + b = -7, and combining with the correct remainder equation 2a - b = 4 gives a = -1.
    • Why not D: Misapplies the Factor Theorem by evaluating f(-1) instead of f(1), treating (x - 1) as though it were (x + 1): f(-1) = -1 + a - b - 6 = 0 gives a - b = 7, and combining with the correct remainder equation 2a - b = 4 gives a = -3.

  7. Question 7Answer: A

    1. Divide x^3/x^2 to start: the first quotient term is x. Multiply: x(x^2 - x - 2) = x^3 - x^2 - 2x.
    2. Subtract: (x^3 - 4x^2 + 2x + 10) - (x^3 - x^2 - 2x) = -3x^2 + 4x + 10.
    3. Divide again: -3x^2/x^2 = -3. Multiply: -3(x^2 - x - 2) = -3x^2 + 3x + 6. Subtract: (-3x^2 + 4x + 10) - (-3x^2 + 3x + 6) = x + 4.
    4. This is the remainder, with degree less than the divisor, so a = 1 and b = 4, giving a + b = 5. So the answer is A.
    • Why not B: Makes a sign slip in the final subtraction of the long division, computing 3x - 4x instead of 4x - 3x for the x-term of the last step, obtaining a remainder of -x + 4 instead of x + 4, giving a + b = -1 + 4 = 3.
    • Why not C: Misfactorises x^2 - x - 2 as (x + 2)(x - 1) instead of the correct (x - 2)(x + 1), and uses the wrong roots x = 1 and x = -2 when applying the remainder theorem: evaluating f(1) = 1 - 4 + 2 + 10 = 9 and treating this directly as a + b.
    • Why not D: Correctly carries out the division to find the quotient x - 3, but then answers using the quotient's coefficients (1 and -3) instead of the remainder's, giving 1 + (-3) = -2.

  8. Question 8Answer: B

    1. Substitute n = 1 into the recurrence: x_2 = 2x_1 + c, so 7 = 2(3) + c = 6 + c, giving c = 1.
    2. Now find x_3: x_3 = 2x_2 + c = 2(7) + 1 = 15.
    3. Then find x_4: x_4 = 2x_3 + c = 2(15) + 1 = 31.
    4. So the answer is B.
    • Why not A: Finds c by computing x_2 - x_1 = 7 - 3 = 4 instead of x_2 - 2x_1 = 1, forgetting that the recurrence doubles the previous term before adding c, then uses this wrong c throughout: x_3 = 2(7) + 4 = 18, x_4 = 2(18) + 4 = 40.
    • Why not C: Correctly finds c = 1 and correctly computes x_3 = 2(7) + 1 = 15, but miscounts the number of steps and reports this value of x_3 as if it were x_4.
    • Why not D: Correctly finds c = 1 and correctly computes x_3 = 15, but forgets to add c on the final step, computing x_4 = 2(15) = 30 instead of 2(15) + 1.

  9. Question 9Answer: C

    1. The original series is 6, 2, 2/3, 2/9, ... with common ratio 1/3. Squaring each term gives 36, 4, 4/9, 4/81, ..., which is itself a geometric series.
    2. Its first term is 6^2 = 36, and its common ratio is (1/3)^2 = 1/9, since squaring a GP with ratio r produces a new GP with ratio r^2.
    3. Since |1/9| < 1, this new series converges, with sum to infinity a/(1 - r) = 36/(1 - 1/9) = 36/(8/9).
    4. 36/(8/9) = 36 x (9/8) = 324/8 = 81/2, so the answer is C.
    • Why not A: Squares the sum to infinity of the original series (6/(1 - 1/3) = 9, then 9^2 = 81) instead of forming the new series of squared terms and summing that series in its own right.
    • Why not B: Correctly identifies the new first term as 6^2 = 36, but keeps the original common ratio 1/3 for the new series instead of using r^2 = 1/9, since squaring every term of a GP with ratio r produces a new GP with ratio r^2, not r.
    • Why not D: Correctly finds the new first term 36 and new common ratio 1/9, but multiplies by (1 - r^2) instead of dividing by it, computing 36 x (8/9) = 32 instead of 36/(8/9).

  10. Question 10Answer: D

    1. The formula for nC2 is n(n - 1)/2, so n(n - 1)/2 = 45, which gives n(n - 1) = 90.
    2. Rearrange to n^2 - n - 90 = 0. Factorise: the two numbers that multiply to -90 and add to -1 are -10 and 9, giving (n - 10)(n + 9) = 0.
    3. So n = 10 or n = -9. Since n must be a positive integer, reject n = -9.
    4. So n = 10, and checking, 10C2 = (10 x 9)/2 = 45, confirming the answer is D.
    • Why not A: Uses n(n + 1)/2 instead of n(n - 1)/2 for nC2, confusing the combinations formula with the triangular-number formula for the sum of the first n integers: n(n + 1) = 90 gives n = 9.
    • Why not B: Uses (n - 1)(n - 2)/2 instead of n(n - 1)/2 for nC2, shifting the index down by one: (n - 1)(n - 2) = 90 gives n = 11.
    • Why not C: Correctly solves n^2 - n - 90 = 0 to get n = 10 or n = -9, but reports the negative root instead of rejecting it, even though n must be a positive count of objects being chosen from.

  11. Question 11Answer: B

    1. For a product of three linear factors (x + a)(x + b)(x + c), the coefficient of x^2 in the expansion is the sum a + b + c.
    2. Here a = 2, b = -1 and c = 5, so the coefficient of x^2 is 2 + (-1) + 5 = 6.
    3. Checking by full expansion: (x + 2)(x - 1) = x^2 + x - 2, and (x^2 + x - 2)(x + 5) = x^3 + 6x^2 + 3x - 10.
    4. The coefficient of x^2 in this expansion is indeed 6, so the answer is B.
    • Why not A: Multiplies the three constants together, 2 x (-1) x 5 = -10, confusing this with the coefficient of x^2; this product is in fact the constant term of the expansion, not the coefficient of x^2.
    • Why not C: Sums the three constants but makes a sign error, treating (x - 1) as though it were (x + 1): 2 + 1 + 5 = 8, instead of using the correct constant -1 from (x - 1).
    • Why not D: Computes the sum of the pairwise products of the three constants, 2x(-1) + (-1)x5 + 5x2 = -2 - 5 + 10 = 3, which actually gives the coefficient of x, instead of summing the three constants directly to get the coefficient of x^2.

  12. Question 12Answer: A

    1. Let the first term be a and common difference d. The third term is a + 2d = 11, and the eighth term is a + 7d = 31.
    2. Subtracting the first equation from the second: 5d = 20, so d = 4. Then a = 11 - 2(4) = 3.
    3. The sum of the first n terms is S_n = (n/2)(2a + (n - 1)d). For n = 20: S_20 = (20/2)(2(3) + 19(4)) = 10 x (6 + 76).
    4. 10 x 82 = 820, so the answer is A.
    • Why not B: Correctly finds the common difference d = 4, but miscalculates the first term using a = 11 - d = 7 instead of a = 11 - 2d = 3, forgetting that the third term is a + 2d, then uses this wrong a in the sum formula to get 10 x (2(7) + 19(4)) = 900.
    • Why not C: Correctly finds a = 3 and d = 4, but uses n instead of (n - 1) in the sum formula, computing 10 x (2(3) + 20(4)) = 10 x 86 = 860 instead of 10 x (2(3) + 19(4)).
    • Why not D: Correctly finds a = 3 and d = 4 and the last term l = a + 19d = 79, then uses the equivalent formula S_n = (n/2)(a + l) but forgets to add the first term a, computing (20/2) x 79 = 790 instead of (20/2) x (3 + 79).

  13. Question 13Answer: C

    1. For two distinct real roots, the discriminant must be positive: (-6)^2 - 4(1)(c) > 0, so 36 - 4c > 0, giving c < 9.
    2. For both roots to be positive, their sum and product must both be positive. The sum of the roots is -(-6)/1 = 6, which is already positive regardless of c.
    3. The product of the roots is c/1 = c, so this requires c > 0.
    4. Combining c < 9 with c > 0 gives 0 < c < 9, so the answer is C.
    • Why not A: Correctly requires the discriminant to be positive for two distinct real roots, giving 36 - 4c > 0 and so c < 9, but does not also check that the roots are positive, missing the requirement that their product c/1 = c must be positive too.
    • Why not B: Misremembers the product-of-roots formula, using -c/a instead of c/a for this equation (confusing it with the sum-of-roots formula, which does carry a minus sign), so requires -c > 0, giving c < 0.
    • Why not D: Correctly forms the discriminant inequality 36 - 4c > 0 but forgets to reverse the inequality sign when dividing both sides by the negative number -4, obtaining c > 9 instead of c < 9.

  14. Question 14Answer: D

    1. Try x = 1: 1 - 7 + 6 = 0, so (x - 1) is a factor by the Factor Theorem.
    2. Divide x^3 - 7x + 6 by (x - 1): the quotient is x^2 + x - 6, since x(x - 1) = x^2 - x, leaving -6x + 6, and -6(x - 1) = -6x + 6 with no remainder.
    3. Factorise the quadratic: x^2 + x - 6 = (x + 3)(x - 2), since 3 and -2 multiply to -6 and add to 1.
    4. So x^3 - 7x + 6 = (x - 1)(x + 3)(x - 2), which written in order is (x - 1)(x - 2)(x + 3), so the answer is D.
    • Why not A: Correctly identifies x = 1 as a root, but makes a sign slip while carrying out the division, obtaining a quotient of x^2 - x - 6 instead of x^2 + x - 6 (flipping the sign of the x-coefficient), which factorises as (x - 3)(x + 2).
    • Why not B: Correctly identifies x = 1 as a root, but makes a sign error on the final step of the division, obtaining a quotient of x^2 + x + 6 instead of x^2 + x - 6 (flipping the sign of the constant term), and since this quadratic has no real factors, reports it unfactorised.
    • Why not C: Correctly reduces to the quadratic factor x^2 + x - 6, but mis-factorises it using the pair 6 and -1 (which multiply to -6 but sum to 5, not the required 1), writing it as (x - 1)(x + 6) instead of (x + 3)(x - 2).

  15. Question 15Answer: B

    1. For |A| = |B|, either A = B or A = -B, so consider both cases.
    2. Case 1: 2x - 5 = x + 4, which gives x = 9.
    3. Case 2: 2x - 5 = -(x + 4) = -x - 4, so 3x = 1, giving x = 1/3.
    4. Checking both: |2(9) - 5| = 13 = |9 + 4|, and |2(1/3) - 5| = 13/3 = |1/3 + 4|, so both are valid and the answer is B.
    • Why not A: Only sets 2x - 5 equal to x + 4, giving x = 9, but forgets that |A| = |B| also allows A = -B, so misses the second case entirely.
    • Why not C: Correctly finds x = 9 from 2x - 5 = x + 4, but expands the second case's bracket incorrectly as 2x - 5 = -x + 4 instead of 2x - 5 = -x - 4, forgetting to distribute the negative sign to both terms inside the bracket, giving x = 3 instead of x = 1/3.
    • Why not D: Misreads |x + 4| as |x - 4| throughout, then correctly applies both cases to this altered equation: 2x - 5 = x - 4 gives x = 1, and 2x - 5 = -(x - 4) gives x = 3.

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