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

Test standard. 15 questions, 15 marks, about 22 minutes.

ESAT Mathematics 2: Algebra, functions and sequences, set 1

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.
  1. 11 mark

    Simplify (27x^6)^(1/3) fully.

    1. A 3x^2
    2. B 3x^18
    3. C 9x^2
    4. D 9x^6
  2. 21 mark

    Rationalise the denominator of 10/(sqrt(5) - 2), and simplify fully.

    1. A 10sqrt(5)
    2. B 10sqrt(5) + 20
    3. C -10sqrt(5) - 20
    4. D (10sqrt(5) + 20)/9
  3. 31 mark

    The equation x^2 - kx + 25 = 0 has equal roots. Find the positive value of k.

    1. A 100
    2. B 5
    3. C 10
    4. D -10
  4. 41 mark

    By completing the square, find the solutions of x^2 - 6x + 4 = 0, giving your answer in the form p +- sqrt(q).

    1. A x = 3 +- sqrt(13)
    2. B x = 6 +- sqrt(5)
    3. C x = 0 or x = 6
    4. D x = 3 +- sqrt(5)
  5. 51 mark

    Find the values of x that satisfy both y = x + 1 and y = x^2 - 5 simultaneously.

    1. A x = 3 or x = -2
    2. B x = -3 or x = 2
    3. C x = 6 or x = -1
    4. D x = 3 (one solution only)
  6. 61 mark

    Solve the inequality x^2 - 5x + 6 <= 0.

    1. A x <= 2 or x >= 3
    2. B 2 <= x <= 3
    3. C 2 < x < 3
    4. D 1 <= x <= 6
  7. 71 mark

    Given that x = 2 is a root of x^3 - 3x^2 - 4x + 12 = 0, factorise the cubic completely.

    1. A (x - 2)^2(x + 3)
    2. B (x - 2)(x + 1)(x - 6)
    3. C (x - 3)(x + 2)^2
    4. D (x - 2)(x - 3)(x + 2)
  8. 81 mark

    Find the remainder when x^3 + 2x^2 - 5x + 1 is divided by (x + 3).

    1. A 31
    2. B 61
    3. C 7
    4. D 6
  9. 91 mark

    The function f is defined for all real x by f(x) = |x - 3|. Given that f(a) = 5, find the possible value(s) of a.

    1. A a = 8 (one solution only)
    2. B a = 8 or a = -2
    3. C a = 8 or a = -8
    4. D a = 3 +- sqrt(5)
  10. 101 mark

    A sequence is defined by u_n = 3n^2 - 2n for n = 1, 2, 3, .... Find u_4.

    1. A 40
    2. B -20
    3. C 136
    4. D 16
  11. 111 mark

    A sequence is defined by the recurrence relation x_1 = 3 and x_(n+1) = 2x_n - 5 for n >= 1. Find x_3.

    1. A 1
    2. B 27
    3. C -18
    4. D -3
  12. 121 mark

    Find the sum of the first 40 positive integers.

    1. A 780
    2. B 800
    3. C 820
    4. D 1640
  13. 131 mark

    Find the sum of the first 6 terms of the geometric series 3 + 6 + 12 + 24 + ...

    1. A 189
    2. B 192
    3. C 93
    4. D 297
  14. 141 mark

    A geometric series has first term 12 and common ratio 1/4. Find the sum to infinity of the series.

    1. A 48/5
    2. B 9
    3. C 48
    4. D 16
  15. 151 mark

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

    1. A 20
    2. B 40
    3. C 80
    4. D 10

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: A

    1. The power 1/3 applies to every factor inside the brackets, so (27x^6)^(1/3) = 27^(1/3) x (x^6)^(1/3).
    2. 27^(1/3) is the number that cubes to give 27, which is 3, since 3^3 = 27.
    3. For the x term, use the law (x^m)^n = x^(mn): (x^6)^(1/3) = x^(6 x 1/3) = x^2.
    4. Combining both parts gives 3x^2, so the answer is A.
    • Why not B: Correctly cube roots 27 to get 3, but then cubes the exponent of x instead of dividing it by 3, giving x^(6x3) = x^18 rather than x^(6/3) = x^2.
    • Why not C: Miscalculates the cube root of 27 as 27/3 = 9 instead of the number whose cube is 27, which is 3.
    • Why not D: Makes both errors at once: treats 27^(1/3) as 27/3 = 9, and separately fails to apply the power 1/3 to the x term at all, leaving x^6 unchanged.

  2. Question 2Answer: B

    1. Multiply the numerator and denominator by the conjugate of the denominator, sqrt(5) + 2.
    2. The denominator becomes (sqrt(5) - 2)(sqrt(5) + 2) = (sqrt(5))^2 - 2^2 = 5 - 4 = 1.
    3. The numerator becomes 10(sqrt(5) + 2) = 10sqrt(5) + 20.
    4. Since the denominator is 1, the fraction simplifies to 10sqrt(5) + 20, so the answer is B.
    • Why not A: Distributes the 10 across the surd term of the numerator but forgets to also multiply the constant term by 10, dropping the '+ 20' part of the expansion entirely.
    • Why not C: Correctly expands the numerator to 10sqrt(5) + 20, but makes a sign error expanding the denominator by difference of two squares, computing (sqrt(5) - 2)(sqrt(5) + 2) as 2^2 - (sqrt(5))^2 = 4 - 5 = -1 instead of the correct 5 - 4 = 1; dividing by -1 flips the sign of the whole answer.
    • Why not D: Applies the difference of two squares to the denominator as (sqrt(5))^2 + 2^2 = 9 instead of the correct (sqrt(5))^2 - 2^2 = 1, adding the squared terms instead of subtracting them.

  3. Question 3Answer: C

    1. For ax^2 + bx + c = 0 to have equal (repeated) roots, the discriminant b^2 - 4ac must equal 0.
    2. Here a = 1, b = -k and c = 25, so the condition is (-k)^2 - 4(1)(25) = 0, which gives k^2 - 100 = 0.
    3. So k^2 = 100, meaning k = 10 or k = -10.
    4. The question asks for the positive value, so the answer is C, k = 10.
    • Why not A: Correctly reaches the equation k^2 = 100 but forgets to take the square root, giving the squared value itself rather than k.
    • Why not B: Uses the condition b^2 = ac instead of the correct b^2 = 4ac for equal roots, omitting the factor of 4 and so understating k^2 as 25 instead of 100.
    • Why not D: Correctly finds that k = 10 or k = -10 solve k^2 = 100, but selects the negative root, missing that the question specifically asks for the positive value.

  4. Question 4Answer: D

    1. Halve the coefficient of x (which is -6) to get -3, and write x^2 - 6x + 4 as (x - 3)^2 - 9 + 4.
    2. Combine the constants: -9 + 4 = -5, so the equation becomes (x - 3)^2 - 5 = 0.
    3. Rearranging gives (x - 3)^2 = 5, so x - 3 = +-sqrt(5).
    4. Therefore x = 3 +- sqrt(5), so the answer is D.
    • Why not A: Combines the constants -9 and +4 incorrectly, computing -9 - 4 = -13 instead of -9 + 4 = -5, so ends up with (x - 3)^2 = 13 rather than (x - 3)^2 = 5.
    • Why not B: Correctly finds the value 5 under the root, but then states the final roots using the original coefficient of x (6) instead of half of it (3), the number that actually belongs inside the completed square.
    • Why not C: Drops the constant term +4 from the original equation, effectively completing the square on x^2 - 6x = 0 alone, which factorises as x(x - 6) = 0 and gives these two rational roots instead.

  5. Question 5Answer: A

    1. Substitute the linear expression for y into the quadratic one: x + 1 = x^2 - 5.
    2. Rearrange to standard form: 0 = x^2 - x - 6, so x^2 - x - 6 = 0.
    3. Factorise: the two numbers that multiply to -6 and add to -1 are -3 and 2, giving (x - 3)(x + 2) = 0.
    4. So x = 3 or x = -2, and the answer is A.
    • Why not B: When rearranging x + 1 = x^2 - 5, moves the x term to the other side without changing its sign, forming x^2 + x - 6 = 0 instead of the correct x^2 - x - 6 = 0.
    • Why not C: Correctly forms x^2 - x - 6 = 0 but picks a factor pair of -6 (namely -6 and 1) without checking that the pair also sums to the middle coefficient -1, instead of the correct pair -3 and 2.
    • Why not D: Finds one root of the quadratic by inspection but stops there, not reporting that a quadratic equation generally has a second solution.

  6. Question 6Answer: B

    1. Factorise the quadratic: the numbers that multiply to 6 and add to 5 are 2 and 3, so x^2 - 5x + 6 = (x - 2)(x - 3).
    2. The graph of y = (x - 2)(x - 3) is an upward-opening parabola with roots at x = 2 and x = 3.
    3. The parabola is below or on the x-axis (y <= 0) only for the x-values between and including the roots.
    4. So the solution is 2 <= x <= 3, and the answer is B.
    • Why not A: Correctly finds the critical values 2 and 3, but selects the region outside them, which is where the upward-opening parabola is positive (x^2 - 5x + 6 >= 0), not where it is negative or zero.
    • Why not C: Uses strict inequalities throughout, forgetting that the original inequality is 'less than or equal to', so the boundary values x = 2 and x = 3 (where the expression equals exactly 0) must be included.
    • Why not D: Misfactorises x^2 - 5x + 6, picking a factor pair of 6 (namely 1 and 6) that does not sum to 5, instead of the correct pair 2 and 3.

  7. Question 7Answer: D

    1. By the Factor Theorem, since x = 2 is a root, (x - 2) is a factor of the cubic.
    2. Dividing x^3 - 3x^2 - 4x + 12 by (x - 2) gives the quotient x^2 - x - 6.
    3. Factorise the quotient: the numbers that multiply to -6 and add to -1 are -3 and 2, so x^2 - x - 6 = (x - 3)(x + 2).
    4. The complete factorisation is therefore (x - 2)(x - 3)(x + 2), so the answer is D.
    • Why not A: Makes a sign slip during the polynomial division and reaches the quadratic quotient x^2 + x - 6 instead of the correct x^2 - x - 6, then factorises that wrong quotient as (x - 2)(x + 3), duplicating the root x = 2.
    • Why not B: Correctly divides to reach the quadratic quotient x^2 - x - 6, but then mis-factorises it using the factor pair 1 and -6 (which multiply to -6 but do not sum to -1), instead of the correct pair -3 and 2.
    • Why not C: Misapplies the Factor Theorem, writing the linear factor for the root x = 2 as (x + 2) instead of the correct (x - 2), while still correctly factorising the quotient x^2 - x - 6 as (x - 3)(x + 2); multiplying these together duplicates the (x + 2) factor to give (x - 3)(x + 2)^2 instead of combining it with (x - 2).

  8. Question 8Answer: C

    1. By the Remainder Theorem, the remainder when a polynomial f(x) is divided by (x + 3) equals f(-3).
    2. Substitute x = -3 into each term: (-3)^3 = -27, 2 x (-3)^2 = 2 x 9 = 18, and -5 x (-3) = 15.
    3. Add these together with the constant term: -27 + 18 + 15 + 1.
    4. -27 + 18 = -9, then -9 + 15 = 6, then 6 + 1 = 7, so the remainder is 7 and the answer is C.
    • Why not A: By the Remainder Theorem, dividing by (x + 3) = (x - (-3)) requires evaluating the polynomial at x = -3, but this substitutes x = 3 instead, using the wrong sign for the root.
    • Why not B: Correctly identifies that x = -3 must be substituted, but miscalculates the odd power (-3)^3 as +27 instead of the correct -27.
    • Why not D: Correctly evaluates the x^3, x^2 and x terms, but forgets to add the constant term +1 of the polynomial at the end.

  9. Question 9Answer: B

    1. The modulus equation |a - 3| = 5 means a - 3 = 5 or a - 3 = -5.
    2. Solving the first case: a - 3 = 5 gives a = 8.
    3. Solving the second case: a - 3 = -5 gives a = -2.
    4. So a = 8 or a = -2, and the answer is B.
    • Why not A: Solves the case a - 3 = 5 to get a = 8, but forgets that |a - 3| = 5 also has a second case, a - 3 = -5, because f is a many-to-one function.
    • Why not C: Correctly finds a = 8 from one case, but assumes the second solution must simply be its negative, giving -8, instead of solving a - 3 = -5 to get the correct value a = -2.
    • Why not D: Squares both sides of the equation but only squares the left-hand side properly, using (a - 3)^2 = 5 instead of the correct (a - 3)^2 = 5^2 = 25.

  10. Question 10Answer: A

    1. Substitute n = 4 into the formula u_n = 3n^2 - 2n.
    2. First find 4^2 = 16, then multiply by 3: 3 x 16 = 48.
    3. Then find 2 x 4 = 8.
    4. Subtract: u_4 = 48 - 8 = 40, so the answer is A.
    • Why not B: Swaps which term is squared, computing 3n - 2n^2 instead of 3n^2 - 2n, giving 3(4) - 2(4)^2 = 12 - 32 = -20.
    • Why not C: Squares the whole coefficient and n together, computing (3n)^2 - 2n instead of 3n^2 - 2n, giving (3 x 4)^2 - 2(4) = 144 - 8 = 136.
    • Why not D: Misreads 4^2 as 4 x 2 = 8 instead of 16, so computes 3(8) - 2(4) = 24 - 8 = 16.

  11. Question 11Answer: D

    1. Apply the recurrence relation x_(n+1) = 2x_n - 5, starting from x_1 = 3.
    2. Find x_2: x_2 = 2x_1 - 5 = 2(3) - 5 = 6 - 5 = 1.
    3. Find x_3 by applying the rule again to x_2: x_3 = 2x_2 - 5 = 2(1) - 5 = 2 - 5 = -3.
    4. So x_3 = -3, and the answer is D.
    • Why not A: Correctly computes x_2 = 2(3) - 5 = 1, but then miscounts the number of steps needed and reports this value of x_2 as x_3 without applying the recurrence a second time.
    • Why not B: Uses the rule x_(n+1) = 2x_n + 5 throughout, with the wrong sign on the 5, giving x_2 = 2(3) + 5 = 11 and then x_3 = 2(11) + 5 = 27.
    • Why not C: Misreads the recurrence as x_(n+1) = 2(x_n - 5), subtracting 5 before doubling instead of after, giving x_2 = 2(3 - 5) = -4 and then x_3 = 2(-4 - 5) = -18.

  12. Question 12Answer: C

    1. The sum of the first n natural numbers is given by the formula n(n + 1)/2.
    2. Here n = 40, so the sum is 40 x 41/2.
    3. 40 x 41 = 1640, and 1640/2 = 820.
    4. So the sum of the first 40 positive integers is 820, and the answer is C.
    • Why not A: Uses the formula n(n - 1)/2 instead of the correct n(n + 1)/2, an off-by-one error that gives 40 x 39/2 = 780.
    • Why not B: Misremembers the sum formula as n^2/2 instead of n(n + 1)/2, giving 40^2/2 = 1600/2 = 800.
    • Why not D: Correctly forms the product n(n + 1) = 40 x 41 = 1640, but forgets to divide by 2 to complete the formula.

  13. Question 13Answer: A

    1. This is a geometric series with first term a = 3, common ratio r = 2, and n = 6 terms.
    2. The sum of the first n terms of a geometric series is S_n = a(r^n - 1)/(r - 1).
    3. Substitute the values: S_6 = 3 x (2^6 - 1)/(2 - 1) = 3 x (64 - 1)/1 = 3 x 63.
    4. 3 x 63 = 189, so the sum of the first 6 terms is 189, and the answer is A.
    • Why not B: Forgets to subtract 1 from r^n in the sum formula, computing a x r^n/(r - 1) = 3 x 64/1 = 192 instead of a x (r^n - 1)/(r - 1).
    • Why not C: Uses n = 5 instead of n = 6 in the formula, an off-by-one error that sums only the first five terms: 3 x (2^5 - 1)/(2 - 1) = 3 x 31 = 93.
    • Why not D: Misapplies the arithmetic series formula S = n/2 x (first term + last term) to this geometric series, using the 6th term 96 to get 6/2 x (3 + 96) = 3 x 99 = 297.

  14. Question 14Answer: D

    1. Since |r| = |1/4| < 1, the geometric series converges and has a sum to infinity.
    2. The formula for the sum to infinity is S = a/(1 - r), where a = 12 and r = 1/4.
    3. Compute 1 - r = 1 - 1/4 = 3/4.
    4. So S = 12/(3/4) = 12 x 4/3 = 48/3 = 16, and the answer is D.
    • Why not A: Uses the formula a/(1 + r) instead of the correct a/(1 - r), a sign error that gives 12/(1 + 1/4) = 12/(5/4) = 48/5.
    • Why not B: Multiplies instead of dividing, computing a x (1 - r) = 12 x (3/4) = 9 rather than a/(1 - r).
    • Why not C: Uses a/r instead of a/(1 - r), forgetting to subtract r from 1 first, giving 12/(1/4) = 48.

  15. Question 15Answer: B

    1. The general term in the expansion of (1 + 2x)^5 is 5Ck x (2x)^k, for k = 0, 1, ..., 5.
    2. The term in x^2 occurs when k = 2, giving 5C2 x (2x)^2.
    3. 5C2 = 5!/(2! x 3!) = (5 x 4)/(2 x 1) = 10, and (2x)^2 = 4x^2.
    4. The coefficient of x^2 is therefore 10 x 4 = 40, so the answer is B.
    • Why not A: Uses 2^1 instead of 2^2 for the (2x)^2 term, forgetting to square the coefficient 2, giving 5C2 x 2 = 10 x 2 = 20.
    • Why not C: Miscalculates 5C2, computing 5 x 4 = 20 instead of dividing by 2! to get the correct value of 10, then multiplies by the correct 2^2 = 4 to give 20 x 4 = 80.
    • Why not D: Finds the correct binomial coefficient 5C2 = 10, but forgets to include the factor of 2^2 = 4 that comes from (2x)^2, using only the coefficient 10 as the final answer.

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