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Admissions tests / TMUA / Paper 1 / Algebra and functions

Test standard. 12 questions, 12 marks, about 45 minutes.

TMUA Paper 1: Algebra and functions, set 1

Indices, surds, quadratics, simultaneous equations, inequalities, polynomials and the factor theorem, applied in contexts that hide which technique is wanted.

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

  • Answer all questions. No calculator.
  • Each question has exactly one correct answer.
  1. 11 mark

    Given that a = 8^(2/3) and b = 27^(-1/3), what is the value of a/b?

    1. A 4/3
    2. B 12
    3. C -12
    4. D 1/12
  2. 21 mark

    Rationalise and simplify (5 + sqrt(3)) / (2 - sqrt(3)).

    1. A 13 - 7 sqrt(3)
    2. B 13
    3. C -13 - 7 sqrt(3)
    4. D 13 + 7 sqrt(3)
  3. 31 mark

    The equation kx^2 + 6x + k = 0, where k is a nonzero constant, has two equal roots. What are the possible values of k?

    1. A k = 3 or k = -3
    2. B k = 6 or k = -6
    3. C k = 3
    4. D k = 9
  4. 41 mark

    The expression x^2 - 6x + 11 can be written in the form (x - p)^2 + q for all real x. What is the minimum value of the expression?

    1. A 11
    2. B -25
    3. C 2
    4. D -20
  5. 51 mark

    The line y = x + 1 intersects the curve y = x^2 - 2x - 3 at two points. What is the sum of the x-coordinates of the two points of intersection?

    1. A 2
    2. B 3
    3. C 1
    4. D -4
  6. 61 mark

    Find the set of values of x for which x^2 - 5x + 6 <= 0.

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

    Solve the inequality 9x - 4 < 3(4x - 1) for x.

    1. A x > -1/3
    2. B x < -1/3
    3. C x > -1
    4. D x > 7/3
  8. 81 mark

    Expand and simplify (2x - 3)(x + 4) - (x - 1)^2.

    1. A x^2 + 3x - 11
    2. B x^2 + 2x - 13
    3. C x^2 + 7x - 13
    4. D x^2 + 5x - 13
  9. 91 mark

    When 2x^3 - 3x^2 - 11x + 6 is divided by (x - 3), the quotient is ax^2 + bx + c. Find the value of a + b + c.

    1. A 9
    2. B 7
    3. C -3
    4. D 3
  10. 101 mark

    The cubic p(x) = x^3 + ax^2 + bx - 6 has (x - 1) as a factor, and leaves a remainder of 4 when divided by (x + 1). What is the value of ab?

    1. A 8
    2. B -24
    3. C 4
    4. D 24
  11. 111 mark

    The function f is defined for all real x by f(x) = |x - 3| + sqrt(4). Given that f(a) = 7 and a < 3, find the value of a.

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

    Factorise 6x^2 + 7x - 3 completely.

    1. A (3x - 1)(2x + 3)
    2. B (3x + 1)(2x - 3)
    3. C (6x - 1)(x + 3)
    4. D (2x - 1)(3x + 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: B

    1. 8^(2/3) means (the cube root of 8) squared: 8^(1/3) = 2, so a = 2^2 = 4.
    2. 27^(-1/3) means 1 divided by the cube root of 27: 27^(1/3) = 3, so b = 1/3.
    3. a/b = 4 / (1/3) = 4 x 3 = 12.
    4. So a/b = 12, which is option B.
    • Why not A: Computes a x b instead of a / b, multiplying the two values rather than dividing.
    • Why not C: Takes 27^(-1/3) to be -1/3, treating the negative exponent as making the result negative rather than taking the reciprocal 1/3.
    • Why not D: Finds b / a instead of a / b, inverting the required ratio.

  2. Question 2Answer: D

    1. Multiply the numerator and denominator by the conjugate of the denominator, 2 + sqrt(3).
    2. The denominator becomes (2 - sqrt(3))(2 + sqrt(3)) = 4 - 3 = 1.
    3. The numerator becomes (5 + sqrt(3))(2 + sqrt(3)) = 10 + 5 sqrt(3) + 2 sqrt(3) + 3 = 13 + 7 sqrt(3).
    4. Since the denominator is 1, the expression simplifies to 13 + 7 sqrt(3), which is option D.
    • Why not A: Makes a sign error expanding the numerator (5 + sqrt(3))(2 + sqrt(3)), treating the two middle terms as though they subtract instead of add to give 7 sqrt(3).
    • Why not B: Drops the surd term after rationalising, incorrectly assuming the sqrt(3) part cancels out.
    • Why not C: Makes a sign error in the difference of two squares, computing (2 - sqrt(3))(2 + sqrt(3)) as -1 instead of +1.

  3. Question 3Answer: A

    1. For a quadratic to have two equal (repeated) roots, its discriminant b^2 - 4ac must equal 0.
    2. Here a = k, b = 6 and c = k, so the discriminant is 6^2 - 4(k)(k) = 36 - 4k^2.
    3. Setting 36 - 4k^2 = 0 gives k^2 = 9, so k = 3 or k = -3.
    4. So the possible values are k = 3 or k = -3, which is option A.
    • Why not B: Uses b^2 - ac = 0 instead of the correct discriminant b^2 - 4ac = 0, dropping the factor of 4.
    • Why not C: Solves k^2 = 9 but takes only the positive square root, missing the solution k = -3.
    • Why not D: Treats the product ac as k instead of k^2 (since a = c = k here), turning the equation into a linear one in k.

  4. Question 4Answer: C

    1. Half the coefficient of x is -3, so x^2 - 6x + 11 = (x - 3)^2 - 9 + 11.
    2. Combine the constants: -9 + 11 = 2, so the expression equals (x - 3)^2 + 2.
    3. Since (x - 3)^2 >= 0 for all real x, the minimum value of the expression is 2, occurring when x = 3.
    4. So the minimum value is 2, which is option C.
    • Why not A: Writes the expression as (x - 3)^2 + 11, forgetting to subtract the 9 that completing the square introduces.
    • Why not B: Takes p = 6 (the full coefficient of x) instead of half of it when completing the square, writing (x - 6)^2 - 25.
    • Why not D: Subtracts the original constant instead of adding it, writing (x - 3)^2 - 20.

  5. Question 5Answer: B

    1. At a point of intersection, x + 1 = x^2 - 2x - 3.
    2. Rearranging gives x^2 - 2x - 3 - x - 1 = 0, that is x^2 - 3x - 4 = 0.
    3. For a quadratic x^2 + px + q = 0, the sum of the roots is -p; here p = -3, so the sum of the x-coordinates is -(-3) = 3.
    4. So the sum of the x-coordinates is 3, which is option B.
    • Why not A: Forgets to move the x-term across before combining the equations, effectively solving x^2 - 2x - 4 = 0.
    • Why not C: Makes a sign error moving the x-term across the equals sign, effectively solving x^2 - x - 4 = 0.
    • Why not D: Calculates the product of the roots (c/a) rather than their sum (-b/a).

  6. Question 6Answer: D

    1. Factorise the quadratic: x^2 - 5x + 6 = (x - 2)(x - 3).
    2. The critical values are x = 2 and x = 3, where the expression equals zero.
    3. Since the coefficient of x^2 is positive, the graph is an upward parabola, so the expression is <= 0 between the roots, where the graph dips below the x-axis.
    4. So the solution is 2 <= x <= 3, which is option D.
    • Why not A: Takes the region outside the two roots, which is the solution for x^2 - 5x + 6 >= 0, rather than between them.
    • Why not B: Reads the roots off the factorised form (x - 2)(x - 3) as -2 and -3 instead of 2 and 3, forgetting to change sign when solving x - 2 = 0 and x - 3 = 0.
    • Why not C: Combines both errors: uses the wrong-signed roots -2 and -3, and also takes the region outside them rather than between them.

  7. Question 7Answer: A

    1. Expand the bracket: 3(4x - 1) = 12x - 3, so the inequality is 9x - 4 < 12x - 3.
    2. Collect the constants on the left and the x-terms on the right: -4 + 3 < 12x - 9x, that is -1 < 3x.
    3. Divide both sides by 3, a positive number, so the inequality sign is unchanged: -1/3 < x, that is x > -1/3.
    4. So the solution is x > -1/3, which is option A.
    • Why not B: Divides both sides of -3x < 1 by -3 without reversing the direction of the inequality.
    • Why not C: Expands 3(4x - 1) as 12x - 1, forgetting to multiply the -1 by 3 as well.
    • Why not D: Moves the -4 across the inequality sign without changing it to +4, as though it were already positive.

  8. Question 8Answer: C

    1. Expand (2x - 3)(x + 4) = 2x^2 + 8x - 3x - 12 = 2x^2 + 5x - 12.
    2. Expand (x - 1)^2 = x^2 - 2x + 1.
    3. Subtract, remembering to change the sign of every term in the second bracket: 2x^2 + 5x - 12 - x^2 + 2x - 1.
    4. Collect like terms to get x^2 + 7x - 13, which is option C.
    • Why not A: Only changes the sign of the first term when subtracting (x - 1)^2, writing -x^2 - 2x + 1 instead of -x^2 + 2x - 1.
    • Why not B: Expands (2x - 3)(x + 4) using only the first and last terms (2x^2 - 12), omitting the two middle terms.
    • Why not D: Expands (x - 1)^2 as x^2 + 1, omitting the middle term -2x, as though squaring a bracket just squares each part.

  9. Question 9Answer: D

    1. Since we are dividing by (x - 3), use synthetic division with the coefficients 2, -3, -11, 6 and the number 3.
    2. Bring down the 2. Multiply by 3 to get 6, add to -3 to get 3. Multiply 3 by 3 to get 9, add to -11 to get -2. Multiply -2 by 3 to get -6, add to 6 to get 0, confirming the division is exact.
    3. The quotient is 2x^2 + 3x - 2, so a = 2, b = 3, c = -2.
    4. a + b + c = 2 + 3 - 2 = 3, which is option D.
    • Why not A: Uses -3 as the synthetic-division number instead of +3, dividing as though by (x + 3) rather than (x - 3).
    • Why not B: Makes a sign error on the constant term of the quotient, giving c = 2 instead of c = -2.
    • Why not C: Makes a sign error on the x-term of the quotient, giving b = -3 instead of b = 3.

  10. Question 10Answer: B

    1. Since (x - 1) is a factor of p(x), the Factor Theorem gives p(1) = 0: 1 + a + b - 6 = 0, so a + b = 5.
    2. Since dividing by (x + 1) leaves remainder 4, the Remainder Theorem gives p(-1) = 4: -1 + a - b - 6 = 4, so a - b = 11.
    3. Adding the two equations: 2a = 16, so a = 8; then b = 5 - 8 = -3.
    4. ab = 8 x (-3) = -24, which is option B.
    • Why not A: Swaps which value of x goes with which condition, using p(-1) = 0 for the factor (x - 1) and p(1) = 4 for the remainder from (x + 1), instead of the correct substitutions.
    • Why not C: Takes the remainder to be -4 instead of +4, giving p(-1) = -4.
    • Why not D: Rearranges a + b = 5 to get b = a - 5 instead of b = 5 - a, a sign slip when transposing.

  11. Question 11Answer: C

    1. sqrt(4) always means the positive square root, so sqrt(4) = 2, not -2.
    2. f(a) = 7 gives |a - 3| + 2 = 7, so |a - 3| = 5.
    3. Since a < 3, we have a - 3 < 0, so |a - 3| = 3 - a (not a - 3). This gives 3 - a = 5.
    4. Solving gives a = -2, which is option C.
    • Why not A: Takes sqrt(4) to be -2, forgetting that sqrt always denotes the positive square root.
    • Why not B: Solves a - 3 = 5 instead of 3 - a = 5, ignoring the given condition a < 3 that fixes which branch of the modulus applies.
    • Why not D: Ignores the +sqrt(4) = 2 term, setting |a - 3| equal to 7 directly instead of 5.

  12. Question 12Answer: A

    1. To factorise 6x^2 + 7x - 3, look for two numbers that multiply to 6 x (-3) = -18 and add to 7: these are 9 and -2.
    2. Split the middle term: 6x^2 + 9x - 2x - 3.
    3. Factor by grouping: 3x(2x + 3) - 1(2x + 3) = (3x - 1)(2x + 3).
    4. Check: (3x - 1)(2x + 3) = 6x^2 + 9x - 2x - 3 = 6x^2 + 7x - 3, which matches, so the answer is option A.
    • Why not B: Uses the correct split numbers 9 and -2 but attaches them to the wrong brackets, which flips the sign of the middle term when expanded.
    • Why not C: Puts the whole leading coefficient 6 into one bracket instead of splitting it as 3 and 2 across both brackets, giving the wrong middle term.
    • Why not D: The bracket (3x + 3) still shares a common factor of 3, so this is not a complete factorisation, and it does not expand back to the original expression either.

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