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Admissions tests / TMUA / Paper 1 / Algebra and functions

Test standard. 12 questions, 12 marks, about 48 minutes.

TMUA Paper 1: Algebra and functions, set 2

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

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

    Simplify (x^4 y^(-6))^(1/2) / (x^(-1) y^2) fully.

    1. A x/y
    2. B x^3/y
    3. C x^3/y^5
    4. D x^3/y^8
  2. 21 mark

    Simplify sqrt(50) + sqrt(18) - sqrt(8) fully.

    1. A 6 sqrt(2)
    2. B sqrt(60)
    3. C 12 sqrt(2)
    4. D 10 sqrt(2)
  3. 31 mark

    The equation 3x^2 + kx + 12 = 0 has no real roots. What is the range of possible values of k?

    1. A k <= -12 or k >= 12
    2. B k < 144
    3. C k < 12
    4. D -12 < k < 12
  4. 41 mark

    By completing the square, solve x^2 + 8x + 10 = 0, giving your answers in the form p +/- sqrt(q).

    1. A x = -8 +/- 3 sqrt(6)
    2. B x = -4 +/- sqrt(6)
    3. C x = 4 +/- sqrt(6)
    4. D x = -4 +/- sqrt(26)
  5. 51 mark

    The line y = mx + 4 is a tangent to the curve y = x^2 + 2x + 7. Find the possible values of m.

    1. A m = 2 +/- 2 sqrt(3)
    2. B m = -2 +/- 2 sqrt(3)
    3. C m = 2 +/- sqrt(3)
    4. D m = 2 +/- 2 sqrt(7)
  6. 61 mark

    Find the set of values of x for which -x^2 + 2x + 8 >= 0.

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

    Solve the inequality 5 - 2x >= 3x - 15 for x.

    1. A x <= 20
    2. B x >= 4
    3. C x <= 4
    4. D x <= 2
  8. 81 mark

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

    1. A 2x^3 + 2x^2 - 15x + 6
    2. B 2x^3 + x^2 - 13x + 6
    3. C 2x^3 + 9x^2 + 7x - 6
    4. D 2x^3 - 9x^2 - 13x + 6
  9. 91 mark

    When 4x^3 - 4x^2 - 5x - 1 is divided by (2x + 1), the quotient is ax^2 + bx + c. Find the value of a + b + c.

    1. A 0
    2. B -1
    3. C 2
    4. D -2
  10. 101 mark

    The cubic p(x) = 2x^3 - x^2 + kx - 10 has (x - 2) as a factor. What is the remainder when p(x) is divided by (x + 1)?

    1. A -12
    2. B -14
    3. C -8
    4. D -10
  11. 111 mark

    Solve the equation |2x - 5| = |x + 4|. What is the sum of the solutions?

    1. A 9
    2. B 12
    3. C 28/3
    4. D 1/3
  12. 121 mark

    The graph of y = x^2 + bx + c has a minimum point at (3, -4). What is the value of b + c?

    1. A 11
    2. B -1
    3. C -13
    4. D 27

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. Apply the outer power 1/2 to each factor inside the bracket: (x^4)^(1/2) = x^2 and (y^(-6))^(1/2) = y^(-3), so the bracket simplifies to x^2 y^(-3).
    2. Divide by x^(-1) y^2 by subtracting exponents: for x, 2 - (-1) = 3; for y, -3 - 2 = -5.
    3. This gives x^3 y^(-5), which is x^3/y^5.
    4. So the fully simplified expression is x^3/y^5, which is option C.
    • Why not A: Divides by adding indices instead of subtracting: computes x^(2 + (-1)) y^(-3 + 2) = x/y rather than subtracting each exponent.
    • Why not B: Misreads the sign of the denominator's y^2, treating it as y^(-2), so computes the y index as -3 - (-2) = -1 instead of -3 - 2 = -5.
    • Why not D: Applies the outer power 1/2 to the x term inside the bracket but not to the y term, leaving y^(-6) instead of halving it to y^(-3), so the y index ends up as -6 - 2 = -8.

  2. Question 2Answer: A

    1. Write each surd in simplified form: sqrt(50) = sqrt(25 x 2) = 5 sqrt(2).
    2. Similarly sqrt(18) = sqrt(9 x 2) = 3 sqrt(2) and sqrt(8) = sqrt(4 x 2) = 2 sqrt(2).
    3. Combine like surds: 5 sqrt(2) + 3 sqrt(2) - 2 sqrt(2) = 6 sqrt(2).
    4. So the expression simplifies to 6 sqrt(2), which is option A.
    • Why not B: Treats sqrt(a) + sqrt(b) as sqrt(a + b), adding the numbers under the root directly to get sqrt(50 + 18 - 8) = sqrt(60), instead of simplifying each surd first.
    • Why not C: Simplifies sqrt(18) as 9 sqrt(2) instead of 3 sqrt(2), taking the whole factor 9 out front rather than its square root.
    • Why not D: Treats the final subtraction as an addition, computing 5 sqrt(2) + 3 sqrt(2) + 2 sqrt(2) instead of 5 sqrt(2) + 3 sqrt(2) - 2 sqrt(2).

  3. Question 3Answer: D

    1. A quadratic has no real roots exactly when its discriminant b^2 - 4ac is negative.
    2. Here a = 3, b = k and c = 12, so the discriminant is k^2 - 4(3)(12) = k^2 - 144.
    3. Setting k^2 - 144 < 0 gives k^2 < 144, which means -12 < k < 12.
    4. So the range of values is -12 < k < 12, which is option D.
    • Why not A: Uses the condition for the equation to HAVE two real roots (discriminant >= 0) instead of no real roots (discriminant < 0), giving the region outside -12 and 12 rather than between them.
    • Why not B: Solves k^2 < 144 by dropping the square root step, treating it as k < 144 directly rather than -12 < k < 12.
    • Why not C: Takes only the positive square root of 144, missing the lower bound and giving k < 12 instead of -12 < k < 12.

  4. Question 4Answer: B

    1. Half the coefficient of x is 4, so x^2 + 8x + 10 = (x + 4)^2 - 16 + 10 = (x + 4)^2 - 6.
    2. Setting (x + 4)^2 - 6 = 0 gives (x + 4)^2 = 6, so x + 4 = +/- sqrt(6).
    3. Solving for x gives x = -4 +/- sqrt(6).
    4. So the solutions are x = -4 +/- sqrt(6), which is option B.
    • Why not A: Uses p = 8 instead of half of 8 when completing the square, writing (x + 8)^2 - 64 + 10 instead of (x + 4)^2 - 16 + 10.
    • Why not C: Completes the square correctly to (x + 4)^2 - 6 = 0 but then forgets to negate p when writing the final solutions, giving x = 4 +/- sqrt(6) instead of x = -4 +/- sqrt(6).
    • Why not D: Combines the constants -16 and +10 with a sign slip, computing -16 - 10 = -26 instead of -16 + 10 = -6.

  5. Question 5Answer: A

    1. At a point where the line meets the curve, mx + 4 = x^2 + 2x + 7, so x^2 + (2 - m)x + 3 = 0.
    2. A line is a tangent to a curve exactly when this equation has a repeated root, so its discriminant is zero: (2 - m)^2 - 4(1)(3) = 0.
    3. This gives (2 - m)^2 = 12, so 2 - m = +/- 2 sqrt(3). Rearranging each case gives m = 2 - 2 sqrt(3) or m = 2 + 2 sqrt(3).
    4. So the possible values are m = 2 +/- 2 sqrt(3), which is option A.
    • Why not B: Makes a sign error moving mx across when forming the quadratic, writing x^2 + (2 + m)x + 3 = 0 instead of x^2 + (2 - m)x + 3 = 0.
    • Why not C: Uses the condition b^2 - ac = 0 instead of the correct discriminant b^2 - 4ac = 0, dropping the factor of 4.
    • Why not D: Forgets to subtract the 4 from the constant term when setting the two expressions equal, using 7 instead of 3 as the constant in the resulting quadratic.

  6. Question 6Answer: D

    1. Multiply both sides by -1, remembering to reverse the inequality: x^2 - 2x - 8 <= 0.
    2. Factorise: x^2 - 2x - 8 = (x - 4)(x + 2), with roots x = 4 and x = -2.
    3. Since the coefficient of x^2 is positive, the expression is <= 0 between the roots.
    4. So the solution is -2 <= x <= 4, which is option D.
    • Why not A: Combines both other errors: uses the wrong-signed root 2 instead of -2, and also takes the region outside the roots rather than between them.
    • Why not B: Reads the roots of (x - 4)(x + 2) as 4 and 2, forgetting that x + 2 = 0 gives x = -2, not x = 2.
    • Why not C: Multiplies both sides by -1 to clear the leading minus sign but forgets to reverse the inequality, so ends up with the region outside the roots instead of between them.

  7. Question 7Answer: C

    1. Collect the x-terms on one side and the constants on the other: 5 + 15 >= 3x + 2x, that is 20 >= 5x.
    2. Divide both sides by 5, a positive number, so the inequality is unchanged: 4 >= x, that is x <= 4.
    3. So the solution is x <= 4, which is option C.
    • Why not A: Combines the x-terms -2x and -3x incorrectly as -x instead of -5x, effectively solving -x >= -20 and then dividing by -1.
    • Why not B: Divides both sides of -5x >= -20 by -5 without reversing the direction of the inequality.
    • Why not D: Makes an arithmetic slip moving the constant, computing -5x >= -15 - 5 as -5x >= -10 instead of -5x >= -20.

  8. Question 8Answer: B

    1. First expand (x + 3)(2x - 1) = 2x^2 - x + 6x - 3 = 2x^2 + 5x - 3.
    2. Multiply this by (x - 2): (2x^2 + 5x - 3)(x - 2) = 2x^3 - 4x^2 + 5x^2 - 10x - 3x + 6.
    3. Collect like terms: 2x^3 + x^2 - 13x + 6.
    4. So the fully expanded expression is 2x^3 + x^2 - 13x + 6, which is option B.
    • Why not A: Drops the cross term -x when expanding (x + 3)(2x - 1), using 2x^2 + 6x - 3 instead of 2x^2 + 5x - 3, before multiplying by (x - 2).
    • Why not C: Misreads the final bracket as (x + 2) instead of (x - 2), flipping the sign of every term that comes from it.
    • Why not D: Combines the two x^2 contributions -4x^2 and +5x^2 with the wrong sign, getting -9x^2 instead of +x^2.

  9. Question 9Answer: D

    1. Write (2x + 1)(ax^2 + bx + c) = 4x^3 - 4x^2 - 5x - 1 and expand the left side: 2a x^3 + (2b + a) x^2 + (2c + b) x + c.
    2. Match the x^3 coefficients: 2a = 4, so a = 2. Match the constant terms: c = -1.
    3. Match the x^2 coefficients: 2b + a = -4, so 2b = -6 and b = -3. (Check the x coefficient: 2c + b = -2 - 3 = -5, which agrees.)
    4. So a = 2, b = -3, c = -1, and a + b + c = -2, which is option D.
    • Why not A: Misreads the sign of the dividend's constant term, taking c = 1 instead of c = -1.
    • Why not B: Drops the a term when matching the x^2 coefficients, solving 2b = -4 instead of 2b + a = -4, giving b = -2 instead of b = -3.
    • Why not C: Misreads the sign of the dividend's x^2 coefficient when setting up the comparison, using +4 instead of -4.

  10. Question 10Answer: A

    1. Since (x - 2) is a factor, the Factor Theorem gives p(2) = 0: 2(8) - 4 + 2k - 10 = 0, so 2 + 2k = 0 and k = -1.
    2. So p(x) = 2x^3 - x^2 - x - 10.
    3. By the Remainder Theorem, the remainder when p(x) is divided by (x + 1) is p(-1) = 2(-1)^3 - (-1)^2 - (-1) - 10.
    4. This equals -2 - 1 + 1 - 10 = -12, which is option A.
    • Why not B: Miscalculates 2^3 as 6 instead of 8 when substituting into the factor theorem equation, giving k = 1 instead of k = -1.
    • Why not C: Treats (-1) raised to an odd power as positive, evaluating (-1)^3 as +1 when computing the remainder.
    • Why not D: Evaluates p(1) instead of p(-1), confusing which value of x corresponds to the factor (x + 1).

  11. Question 11Answer: C

    1. An equation of the form |A| = |B| holds exactly when A = B or A = -B, so consider both cases.
    2. Case 1: 2x - 5 = x + 4 gives x = 9.
    3. Case 2: 2x - 5 = -(x + 4) gives 2x - 5 = -x - 4, so 3x = 1 and x = 1/3.
    4. The sum of the two solutions is 9 + 1/3 = 28/3, which is option C.
    • Why not A: Only finds the solution from the case 2x - 5 = x + 4, missing the second case entirely.
    • Why not B: Finds both solutions but makes a sign error distributing the negative in the second case, writing 2x - 5 = -x + 4 instead of 2x - 5 = -x - 4, giving x = 3 instead of x = 1/3, and so a sum of 9 + 3 = 12.
    • Why not D: Only finds the solution from the case 2x - 5 = -(x + 4), missing the first case entirely.

  12. Question 12Answer: B

    1. A quadratic with minimum point (h, k) can be written as y = (x - h)^2 + k, so here y = (x - 3)^2 - 4.
    2. Expand: (x - 3)^2 - 4 = x^2 - 6x + 9 - 4 = x^2 - 6x + 5.
    3. Comparing with y = x^2 + bx + c gives b = -6 and c = 5, so b + c = -1.
    4. So b + c = -1, which is option B.
    • Why not A: Writes the vertex form with the wrong sign, using (x + 3)^2 - 4 instead of (x - 3)^2 - 4.
    • Why not C: Expands (x - 3)^2 as x^2 - 9, forgetting the middle term -6x, as though squaring a bracket simply squares each part.
    • Why not D: Swaps the vertex coordinates, using -4 as the horizontal shift and 3 as the vertical shift, giving (x + 4)^2 + 3 instead of (x - 3)^2 - 4.

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