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Admissions tests / TMUA / Paper 1 / Algebra and functions

Demanding. 12 questions, 12 marks, about 51 minutes.

TMUA Paper 1: Algebra and functions, set 3

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

    Solve for n: 27^n = 9^(n+1) x 3.

    1. A n = 0
    2. B n = 3
    3. C n = 1
    4. D n = 2
  2. 21 mark

    Simplify 1/(sqrt(7) - sqrt(5)) - 1/(sqrt(7) + sqrt(5)).

    1. A 0
    2. B 2 sqrt(5)
    3. C sqrt(7)
    4. D sqrt(5)
  3. 31 mark

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

    1. A x^3 - 7x - 6
    2. B x^3 - 6x - 6
    3. C x^3 - 5x - 6
    4. D x^3 + 5x + 6
  4. 41 mark

    The equation x^2 + (k - 2)x + (k + 1) = 0 has two distinct real roots for a range of values of k. Which of the following describes that range?

    1. A 0 < k < 8
    2. B k > 8
    3. C k < 0 or k > 8
    4. D k <= 0 or k >= 8
  5. 51 mark

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

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

    Find the value of n for which the line y = 2x + n is tangent to the curve y = x^2 - 4x + 7.

    1. A n = -29
    2. B n = -2
    3. C n = 2
    4. D n = 3
  7. 71 mark

    When x^4 + x^3 - 7x^2 - x + 6 is divided by x^2 - 1, the quotient is x^2 + px + q. Find the value of p + q.

    1. A -5
    2. B -7
    3. C -6
    4. D 7
  8. 81 mark

    When the polynomial f(x) = x^4 - 3x^3 + 2x^2 + ax + b is divided by (x - 1)(x + 1), the remainder is 5x - 3. Find the value of a - b.

    1. A 8
    2. B 12
    3. C -14
    4. D 14
  9. 91 mark

    How many real solutions does the equation |3x - 4| = x - 2 have?

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

    Given that 3x^2 - 12x + 7 = a(x + b)^2 + c for all real x, where a, b and c are constants, find the value of a + b + c.

    1. A 4
    2. B 0
    3. C -4
    4. D -8
  11. 111 mark

    Given that sqrt(48) = a sqrt(3), where a is a positive integer, and that (2 + sqrt(3))^2 = b + c sqrt(3), where b and c are integers, find the value of a + b + c.

    1. A 11
    2. B 15
    3. C 13
    4. D 27
  12. 121 mark

    Given that 3^n = 7, find the value of 9^n - 2(3^(n+1)).

    1. A -35
    2. B 29
    3. C 7
    4. D 26

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. Write every term with base 3: 27^n = (3^3)^n = 3^(3n).
    2. 9^(n+1) x 3 = (3^2)^(n+1) x 3^1 = 3^(2(n+1)) x 3^1 = 3^(2n+2+1) = 3^(2n+3).
    3. Since the bases match, equate the exponents: 3n = 2n + 3, so n = 3.
    4. So n = 3, which is option B.
    • Why not A: Treats 27^n as 3^(n+3), adding the exponent 3 instead of multiplying by it (27 = 3^3 means the exponent triples, not increases by 3), giving n + 3 = 2n + 3 and so n = 0.
    • Why not C: Converts 9^(n+1) to 3^(n+1) instead of 3^(2(n+1)), forgetting that 9 = 3^2 doubles the exponent, giving 3n = n + 2 and so n = 1.
    • Why not D: Ignores the extra factor of 3 on the right-hand side and solves 27^n = 9^(n+1) as if the x 3 were not there, giving 3n = 2n + 2 and so n = 2.

  2. Question 2Answer: D

    1. Rationalise each fraction separately using the difference of two squares. For the first: 1/(sqrt(7)-sqrt(5)) = (sqrt(7)+sqrt(5)) / ((sqrt(7))^2 - (sqrt(5))^2) = (sqrt(7)+sqrt(5))/2.
    2. For the second: 1/(sqrt(7)+sqrt(5)) = (sqrt(7)-sqrt(5)) / ((sqrt(7))^2 - (sqrt(5))^2) = (sqrt(7)-sqrt(5))/2.
    3. Subtract: (sqrt(7)+sqrt(5))/2 - (sqrt(7)-sqrt(5))/2 = [(sqrt(7)+sqrt(5)) - (sqrt(7)-sqrt(5))] / 2 = (2 sqrt(5))/2.
    4. This simplifies to sqrt(5), which is option D.
    • Why not A: Assumes 1/(sqrt(7) - sqrt(5)) and 1/(sqrt(7) + sqrt(5)) are equal, since both denominators involve the same two surds, and so concludes the difference is 0 without rationalising either fraction.
    • Why not B: Rationalises both fractions correctly to get (sqrt(7)+sqrt(5))/2 and (sqrt(7)-sqrt(5))/2, but forgets to divide the final numerator by 2 when combining them, giving 2 sqrt(5) instead of sqrt(5).
    • Why not C: After rationalising, subtracts the two numerators the wrong way round, treating the surds that should cancel (the sqrt(7) terms) as the ones that survive, and the ones that should survive (the sqrt(5) terms) as the ones that cancel.

  3. Question 3Answer: A

    1. First expand two of the brackets: (x + 2)(x - 3) = x^2 - 3x + 2x - 6 = x^2 - x - 6.
    2. Now multiply by the remaining bracket: (x^2 - x - 6)(x + 1) = x^3 + x^2 - x^2 - x - 6x - 6.
    3. The x^2 terms cancel (x^2 - x^2 = 0), and the x terms combine: -x - 6x = -7x.
    4. So the expansion simplifies to x^3 - 7x - 6, which is option A.
    • Why not B: Expands (x^2 - x - 6)(x + 1) but drops the product of -x and 1 (one of the four terms in the second stage of expansion), giving x^3 - 6x - 6 instead of the full x^3 - 7x - 6.
    • Why not C: Makes a sign error on the -x x 1 term in the second stage of expansion, treating it as +x instead of -x, giving x^3 - 5x - 6.
    • Why not D: Makes a sign error in the first stage, expanding (x + 2)(x - 3) as x^2 - x + 6 instead of x^2 - x - 6 (taking 2 x (-3) to be +6), which then carries through the rest of the expansion to give x^3 + 5x + 6.

  4. Question 4Answer: C

    1. For two distinct real roots, the discriminant b^2 - 4ac must be strictly greater than 0. Here a=1, b=(k-2), c=(k+1).
    2. Discriminant = (k-2)^2 - 4(k+1) = k^2 - 4k + 4 - 4k - 4 = k^2 - 8k.
    3. Require k^2 - 8k > 0, that is k(k - 8) > 0. This product is positive when both factors are positive (k > 8) or both are negative (k < 0).
    4. So the range is k < 0 or k > 8, which is option C.
    • Why not A: Finds the correct critical values k = 0 and k = 8 but selects the region between them, as if the condition were for NO real roots (discriminant < 0) rather than two distinct real roots (discriminant > 0).
    • Why not B: Solves k(k - 8) > 0 by keeping only the branch k > 8, forgetting that a product of two negative factors is also positive, so k < 0 satisfies the inequality too.
    • Why not D: Uses discriminant >= 0 instead of the strict inequality > 0 that two DISTINCT roots require, so wrongly includes the boundary values k = 0 and k = 8, which actually give a repeated (not distinct) root.

  5. Question 5Answer: D

    1. Factorise: 2x^2 + x - 6 = (2x - 3)(x + 2). Check: (2x-3)(x+2) = 2x^2 + 4x - 3x - 6 = 2x^2 + x - 6, which matches.
    2. The critical values are where each factor is zero: 2x - 3 = 0 gives x = 3/2, and x + 2 = 0 gives x = -2.
    3. Since the coefficient of x^2 is positive (2), the graph is an upward parabola, so the expression is >= 0 outside the roots, i.e. at or beyond them.
    4. So the solution is x <= -2 or x >= 3/2, which is option D.
    • Why not A: Finds the correct critical values x = -2 and x = 3/2 but takes the region BETWEEN them, which is the solution to 2x^2 + x - 6 <= 0, the opposite inequality to the one asked.
    • Why not B: Factorises correctly as (2x - 3)(x + 2) but then solves 2x - 3 = 0 as x = 3, forgetting to divide by the coefficient 2, so uses the wrong critical value 3 instead of 3/2.
    • Why not C: Makes a sign error factorising, using (2x + 3)(x - 2) instead of (2x - 3)(x + 2) (which expands to 2x^2 - x - 6, not the required 2x^2 + x - 6), giving the wrong-signed critical values -3/2 and 2.

  6. Question 6Answer: B

    1. At a point of tangency, the line meets the curve exactly once, so x^2 - 4x + 7 = 2x + n has a repeated root. Rearranging: x^2 - 6x + (7 - n) = 0.
    2. A repeated root means the discriminant is zero: (-6)^2 - 4(1)(7-n) = 0.
    3. This gives 36 - 28 + 4n = 0, that is 8 + 4n = 0.
    4. Solving gives n = -2, which is option B.
    • Why not A: Uses the condition b^2 - ac = 0 instead of the correct discriminant b^2 - 4ac = 0 (dropping the factor of 4), giving 36 - (7-n) = 0 and so n = -29.
    • Why not C: Makes a sign error rearranging the equations, writing x^2 - 6x + (7+n) = 0 instead of x^2 - 6x + (7-n) = 0, which reverses the sign of n in the final equation and gives n = 2.
    • Why not D: Uses the curve's original coefficient of x, -4, as the value of b in the discriminant, instead of recomputing b = -6 after combining the line and curve equations, giving 16 - 4(7-n) = 0 and so n = 3.

  7. Question 7Answer: A

    1. Divide x^4 + x^3 - 7x^2 - x + 6 by x^2 - 1 using long division. x^4 / x^2 = x^2; multiplying back gives x^4 - x^2, and subtracting leaves x^3 - 6x^2 - x + 6.
    2. x^3 / x^2 = x; multiplying back gives x^3 - x, and subtracting leaves -6x^2 + 0x + 6, i.e. -6x^2 + 6.
    3. -6x^2 / x^2 = -6; multiplying back gives -6x^2 + 6, and subtracting leaves 0, confirming the division is exact.
    4. So the quotient is x^2 + x - 6, giving p = 1 and q = -6, so p + q = -5, which is option A.
    • Why not B: Misreads the divisor as x^2 + 1 instead of x^2 - 1, and stops once a quadratic term is reached without checking that the remainder is zero, arriving at the quotient x^2 + x - 8 and so p + q = 1 + (-8) = -7.
    • Why not C: Correctly finds the quotient x^2 + x - 6 (so p=1, q=-6) but then multiplies p and q instead of adding them, giving p x q = -6.
    • Why not D: Makes a sign error in the final subtraction step, reporting q as +6 instead of -6, giving p + q = 1 + 6 = 7.

  8. Question 8Answer: D

    1. Since the divisor x^2 - 1 has roots x=1 and x=-1, the Remainder Theorem gives f(1) equal to the remainder evaluated at x=1, and f(-1) equal to the remainder evaluated at x=-1.
    2. f(1) = 1 - 3 + 2 + a + b = a + b. The remainder at x=1 is 5(1) - 3 = 2. So a + b = 2.
    3. f(-1) = 1 + 3 + 2 - a + b = 6 - a + b. The remainder at x=-1 is 5(-1) - 3 = -8. So 6 - a + b = -8, that is b - a = -14, i.e. a - b = 14.
    4. So a - b = 14, which is option D.
    • Why not A: When evaluating f(-1), treats (-1)^3 as +1 instead of -1 (an odd-power sign error), so computes -3(-1)^3 as -3 instead of +3. This gives the wrong equation for f(-1) and leads to a - b = 8 instead of 14.
    • Why not B: Drops the 2x^2 term when substituting into f(1) and f(-1) (evaluating only x^4 - 3x^3 + ax + b), which changes both simultaneous equations and gives a - b = 12.
    • Why not C: Sets up the two equations correctly but reports b - a instead of the value asked for, a - b, giving the negative of the correct answer.

  9. Question 9Answer: A

    1. Since |3x-4| >= 0 for every real x, any solution must also make the right-hand side non-negative: x - 2 >= 0, that is x >= 2.
    2. For every x >= 2, 3x - 4 >= 3(2) - 4 = 2 > 0, so on this whole domain the modulus opens the same way: |3x-4| = 3x - 4.
    3. The equation becomes 3x - 4 = x - 2, giving 2x = 2, so x = 1. But x = 1 does not satisfy the requirement x >= 2 found in the first step.
    4. So no value of x can satisfy both the equation and the condition x >= 2 that any solution must meet: the equation has 0 real solutions, which is option A.
    • Why not B: Solves 3x - 4 = x - 2 to get x = 1 and reports it as a solution, without checking whether it satisfies the necessary condition x - 2 >= 0 (since |3x-4| can never be negative); substituting shows x - 2 = -1 there, so it is not actually valid.
    • Why not C: Solves both cases the modulus could split into (3x-4=x-2 giving x=1, and -(3x-4)=x-2 giving x=3/2) and, without checking either against x - 2 >= 0, believes both are valid solutions.
    • Why not D: Mistakes the equation for an inequality such as |3x - 4| <= x - 2, which could hold over a range of x, rather than recognising that an equation between two expressions can have only a finite number of solutions here.

  10. Question 10Answer: C

    1. Factor the leading coefficient out of the first two terms: 3x^2 - 12x + 7 = 3(x^2 - 4x) + 7.
    2. Complete the square inside the bracket: x^2 - 4x = (x - 2)^2 - 4. So 3(x^2-4x) + 7 = 3[(x-2)^2 - 4] + 7 = 3(x-2)^2 - 12 + 7 = 3(x-2)^2 - 5.
    3. Writing (x - 2) as (x + b) gives b = -2. So a = 3, b = -2, c = -5.
    4. a + b + c = 3 + (-2) + (-5) = -4, which is option C.
    • Why not A: Completes the square as 3(x-2)^2 - 4 + 7 instead of 3(x-2)^2 - 12 + 7, forgetting to multiply the subtracted 4 by the factor of 3 that was taken out first. This gives c = 3 instead of -5, and a total of 3 - 2 + 3 = 4.
    • Why not B: Correctly reaches 3(x-2)^2 - 5 but then writes this in the form a(x+b)^2+c with b = +2 instead of b = -2, forgetting to change the sign when converting (x - 2) into (x + b) form. This gives a total of 3 + 2 - 5 = 0.
    • Why not D: Finds b using the coefficient of x directly (-12 / 2 = -6) without first dividing by the leading coefficient 3, instead of the correct b = -12 / (2 x 3) = -2. This gives a total of 3 - 6 - 5 = -8.

  11. Question 11Answer: B

    1. 48 = 16 x 3, and 16 is a perfect square, so sqrt(48) = sqrt(16) x sqrt(3) = 4 sqrt(3). So a = 4.
    2. Expand (2+sqrt(3))^2 = 2^2 + 2(2)(sqrt(3)) + (sqrt(3))^2 = 4 + 4 sqrt(3) + 3.
    3. Collect the rational and irrational parts: 4 + 3 = 7, and the sqrt(3) coefficient is 4. So (2+sqrt(3))^2 = 7 + 4 sqrt(3), giving b=7 and c=4.
    4. a + b + c = 4 + 7 + 4 = 15, which is option B.
    • Why not A: Expands (2+sqrt(3))^2 by squaring each term separately, as if it were (p+q)^2 = p^2+q^2, forgetting the cross term entirely: 4 + 3 = 7, so b=7 but c=0. With the correct a=4, this gives a total of 4+7+0=11.
    • Why not C: Expands (2+sqrt(3))^2 but forgets the factor of 2 in the cross term 2 x 2 x sqrt(3), writing it as just 2 sqrt(3) instead of 4 sqrt(3), giving c=2 instead of c=4. With the correct a=4 and b=7, this gives a total of 4+7+2=13.
    • Why not D: Writes sqrt(48) = sqrt(16 x 3) but then forgets to take the square root of 16, using 16 itself as the coefficient a instead of sqrt(16)=4, giving a=16 instead of 4. With the correct b=7 and c=4, this gives a total of 16+7+4=27.

  12. Question 12Answer: C

    1. 9^n = (3^2)^n = (3^n)^2 = 7^2 = 49, using the given value 3^n = 7.
    2. 3^(n+1) = 3^n x 3^1 = 7 x 3 = 21.
    3. So 9^n - 2(3^(n+1)) = 49 - 2(21) = 49 - 42.
    4. This equals 7, which is option C.
    • Why not A: Treats 9^n as equal to 3^n, forgetting that 9 = 3^2 means 9^n = (3^n)^2, not the same value as 3^n. Using 9^n = 7 (instead of 49) and 3^(n+1)=21 gives 7 - 2(21) = 7 - 42 = -35.
    • Why not B: Treats 3^(n+1) as 3^n + 3^1 = 7 + 3 = 10, adding the powers instead of using 3^(n+1) = 3^n x 3 = 21. With the correct 9^n=49, this gives 49 - 2(10) = 29.
    • Why not D: Correctly finds 9^n = 49 and 3^(n+1) = 21, but then evaluates 49 - 2(21) by subtracting 2 and 21 separately rather than multiplying 2 by 21 first, computing 49 - 2 - 21 = 26 instead of 49 - 42 = 7.

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