TMUA Paper 2: Reasoning with algebra and number, set 2

• Answer all questions. No calculator is allowed.
• Each question has exactly one correct answer.
• This strand tests reasoning about algebra and number results, not just computing them: check every statement or option against the working, rather than trusting a first impression.

1. 11 mark

   Consider the following four statements, where x and y are positive real numbers.

   (i) x^(-1) = 1/x

   (ii) x^(1/2) always denotes the positive square root of x

   (iii) (xy)^(-1) = x^(-1) + y^(-1)

   (iv) x^0 = 0

   How many of these four statements are true for all positive real x and y?

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

   Consider the statement: 'For every positive integer n, the value of 6n + 1 is a prime number.' Which of the following values of n provides a counterexample to this statement?

   1. A n = 4
   2. B n = 1
   3. C n = 2
   4. D n = 3
3. 31 mark

   For real numbers x, let P be the statement 'x is a root of x^2 - 5x + 6 = 0' and let Q be the statement 'x = 2 or x = 3'. Which of the following is true?

   1. A P is sufficient for Q but not necessary
   2. B P is neither necessary nor sufficient for Q
   3. C P is necessary for Q but not sufficient
   4. D P is necessary and sufficient for Q
4. 41 mark

   Let f(x) = x^2 for all real x, and let g(x) = sqrt(x) for x >= 0. Which of the following statements is true?

   1. A f is a one-to-one mapping and g is a many-to-one mapping
   2. B f is a many-to-one mapping and g is a one-to-one mapping
   3. C Both f and g are one-to-one mappings
   4. D Both f and g are many-to-one mappings
5. 51 mark

   Consider the inequality (x-3)^2 > 0. Which of the following describes the complete solution set for real x?

   1. A x > 3
   2. B All real x
   3. C No real x satisfies this inequality
   4. D x < 3 or x > 3 (that is, every real x except x = 3)
6. 61 mark

   A sequence is defined by a_1 = 2 and a_n = a_(n-1) + 3n for n >= 2. Which of the following is a correct expression for a_n, for every positive integer n?

   1. A a_n = 3n(n+1)/2 - 1
   2. B a_n = 6n - 4
   3. C a_n = 3n(n+1)/2 + 2
   4. D a_n = 3n(n+1)/2 - 3
7. 71 mark

   y is inversely proportional to x^2. When x=2, y=36. What is y when x=4?

   1. A 18
   2. B 144
   3. C 9
   4. D 1/9
8. 81 mark

   A length is measured as 8.4 cm, correct to the nearest 0.1 cm. Which of the following is the correct error interval for the true length L?

   1. A 8.3 <= L < 8.5
   2. B 8.35 <= L < 8.45
   3. C 8.35 < L < 8.45
   4. D 8.4 <= L < 8.45
9. 91 mark

   Let m and n be integers. How many of the following four statements are true for all integers m and n?

   (1) If m and n are both even, then m+n is even.

   (2) If m and n are both odd, then m+n is even.

   (3) If m is even and n is odd, then mn is even.

   (4) If m and n are both odd, then mn is odd.

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

    For which values of the constant k does the equation x^2 + kx + (k+3) = 0 have two distinct real roots?

    1. A -2 < k < 6
    2. B k <= -2 or k >= 6
    3. C k > 6
    4. D k < -2 or k > 6
11. 111 mark

    Consider the simultaneous equations y = x and y = (x-2)^2. How many of the following statements about the solutions to this system are true?

    (1) The system has exactly two distinct solutions.

    (2) One solution has x = 1.

    (3) One solution has y = 2.

    (4) The sum of the x-coordinates of the two solutions is 5.

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

    Let a and b be positive integers. Which of the following statements is true for all such a and b?

    1. A HCF(a,b) + LCM(a,b) = a + b
    2. B HCF(a,b) x LCM(a,b) = a x b
    3. C If HCF(a,b) = 1, then a and b are both prime numbers
    4. D LCM(a,b) <= (a x b)/2 for all positive integers a and b

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. Statement (i): x^(-1) = 1/x is the standard definition of a negative exponent, true for all nonzero x.
   2. Statement (ii): by convention, x^(1/2) always denotes the positive square root of x for x >= 0, never the negative root, so this is true.
   3. Statement (iii): test x=2, y=3. (xy)^(-1) = 6^(-1) = 1/6, but x^(-1) + y^(-1) = 1/2 + 1/3 = 5/6, which is not 1/6, so statement (iii) is false. The correct identity is (xy)^(-1) = x^(-1) * y^(-1), a product, not a sum.
   4. Statement (iv): x^0 = 1 for every nonzero x (for example 5^0 = 1), not 0, so statement (iv) is false.
   5. Exactly two of the four statements, (i) and (ii), are true, so the answer is C.
   • Why not A: Wrongly accepts statement (iii) by confusing the reciprocal-of-a-product rule with the additive law of logarithms, log(xy) = log(x) + log(y), and carries that additive pattern over to reciprocals; also wrongly accepts statement (iv) by extrapolating the pattern x^2, x^1, x^0 as an arithmetic sequence that decreases by x each step (giving x^0 = x^1 - x = 0), rather than a geometric one that divides by x each step. Combined with correctly accepting (i) and (ii), this counts all four as true.
   • Why not B: Wrongly accepts statement (iii) alone, using the same log-addition confusion described above, while correctly rejecting statement (iv); this counts (i), (ii) and (iii) as true, giving 3.
   • Why not D: Wrongly rejects statement (ii), believing sqrt(x) could denote either the positive or the negative root and ignoring the stated convention that x^(1/2) always means the positive root only; this leaves only statement (i) counted as true, giving 1.

2. Question 2Answer: A

   1. Evaluate 6n+1 at n=4: 6(4)+1 = 24+1 = 25.
   2. Recognise 25 = 5^2, which is composite, not prime.
   3. The statement claims 6n+1 is always prime, but n=4 gives a composite number, so n=4 is a counterexample.
   4. Checking the other options confirms they do not work: n=1 gives 7 (prime), n=2 gives 13 (prime) and n=3 gives 19 (prime), so only n=4 breaks the pattern, giving answer A.
   • Why not B: Drops the constant term entirely, computing 6n+1 at n=1 as just 6(1)=6 instead of 6(1)+1=7; 6 is even and clearly composite, wrongly making n=1 look like a counterexample, when 6(1)+1=7 is actually prime.
   • Why not C: Misreads the notation '6n' as '6+n' (confusing ab, meaning a times b, with a+b), computing 6n+1 at n=2 as 6+2+1=9 instead of 6(2)+1=13; 9 = 3^2 is composite, wrongly flagging n=2, when the correct value 13 is prime.
   • Why not D: Misapplies the distributive law, reading '6n+1' as '6(n+1)' as though the +1 sat inside brackets that were never there, computing 6(3+1)=24 instead of 6(3)+1=19; 24 is composite, wrongly flagging n=3, when the correct value 19 is prime.

3. Question 3Answer: D

   1. Factorise x^2 - 5x + 6 by finding two numbers that multiply to 6 and add to -5: these are -2 and -3, so x^2 - 5x + 6 = (x-2)(x-3).
   2. So x^2 - 5x + 6 = 0 exactly when x=2 or x=3: P holds if and only if Q holds.
   3. Since P implies Q (any root of the equation is 2 or 3) and Q implies P (both 2 and 3 do satisfy the equation), P is both necessary and sufficient for Q.
   4. So the answer is D.
   • Why not A: Assumes the quadratic might have a root beyond 2 and 3 that the factorisation missed, treating a quadratic as if it could have more than two roots; a quadratic has at most two roots, and here they are exactly 2 and 3, so Q does guarantee P, making P necessary too, not merely sufficient.
   • Why not B: Factorises x^2 - 5x + 6 by picking a factor pair of 6 without checking that it also sums to 5, choosing 1 and 6 (since 1 x 6 = 6) instead of 2 and 3 (since 2 x 3 = 6 and 2+3=5); this wrongly gives roots x=1 and x=6, which share nothing with Q's values 2 and 3, making P and Q look completely unrelated.
   • Why not C: Correctly finds that x=2 and x=3 satisfy the equation, but wrongly suspects the factorisation could have introduced an error, doubting that every root of the original equation is genuinely captured by 2 and 3, and so doubts sufficiency even though it holds.

4. Question 4Answer: B

   1. f(x) = x^2 is defined for all real x. Take x=2 and x=-2: f(2)=4 and f(-2)=4, so two different inputs give the same output, meaning f is many-to-one.
   2. g(x) = sqrt(x) is defined for x >= 0, and by convention always denotes the positive square root, so it is strictly increasing and never gives two different outputs from two different inputs.
   3. So f is many-to-one and g is one-to-one, which is option B.
   • Why not A: Reverses which function is which: believes squaring is one-to-one because it 'can be undone' by a square root, ignoring that f(-2) = f(2) = 4 shows two different inputs give the same output; and believes g is many-to-one on the grounds that every positive number has two square roots, forgetting that sqrt(x) is defined here to always mean the positive root only.
   • Why not C: Assumes f(x)=x^2 is one-to-one by picturing only the restricted case x >= 0, forgetting the stated domain is all real x; since f(-2) = f(2) = 4, two distinct inputs map to the same output, so f is many-to-one, not one-to-one.
   • Why not D: Believes g is many-to-one on the grounds that every positive number has two square roots, positive and negative, forgetting that sqrt(x) is defined here to always mean the positive root only; since g is then single-valued and strictly increasing on x >= 0, each output comes from exactly one input, making g one-to-one.

5. Question 5Answer: D

   1. (x-3)^2 is a square, so it is always >= 0 for every real x, with equality exactly when x=3.
   2. For every other value of x, (x-3)^2 is strictly positive, since a nonzero real number squared is always positive.
   3. So (x-3)^2 > 0 holds for every real x except x=3, that is, x < 3 or x > 3, which is option D.
   • Why not A: Drops the square and treats (x-3)^2 > 0 as though it were the linear inequality x - 3 > 0, giving only x > 3; this ignores that squaring makes any nonzero value of x-3, whether positive or negative, produce a positive result.
   • Why not B: Treats '(x-3)^2 > 0' loosely as 'not negative', which (x-3)^2 >= 0 always satisfies, and misses that the inequality is strict; at x=3 the expression equals exactly 0, which does not satisfy '> 0', so x=3 must be excluded.
   • Why not C: Confuses the equation (x-3)^2 = 0 having a repeated root (zero discriminant) with the inequality (x-3)^2 > 0 being impossible to satisfy; in fact the expression is positive everywhere except exactly at that repeated root, not nowhere.

6. Question 6Answer: A

   1. Compute initial terms directly: a_1=2, a_2=a_1+3(2)=2+6=8, a_3=a_2+3(3)=8+9=17, a_4=a_3+3(4)=17+12=29.
   2. Telescoping the recurrence from a_1 to a_n: a_n - a_1 = sum from k=2 to n of 3k = 3 * (sum from k=2 to n of k) = 3 * (n(n+1)/2 - 1).
   3. So a_n = 2 + 3n(n+1)/2 - 3 = 3n(n+1)/2 - 1.
   4. Check against the computed terms: n=1 gives 3(1)(2)/2 - 1 = 3 - 1 = 2 (matches a_1); n=4 gives 3(4)(5)/2 - 1 = 30 - 1 = 29 (matches a_4).
   5. So the correct formula is a_n = 3n(n+1)/2 - 1, option A.
   • Why not B: Assumes the sequence is arithmetic using only the first common difference, a_2 - a_1 = 8 - 2 = 6, without checking further terms; the next difference is a_3 - a_2 = 17 - 8 = 9, not 6, so the sequence is not arithmetic and this formula only matches the first couple of terms before failing.
   • Why not C: When summing the increments from k=2 to n, writes the sum of k from 2 to n as the full sum n(n+1)/2 instead of n(n+1)/2 - 1, forgetting to remove the k=1 term that the full sum formula includes; this makes the resulting formula exactly 3 too high for every n.
   • Why not D: Correctly computes the total increase from a_1 to a_n as 3n(n+1)/2 - 3, but then forgets to add back the starting value a_1 = 2, presenting the increase itself as if it were a_n.

7. Question 7Answer: C

   1. y inversely proportional to x^2 means y = k/x^2 for some constant k.
   2. Using x=2, y=36: 36 = k/4, so k = 144.
   3. So y = 144/x^2. At x=4: y = 144/16 = 9.
   4. The answer is C.
   • Why not A: Treats the relationship as simple inverse proportion, y proportional to 1/x, instead of inverse-square; finds a constant using k = y*x = 36*2 = 72, then computes y = 72/4 = 18 at x=4, instead of correctly using k = y*x^2.
   • Why not B: Does not invert the relationship at all, treating y as though directly proportional to x^2; finds a constant using k = y/x^2 = 36/4 = 9, then computes y = 9*(4^2) = 144 at x=4, which increases with x instead of decreasing as an inverse relationship should.
   • Why not D: Correctly computes the constant k = y*x^2 = 36*4 = 144 from the given data, but then applies the relationship upside down when finding the new y, computing x^2/k = 16/144 = 1/9 instead of k/x^2 = 144/16 = 9.

8. Question 8Answer: B

   1. A value rounded to the nearest 0.1 could have been any true value within half of 0.1, that is 0.05, either side of the rounded value.
   2. The rounded value is 8.4, so the true length L satisfies 8.4 - 0.05 <= L < 8.4 + 0.05, that is 8.35 <= L < 8.45.
   3. The lower bound is included, since a true length of exactly 8.35 rounds up to 8.4, while the upper bound is excluded, since a true length of exactly 8.45 would round to 8.5, not 8.4.
   4. So the correct error interval is 8.35 <= L < 8.45, option B.
   • Why not A: Uses the full rounding unit of 0.1 as the half-interval on each side, instead of half of it (0.05); the measurement is correct to the nearest 0.1 cm, so the true value can differ from 8.4 by at most half of 0.1, not the whole 0.1.
   • Why not C: Uses the correct half-interval of 0.05 on each side but makes both inequality signs strict, forgetting the standard convention that the lower bound of an error interval is included: a true length of exactly 8.35 itself rounds to 8.4, so it must be included, not excluded.
   • Why not D: Only adjusts the upper bound down by the half-interval and leaves the lower bound at the rounded value itself, as though the true length could only have been an underestimate of 8.4 and never an overestimate.

9. Question 9Answer: A

   1. Statement (1): if m=2j and n=2k for integers j,k, then m+n=2(j+k), which is even. True.
   2. Statement (2): if m=2j+1 and n=2k+1, then m+n=2j+2k+2=2(j+k+1), which is even (for example, 3+5=8). True.
   3. Statement (3): if m=2j, then mn=2jn for any integer n, which is always even regardless of whether n is odd. True.
   4. Statement (4): if m=2j+1 and n=2k+1, then mn=(2j+1)(2k+1)=4jk+2j+2k+1=2(2jk+j+k)+1, which is odd (for example, 3x5=15). True.
   5. All four statements are true, so the answer is A.
   • Why not B: Wrongly rejects statement (2), confusing the addition rule for two odd numbers with the multiplication rule (odd x odd = odd) and assuming addition should behave the same way; in fact 3+5=8, an even number, so two odd numbers always sum to an even number.
   • Why not C: In addition to the error above, also wrongly rejects statement (3), doubting that a single even factor is enough to guarantee an even product; but m even means m=2k for some integer k, so mn=2kn is always a multiple of 2, regardless of whether n is odd or even.
   • Why not D: In addition to both errors above, also wrongly rejects statement (4), mistakenly believing a product of two odd numbers could sometimes come out even; but an odd number has no factor of 2 to contribute, so a product of two odd numbers can never gain one either, and is always odd.

10. Question 10Answer: D

    1. For x^2+kx+(k+3)=0 to have two distinct real roots, its discriminant must be strictly positive: k^2 - 4(k+3) > 0.
    2. Expand: k^2 - 4k - 12 > 0.
    3. Factorise: (k-6)(k+2) > 0.
    4. This upward-opening parabola in k is positive outside its roots, so the solution is k < -2 or k > 6.
    5. The answer is D.
    • Why not A: Solves the discriminant inequality (k-6)(k+2) > 0 by picking the region BETWEEN the two roots of this factorised expression, as if solving '< 0' instead of '> 0'; the expression (k-6)(k+2) is an upward-opening parabola in k, so it is negative between its roots and positive outside them.
    • Why not B: Correctly identifies the boundary values k=-2 and k=6, but includes them with '<=' and '>='; at these exact values the discriminant equals zero, giving a single repeated root rather than two distinct ones, so the boundary values must be excluded.
    • Why not C: Only keeps the branch k>6 from the factorised inequality (k-6)(k+2)>0, missing that the product of two negative numbers is also positive; when both factors are negative, that is k<-2, the inequality (k-6)(k+2)>0 is also satisfied.

11. Question 11Answer: C

    1. Substitute y=x into y=(x-2)^2: x = (x-2)^2 = x^2-4x+4, which rearranges to x^2-5x+4=0.
    2. Factorise: (x-1)(x-4)=0, so x=1 or x=4, giving solutions (1,1) and (4,4).
    3. Statement (1): the discriminant of x^2-5x+4=0 is 25-16=9>0, confirming two distinct real solutions. True.
    4. Statement (2): x=1 is one of the solutions, the point (1,1). True.
    5. Statement (3): the y-coordinates of the two solutions are 1 and 4; neither is 2, so this is false.
    6. Statement (4): the sum of the x-coordinates is 1+4=5, matching the sum-of-roots formula -b/a=5 for x^2-5x+4=0. True.
    7. Exactly three of the four statements, (1), (2) and (4), are true, so the answer is C.
    • Why not A: Miscalculates y at the x=1 solution using the wrong constant term for the curve, expanding (x-2)^2 as if it were x^2-4x+5 rather than x^2-4x+4; at x=1 this wrongly gives y=1-4+5=2, appearing to confirm statement (3) as well as the three genuinely true statements.
    • Why not B: Uses the wrong sign convention for the sum of the roots of x^2-5x+4=0, computing the sum as +b/a = -5 instead of -b/a = 5; since -5 does not match the claimed value of 5 in statement (4), wrongly rejects it while correctly accepting (1) and (2).
    • Why not D: Makes the sign-convention error above, wrongly rejecting statement (4), and separately checks statement (2) only against the solution (4,4), finding its x-coordinate is 4 rather than 1, and wrongly concludes from this single check that no solution has x=1, missing that the other solution, (1,1), does; only statement (1) survives.

12. Question 12Answer: B

    1. Test option B with a=4, b=6: HCF(4,6)=2 and LCM(4,6)=12. HCF x LCM = 2 x 12 = 24, and a x b = 4 x 6 = 24, which match.
    2. This is not a coincidence: HCF(a,b) x LCM(a,b) = a x b is a standard identity that holds for all positive integers a and b.
    3. Test option A with the same values: HCF+LCM=2+12=14, but a+b=10, so A is false.
    4. Test option C with a=4, b=9: HCF(4,9)=1, but neither 4 nor 9 is prime, so C is false.
    5. Test option D with a=b=1: LCM(1,1)=1, but (a x b)/2 = 1/2, and 1 is not <= 1/2, so D is false already at this simplest case.
    6. Only option B holds for all positive integers a and b, so the answer is B.
    • Why not A: Assumes the multiplicative identity HCF(a,b) x LCM(a,b) = a x b has a similar additive version; testing a=4, b=6 disproves this, since HCF(4,6)=2 and LCM(4,6)=12 give HCF+LCM=14, but a+b=10, and 14 is not equal to 10.
    • Why not C: Confuses HCF(a,b)=1 (a and b share no common factor, that is they are coprime) with a and b each being prime; a=4 and b=9 are coprime, since HCF(4,9)=1, but neither 4 nor 9 is a prime number.
    • Why not D: Assumes an arbitrary upper bound on the LCM relative to the product with no basis; testing a=b=1 disproves this immediately, since LCM(1,1)=1, but (a x b)/2 = 1/2, and 1 is not <= 1/2.