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Algebra -- Diagnostic Tests

Algebra — Diagnostic Tests

Unit Tests

UT-1: Equations and Inequalities

Question:

(a) Solve the following equations: (i) 3(2x1)+4(x+5)=2(5x3)3(2x - 1) + 4(x + 5) = 2(5x - 3) (ii) x+32x15=2x+13\frac{x + 3}{2} - \frac{x - 1}{5} = \frac{2x + 1}{3}

(b) Solve the quadratic equation 2x27x+3=02x^2 - 7x + 3 = 0 by: (i) factorisation, (ii) the quadratic formula. Verify that both methods give the same solutions.

(c) Solve the inequality 3x211x403x^2 - 11x - 4 \leq 0.

(d) Solve the simultaneous equations: 2x+3y=122x + 3y = 12 and 5x2y=115x - 2y = 11.

Solution:

(a)

(i) 3(2x1)+4(x+5)=2(5x3)3(2x - 1) + 4(x + 5) = 2(5x - 3)

6x3+4x+20=10x66x - 3 + 4x + 20 = 10x - 6 10x+17=10x610x + 17 = 10x - 6 17=617 = -6

This is a contradiction, so there is no solution.

(ii) x+32x15=2x+13\frac{x + 3}{2} - \frac{x - 1}{5} = \frac{2x + 1}{3}

LCM of 2, 5, 3 = 30:

15(x+3)6(x1)=10(2x+1)15(x + 3) - 6(x - 1) = 10(2x + 1) 15x+456x+6=20x+1015x + 45 - 6x + 6 = 20x + 10 9x+51=20x+109x + 51 = 20x + 10 41=11x41 = 11x x=4111x = \frac{41}{11}

(b)

(i) Factorising 2x27x+3=02x^2 - 7x + 3 = 0: Looking for two numbers that multiply to 2×3=62 \times 3 = 6 and add to 7-7: 6-6 and 1-1.

2x26xx+3=02x^2 - 6x - x + 3 = 0 2x(x3)1(x3)=02x(x - 3) - 1(x - 3) = 0 (2x1)(x3)=0(2x - 1)(x - 3) = 0

x=12x = \frac{1}{2} or x=3x = 3.

(ii) Using the quadratic formula x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} with a=2a = 2, b=7b = -7, c=3c = 3:

x=7±49244=7±254=7±54x = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm \sqrt{25}}{4} = \frac{7 \pm 5}{4}

x=3x = 3 or x=12x = \frac{1}{2}. Both methods agree.

(c) 3x211x403x^2 - 11x - 4 \leq 0. Factorising: 3x212x+x4=3x(x4)+1(x4)=(3x+1)(x4)03x^2 - 12x + x - 4 = 3x(x - 4) + 1(x - 4) = (3x + 1)(x - 4) \leq 0.

Roots at x=13x = -\frac{1}{3} and x=4x = 4. The quadratic opens upward (a=3>0a = 3 > 0), so it is negative between the roots.

Solution: 13x4-\frac{1}{3} \leq x \leq 4, or x[13,4]x \in \left[-\frac{1}{3}, 4\right].

(d) 2x+3y=122x + 3y = 12 … (1), 5x2y=115x - 2y = 11 … (2)

Multiply (1) by 2: 4x+6y=244x + 6y = 24 … (3) Multiply (2) by 3: 15x6y=3315x - 6y = 33 … (4)

Add (3) and (4): 19x=5719x = 57, so x=3x = 3.

Substitute into (1): 2(3)+3y=122(3) + 3y = 12, 6+3y=126 + 3y = 12, y=2y = 2.

Solution: x=3x = 3, y=2y = 2.


UT-2: Sequences and Series

Question:

(a) An arithmetic sequence has first term a=5a = 5 and common difference d=3d = 3. Find: (i) the 20th term, (ii) the sum of the first 20 terms.

(b) A geometric sequence has first term a=2a = 2 and common ratio r=0.5r = 0.5. Find: (i) the 8th term, (ii) the sum to infinity.

(c) The nnth term of a sequence is given by un=3n2nu_n = 3n^2 - n. Find the first four terms and determine whether this is an arithmetic, geometric, or neither type of sequence.

(d) A ball is dropped from a height of 2m2\,\text{m}. Each time it bounces, it reaches 34\frac{3}{4} of its previous height. Calculate the total vertical distance travelled by the ball before it comes to rest.

Solution:

(a) a=5a = 5, d=3d = 3.

(i) un=a+(n1)du_n = a + (n - 1)d. u20=5+19×3=5+57=62u_{20} = 5 + 19 \times 3 = 5 + 57 = 62.

(ii) Sn=n2(2a+(n1)d)=202(10+57)=10×67=670S_n = \frac{n}{2}(2a + (n - 1)d) = \frac{20}{2}(10 + 57) = 10 \times 67 = 670.

(b) a=2a = 2, r=0.5r = 0.5.

(i) un=arn1u_n = ar^{n-1}. u8=2×(0.5)7=2×1128=2128=164u_8 = 2 \times (0.5)^7 = 2 \times \frac{1}{128} = \frac{2}{128} = \frac{1}{64}.

(ii) Since r<1|r| < 1, the sum to infinity exists: S=a1r=210.5=20.5=4S_\infty = \frac{a}{1 - r} = \frac{2}{1 - 0.5} = \frac{2}{0.5} = 4.

(c) un=3n2nu_n = 3n^2 - n.

u1=31=2u_1 = 3 - 1 = 2 u2=122=10u_2 = 12 - 2 = 10 u3=273=24u_3 = 27 - 3 = 24 u4=484=44u_4 = 48 - 4 = 44

This is neither arithmetic nor geometric. The differences are 102=810 - 2 = 8, 2410=1424 - 10 = 14, 4424=2044 - 24 = 20 (not constant, so not arithmetic). The ratios 10/2=510/2 = 5, 24/10=2.424/10 = 2.4, 44/241.8344/24 \approx 1.83 are not constant either, so not geometric. It is a quadratic sequence.

(d) The ball drops 2m2\,\text{m} (down), bounces to 34×2=1.5m\frac{3}{4} \times 2 = 1.5\,\text{m} (up), drops 1.5m1.5\,\text{m} (down), bounces to 1.5×34=1.125m1.5 \times \frac{3}{4} = 1.125\,\text{m} (up), and so on.

The total distance = initial drop + total of all up-and-down bounces.

Total distance =2+2×(1.5+1.125+0.84375+)= 2 + 2 \times \left(1.5 + 1.125 + 0.84375 + \ldots\right)

The bounce heights form a geometric series: a=1.5a = 1.5, r=34r = \frac{3}{4}.

Sum to infinity of bounce heights: S=1.510.75=1.50.25=6S_\infty = \frac{1.5}{1 - 0.75} = \frac{1.5}{0.25} = 6.

Total distance =2+2×6=14m= 2 + 2 \times 6 = 14\,\text{m}.


UT-3: Algebraic Fractions and Proofs

Question:

(a) Simplify the algebraic fraction x29x2+5x+6÷x+3x+2\frac{x^2 - 9}{x^2 + 5x + 6} \div \frac{x + 3}{x + 2}.

(b) Solve the equation 2x1+3x+2=1\frac{2}{x - 1} + \frac{3}{x + 2} = 1. Identify any values of xx that are excluded from the solution.

(c) Prove that the sum of any three consecutive integers is always a multiple of 3.

(d) Prove that for any even integer nn, n2n^2 is always a multiple of 4.

Solution:

(a) x29x2+5x+6÷x+3x+2=x29x2+5x+6×x+2x+3\frac{x^2 - 9}{x^2 + 5x + 6} \div \frac{x + 3}{x + 2} = \frac{x^2 - 9}{x^2 + 5x + 6} \times \frac{x + 2}{x + 3}

Factorising: x29=(x3)(x+3)x^2 - 9 = (x - 3)(x + 3), x2+5x+6=(x+2)(x+3)x^2 + 5x + 6 = (x + 2)(x + 3).

=(x3)(x+3)(x+2)(x+3)×x+2x+3=(x3)(x+2)(x+3)2= \frac{(x - 3)(x + 3)}{(x + 2)(x + 3)} \times \frac{x + 2}{x + 3} = \frac{(x - 3)(x + 2)}{(x + 3)^2}

(b) 2x1+3x+2=1\frac{2}{x - 1} + \frac{3}{x + 2} = 1

Multiply through by (x1)(x+2)(x - 1)(x + 2):

2(x+2)+3(x1)=(x1)(x+2)2(x + 2) + 3(x - 1) = (x - 1)(x + 2) 2x+4+3x3=x2+x22x + 4 + 3x - 3 = x^2 + x - 2 5x+1=x2+x25x + 1 = x^2 + x - 2 x24x3=0x^2 - 4x - 3 = 0

x=4±16+122=4±282=4±272=2±7x = \frac{4 \pm \sqrt{16 + 12}}{2} = \frac{4 \pm \sqrt{28}}{2} = \frac{4 \pm 2\sqrt{7}}{2} = 2 \pm \sqrt{7}

Excluded values: x1x \neq 1 and x2x \neq -2 (these make denominators zero). Since 2+74.652 + \sqrt{7} \approx 4.65 and 270.652 - \sqrt{7} \approx -0.65, neither is excluded.

Solutions: x=2+7x = 2 + \sqrt{7} and x=27x = 2 - \sqrt{7}.

(c) Let three consecutive integers be nn, n+1n + 1, and n+2n + 2.

Sum =n+(n+1)+(n+2)=3n+3=3(n+1)= n + (n + 1) + (n + 2) = 3n + 3 = 3(n + 1).

Since n+1n + 1 is an integer, 3(n+1)3(n + 1) is a multiple of 3. Therefore, the sum of any three consecutive integers is always a multiple of 3.

(d) Let nn be any even integer. Then n=2kn = 2k for some integer kk.

n2=(2k)2=4k2n^2 = (2k)^2 = 4k^2

Since k2k^2 is an integer, 4k24k^2 is a multiple of 4. Therefore, the square of any even integer is always a multiple of 4.


Integration Tests

IT-1: Applied Sequences and Equations

Question:

(a) A company’s profits increase by £5000\pounds 5000 each year. In the first year, profits are £20000\pounds 20000. In which year will the profits first exceed £100000\pounds 100000?

(b) The half-life of a radioactive substance is 8 days. If a sample initially contains 640mg640\,\text{mg}, calculate the amount remaining after 40 days.

(c) A geometric sequence has first term 3 and common ratio rr. The sum of the first 4 terms is 255. Find the value of rr.

(d) The equation x2+px+q=0x^2 + px + q = 0 has roots α\alpha and β\beta. Express the following in terms of pp and qq: (i) α2+β2\alpha^2 + \beta^2, (ii) 1α+1β\frac{1}{\alpha} + \frac{1}{\beta}.

Solution:

(a) This is an arithmetic sequence with a=20000a = 20000 and d=5000d = 5000. We need un>100000u_n > 100000.

a+(n1)d>100000a + (n - 1)d > 100000 20000+(n1)(5000)>10000020000 + (n - 1)(5000) > 100000 (n1)(5000)>80000(n - 1)(5000) > 80000 n1>16n - 1 > 16 n>17n > 17

Profits first exceed £100000\pounds 100000 in year 18.

(b) After 40 days, the number of half-lives elapsed =40/8=5= 40 / 8 = 5.

Remaining amount =640×(12)5=640×132=20mg= 640 \times \left(\frac{1}{2}\right)^5 = 640 \times \frac{1}{32} = 20\,\text{mg}.

(c) S4=a(r41)r1=255S_4 = \frac{a(r^4 - 1)}{r - 1} = 255. With a=3a = 3:

3(r41)r1=255\frac{3(r^4 - 1)}{r - 1} = 255

Since r41=(r21)(r2+1)=(r1)(r+1)(r2+1)r^4 - 1 = (r^2 - 1)(r^2 + 1) = (r - 1)(r + 1)(r^2 + 1):

3(r1)(r+1)(r2+1)r1=255\frac{3(r - 1)(r + 1)(r^2 + 1)}{r - 1} = 255 3(r+1)(r2+1)=2553(r + 1)(r^2 + 1) = 255 (r+1)(r2+1)=85(r + 1)(r^2 + 1) = 85

Since 85 = 5×175 \times 17, try r+1=5r + 1 = 5: r=4r = 4. Then r2+1=17r^2 + 1 = 17. So 5×17=855 \times 17 = 85. This works.

Therefore r=4r = 4.

(d) By Vieta’s formulas: α+β=p\alpha + \beta = -p and αβ=q\alpha\beta = q.

(i) α2+β2=(α+β)22αβ=(p)22q=p22q\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (-p)^2 - 2q = p^2 - 2q.

(ii) 1α+1β=α+βαβ=pq\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-p}{q}.


IT-2: Advanced Algebra and Problem Solving

Question:

(a) Find the values of kk for which the quadratic equation x2+2kx+k2+8=0x^2 + 2kx + k^2 + 8 = 0 has: (i) two distinct real roots, (ii) one repeated root, (iii) no real roots.

(b) Prove algebraically that the product of two consecutive even numbers is always even.

(c) The first three terms of a geometric sequence are xx, x+4x + 4, and x+12x + 12. Find the value of xx and the common ratio.

(d) A rectangle has length (x+5)cm(x + 5)\,\text{cm} and width (x3)cm(x - 3)\,\text{cm}. The area is 75cm275\,\text{cm}^2. Find xx and hence the perimeter of the rectangle.

Solution:

(a) Discriminant Δ=b24ac=(2k)24(1)(k2+8)=4k24k232=32\Delta = b^2 - 4ac = (2k)^2 - 4(1)(k^2 + 8) = 4k^2 - 4k^2 - 32 = -32.

Since Δ=32<0\Delta = -32 < 0 for all values of kk, the equation always has no real roots regardless of the value of kk.

(i) No value of kk gives two distinct real roots. (ii) No value of kk gives one repeated root. (iii) All values of kk give no real roots.

(b) Let two consecutive even numbers be 2n2n and 2(n+1)2(n + 1), where nn is an integer.

Product =2n×2(n+1)=4n(n+1)= 2n \times 2(n + 1) = 4n(n + 1).

Since nn and n+1n + 1 are consecutive integers, one of them is always even, so n(n+1)n(n + 1) is even. Let n(n+1)=2mn(n + 1) = 2m for some integer mm.

Product =4×2m=8m= 4 \times 2m = 8m, which is a multiple of 2 (even). Therefore, the product of two consecutive even numbers is always even.

(c) For a geometric sequence, u2u1=u3u2\frac{u_2}{u_1} = \frac{u_3}{u_2}:

x+4x=x+12x+4\frac{x + 4}{x} = \frac{x + 12}{x + 4}

(x+4)2=x(x+12)(x + 4)^2 = x(x + 12) x2+8x+16=x2+12xx^2 + 8x + 16 = x^2 + 12x 16=4x16 = 4x x=4x = 4

The terms are: 4,8,164, 8, 16. Common ratio r=8/4=2r = 8/4 = 2.

(d) Area =(x+5)(x3)=75= (x + 5)(x - 3) = 75

x2+2x15=75x^2 + 2x - 15 = 75 x2+2x90=0x^2 + 2x - 90 = 0 (x+10)(x8)=0(x + 10)(x - 8) = 0

x=8x = 8 (rejecting x=10x = -10 since dimensions cannot be negative).

Length =8+5=13cm= 8 + 5 = 13\,\text{cm}. Width =83=5cm= 8 - 3 = 5\,\text{cm}.

Perimeter =2(13+5)=2(18)=36cm= 2(13 + 5) = 2(18) = 36\,\text{cm}.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Common Pitfalls

  • Forgetting to check excluded values when solving equations with algebraic fractions (values that make denominators zero).
  • Sign errors when expanding brackets with negative terms, particularly double negatives.
  • Confusing the formula for arithmetic series sum (Sn=n2[2a+(n1)d]S_n = \frac{n}{2}[2a + (n-1)d]) with the geometric series sum.
  • Applying the sum to infinity formula when r1|r| \geq 1 — the sum to infinity only exists when r<1|r| < 1.
  • In proof questions, failing to define the variable (e.g., “let nn be an integer”) at the start of the proof.