QUESTIONS
SECTION A (20%)
1. Given that y^(y²) = 16, find the value of y.
2. Find the sum to infinity of the series 20 + 4 + 4/5 + ...
3. Is the binary operation defined by a*b = a−b−ab, commutative?
4. Express (−1−i√3)⁵⁵ in the form a+ib.
5. Without using tables or calculator, calculate cot 15°.
SECTION B: ALGEBRA
6a. Given that P = (1 1 1; −1 1 0; 1 0 3), Q = (2 0 −1; 2 1 0; −1 1 1), show that |PQ| = |P||Q|.
6b. Using mathematical induction, show that Σᵣ₌₁ⁿ(2r−1) = n².
6c. Find the value b for which the equation bx² + bx + 1 = 0 to have equal roots.
7a. Find the set of values of x for which 4^(x+3) = 3^(x+4).
7b. The expression px³+qx²−x+r is divisible by (x²−1) and has a remainder 0 when divided by (x+2). Find the values of p, q and r.
8a. Resolve (x−1)/[(x²−4)(x−3)] into partial fractions. Hence, obtain its binomial expansion up to the term in x².
8b. If α and β are the roots of the equation ax²+x+b=0, form the equation whose roots are 1/(α+2) and 1/(β+2) in terms of a and b.
8c. If one of the roots of equation ax²+x+c=0 is trice the other, prove that 16ac=3.
SECTION C: TRIGONOMETRY AND COMPLEX NUMBERS
9a. Find the roots of the equation z³−1=0, where z=x+iy and show that their sum equal zero.
9b. Given that cosnθ+i sinnθ=(cosθ+i sinθ)ⁿ, express cos5θ in terms of cosθ.
10a(i). If z₁=3+i and z₂=2−i, find the value of z₁²−z₂² in the form z=a+ib.
10a(ii). Simplify y²−1/y², given that y=1+i.
10b. Given that tan(A+B)=1 and tanA=x, find the value of tanB.
11a. Using the expressions for cos(A−B), sin(A−B), obtain an expression for cot(A−B). Hence find the value of A−B if cotA=1/5 and cotB=1.
11b. Express 6cosx−8sinx in terms of Rcos(x+α) and hence solve the equation 6cosx−8sinx=5 for 0°≤x≤180°.
ANSWERS
Question 1
y^(y²) = 16
Try y=2: 2^(2²)=2⁴=16 ✓
y = 2
Question 2
Series: 20+4+4/5+...
Common ratio r = 4/20 = 1/5
Sum to infinity S = a/(1−r) = 20/(1−1/5) = 20/(4/5) = 25
Question 3
ab = a−b−ab
ba = b−a−ba = b−a−ab
Since a−b−ab ≠ b−a−ab (unless a=b),
The operation is NOT commutative (ab ≠ ba in general)
Question 4
−1−i√3 in polar form:
r = √(1+3) = 2
θ = arg = π + arctan(√3/1) in third quadrant = π+π/3 = 4π/3 (or equivalently −2π/3)
Using θ = −2π/3 (or 240°):
(−1−i√3)⁵⁵ = 2⁵⁵[cos(55×(−2π/3)) + i sin(55×(−2π/3))]
55×(−2π/3) = −110π/3
Reduce mod 2π: −110π/3 = −36×2π + (−110π/3+72π) = −110π/3+216π/3 = 106π/3
106π/3 mod 2π: 106π/3 − 34π = 106π/3−102π/3 = 4π/3
cos(4π/3) = −1/2; sin(4π/3) = −√3/2
2⁵⁵ = 3.6029×10¹⁶
(−1−i√3)⁵⁵ = 2⁵⁵(−1/2 − i√3/2) = −2⁵⁴ − i·2⁵⁴√3
≈ −1.801×10¹⁶ − 3.121×10¹⁶ i
Question 5
cot15° = cos15°/sin15°
Using cot15° = cot(45°−30°) = (cot45°cot30°+1)/(cot30°−cot45°) = (1×√3+1)/(√3−1) = (√3+1)/(√3−1)
Rationalize: multiply by (√3+1)/(√3+1):
= (√3+1)²/[(√3)²−1²] = (3+2√3+1)/2 = (4+2√3)/2 = 2+√3
cot15° = 2+√3 ≈ 3.732
Question 6
a) Show |PQ|=|P||Q|
P = |1 1 1; −1 1 0; 1 0 3|
|P| = 1(1×3−0×0) − 1(−1×3−0×1) + 1(−1×0−1×1)
= 1(3) − 1(−3) + 1(−1)
= 3+3−1 = 5
Q = |2 0 −1; 2 1 0; −1 1 1|
|Q| = 2(1×1−0×1) − 0(2×1−0×(−1)) + (−1)(2×1−1×(−1))
= 2(1) − 0 + (−1)(2+1)
= 2 − 3 = −1
|P||Q| = 5×(−1) = −5
PQ = P×Q:
Row1: [1×2+1×2+1×(−1), 1×0+1×1+1×1, 1×(−1)+1×0+1×1] = [3, 2, 0]
Row2: [−1×2+1×2+0×(−1), −1×0+1×1+0×1, −1×(−1)+1×0+0×1] = [0, 1, 1]
Row3: [1×2+0×2+3×(−1), 1×0+0×1+3×1, 1×(−1)+0×0+3×1] = [−1, 3, 2]
PQ = |3 2 0; 0 1 1; −1 3 2|
|PQ| = 3(1×2−1×3) − 2(0×2−1×(−1)) + 0(0×3−1×(−1))
= 3(2−3) − 2(0+1) + 0
= 3(−1) − 2(1) + 0
= −3−2 = −5
|PQ| = −5 = |P||Q| = 5×(−1) = −5 ✓ (verified equal)
b) Prove by induction: Σᵣ₌₁ⁿ(2r−1) = n²
Base case (n=1): LHS = 2(1)−1 = 1; RHS = 1² = 1. True.
Inductive step: Assume true for n=k: Σᵣ₌₁ᵏ(2r−1) = k²
For n=k+1: Σᵣ₌₁ᵏ⁺¹(2r−1) = Σᵣ₌₁ᵏ(2r−1) + (2(k+1)−1)
= k² + 2k+2−1 = k²+2k+1 = (k+1)²
Since it holds for n=k+1 whenever true for n=k, and it's true for n=1, by mathematical induction, Σᵣ₌₁ⁿ(2r−1)=n² for all positive integers n. ✓
c) Value of b for equal roots of bx²+bx+1=0
For equal roots, discriminant = 0:
b²−4(b)(1) = 0
b²−4b = 0
b(b−4) = 0
b = 0 or b = 4 (b=0 makes it not quadratic, so b = 4)
Question 7
a) Solve 4^(x+3) = 3^(x+4)
Take logs: (x+3)ln4 = (x+4)ln3
x ln4 + 3ln4 = x ln3 + 4ln3
x(ln4−ln3) = 4ln3−3ln4
x = (4ln3−3ln4)/(ln4−ln3)
ln4=1.3863, ln3=1.0986
Numerator: 4(1.0986)−3(1.3863) = 4.3944−4.1589 = 0.2355
Denominator: 1.3863−1.0986 = 0.2877
x = 0.2355/0.2877 ≈ 0.818
b) px³+qx²−x+r divisible by (x²−1), remainder 0 when divided by (x+2)
Since divisible by (x²−1)=(x−1)(x+1), both x=1 and x=−1 are roots:
f(1) = p+q−1+r = 0 → p+q+r = 1 ... (i)
f(−1) = −p+q+1+r = 0 → −p+q+r = −1 ... (ii)
f(−2) = 0 (remainder 0 when divided by x+2):
−8p+4q+2+r = 0 → −8p+4q+r = −2 ... (iii)
From (i)−(ii): 2p = 2 → p = 1
Substitute p=1 into (i): 1+q+r=1 → q+r=0 → r=−q ... (iv)
Substitute p=1, r=−q into (iii): −8+4q−q = −2 → 3q = 6 → q = 2
From (iv): r = −2
p=1, q=2, r=−2
Question 8
a) Resolve (x−1)/[(x²−4)(x−3)] into partial fractions
(x²−4)(x−3) = (x−2)(x+2)(x−3)
(x−1)/[(x−2)(x+2)(x−3)] = A/(x−2) + B/(x+2) + C/(x−3)
x−1 = A(x+2)(x−3) + B(x−2)(x−3) + C(x−2)(x+2)
At x=2: 1 = A(4)(−1) = −4A → A = −1/4
At x=−2: −3 = B(−4)(−5) = 20B → B = −3/20
At x=3: 2 = C(1)(5) = 5C → C = 2/5
Partial fractions: (x−1)/[(x²−4)(x−3)] = −1/[4(x−2)] − 3/[20(x+2)] + 2/[5(x−3)]
Binomial expansion up to x²:
−1/[4(x−2)] = 1/[4(2−x)] = (1/8)(1−x/2)⁻¹ = (1/8)[1+x/2+x²/4+...] = 1/8+x/16+x²/32+...
−3/[20(x+2)] = −3/[40(1+x/2)] = −(3/40)[1−x/2+x²/4−...] = −3/40+3x/80−3x²/160+...
2/[5(x−3)] = −2/[15(1−x/3)] = −(2/15)[1+x/3+x²/9+...] = −2/15−2x/45−2x²/135+...
Sum constant terms: 1/8−3/40−2/15 = 15/120−9/120−16/120 = −10/120 = −1/12
Sum x coefficients: 1/16+3/80−2/45
Convert to common denom 720: 45/720+27/720−32/720 = 40/720 = 1/18
Sum x² coefficients: 1/32−3/160−2/135
Convert to common denom (approx): 1/32=0.03125, 3/160=0.01875, 2/135=0.01481
0.03125−0.01875−0.01481 = −0.00231 ≈ −1/433 (approximately −0.0023)
f(x) ≈ −1/12 + x/18 − 0.0023x² + ...
b) Roots α,β of ax²+x+b=0; form equation with roots 1/(α+2), 1/(β+2)
α+β = −1/a; αβ = b/a
Sum of new roots = 1/(α+2)+1/(β+2) = [(β+2)+(α+2)]/[(α+2)(β+2)] = (α+β+4)/(αβ+2(α+β)+4)
= (−1/a+4)/(b/a+2(−1/a)+4) = [(4a−1)/a]/[(b−2+4a)/a] = (4a−1)/(4a+b−2)
Product of new roots = 1/[(α+2)(β+2)] = 1/(αβ+2(α+β)+4) = 1/(b/a−2/a+4) = a/(b−2+4a)
New equation: x² − [sum]x + [product] = 0
(4a+b−2)x² − (4a−1)x + a = 0
c) If one root is thrice the other for ax²+x+c=0, prove 16ac=3
Let roots be β and 3β.
Sum: β+3β = 4β = −1/a → β = −1/(4a)
Product: β×3β = 3β² = c/a → β² = c/(3a)
From sum: β² = 1/(16a²)
Equate: 1/(16a²) = c/(3a)
3a = 16a²c
3 = 16ac
16ac = 3 ✓ (proved)
Question 9
a) Find roots of z³−1=0
z³=1 → z = 1^(1/3) = cos(2kπ/3)+i sin(2kπ/3), k=0,1,2
k=0: z₁ = 1
k=1: z₂ = cos(2π/3)+i sin(2π/3) = −1/2+i√3/2
k=2: z₃ = cos(4π/3)+i sin(4π/3) = −1/2−i√3/2
Roots: z=1, z=−1/2+i√3/2, z=−1/2−i√3/2
Sum = 1+(−1/2+i√3/2)+(−1/2−i√3/2) = 1−1/2−1/2 = 0 ✓ (verified)
b) Express cos5θ in terms of cosθ
Using De Moivre's theorem: cos5θ+isin5θ = (cosθ+isinθ)⁵
Expand using binomial theorem:
(c+is)⁵ = c⁵+5c⁴(is)+10c³(is)²+10c²(is)³+5c(is)⁴+(is)⁵
= c⁵+5ic⁴s−10c³s²−10ic²s³+5cs⁴+is⁵
Real part: c⁵−10c³s²+5cs⁴ = cos5θ
Using s²=1−c²:
cos5θ = c⁵−10c³(1−c²)+5c(1−c²)²
= c⁵−10c³+10c⁵+5c(1−2c²+c⁴)
= c⁵−10c³+10c⁵+5c−10c³+5c⁵
= 16c⁵−20c³+5c
cos5θ = 16cos⁵θ − 20cos³θ + 5cosθ
Question 10
a)(i) z₁=3+i, z₂=2−i; find z₁²−z₂²
z₁² = (3+i)² = 9+6i+i² = 9+6i−1 = 8+6i
z₂² = (2−i)² = 4−4i+i² = 4−4i−1 = 3−4i
z₁²−z₂² = (8+6i)−(3−4i) = 5+10i
z₁²−z₂² = 5+10i
(ii) Simplify y²−1/y², given y=1+i
y² = (1+i)² = 1+2i+i² = 1+2i−1 = 2i
1/y² = 1/(2i) = −i/2 (multiplying by i/i: i/(2i²)=i/(−2)=−i/2)
y²−1/y² = 2i−(−i/2) = 2i+i/2 = 5i/2
y²−1/y² = 5i/2 = 2.5i
b) tan(A+B)=1, tanA=x; find tanB
tan(A+B) = (tanA+tanB)/(1−tanA tanB) = 1
(x+tanB) = 1−x tanB
tanB+x tanB = 1−x
tanB(1+x) = 1−x
tanB = (1−x)/(1+x)
Question 11
a) Expression for cot(A−B), and value of A−B
cos(A−B) = cosA cosB + sinA sinB
sin(A−B) = sinA cosB − cosA sinB
cot(A−B) = cos(A−B)/sin(A−B) = (cosA cosB+sinA sinB)/(sinA cosB−cosA sinB)
Divide numerator and denominator by sinA sinB:
cot(A−B) = (cotB cotA·... )
Simplify by dividing num/denom by (sinAsinB):
= (cotA cotB+1)/(cotB−cotA)... let's redo properly:
Divide num and denom by sinA sinB:
Numerator: cosAcosB/sinAsinB + 1 = cotA cotB + 1
Denominator: cosB/sinB − cosA/sinA = cotB − cotA
cot(A−B) = (cotA cotB + 1)/(cotB − cotA)
Given cotA=1/5, cotB=1:
cot(A−B) = [(1/5)(1)+1]/(1−1/5) = (6/5)/(4/5) = 6/4 = 3/2
A−B = cot⁻¹(3/2)
A−B = arctan(2/3) ≈ 33.69°
b) Express 6cosx−8sinx as Rcos(x+α)
Rcos(x+α) = R(cosx cosα − sinx sinα) = Rcosα cosx − Rsinα sinx
Comparing: Rcosα = 6; Rsinα = 8
R = √(6²+8²) = √100 = R=10
tanα = 8/6 = 4/3 → α = arctan(4/3) ≈ 53.13°
6cosx−8sinx = 10cos(x+53.13°)
Solve 6cosx−8sinx=5 for 0°≤x≤180°:
10cos(x+53.13°) = 5
cos(x+53.13°) = 0.5
x+53.13° = 60° or x+53.13° = 300° (or −60°)
x = 60°−53.13° = 6.87°
x = 300°−53.13° = 246.87° (out of range 0–180°)
x = −60°−53.13° = −113.13° (invalid, add 360°: 246.87°, still out of range)
Check other solution: x+53.13°=−60°+360°=300° already considered.
x ≈ 6.87° (only solution within 0°≤x≤180°)