1. [Maximum mark: 3]
It is given that θ is an acute angle in degrees such that sinθ = 2/3.
Find the exact value of sin(θ + 60°).
▶️Answer/Explanation
Step 1: Find cosθ using Pythagorean identity:
√(1 – (2/3)²) = √5/3
Step 2: Apply sine addition formula:
sin(θ + 60°) = sinθcos60° + cosθsin60°
Step 3: Substitute values:
= (2/3)(1/2) + (√5/3)(√3/2) = 1/3 + √15/6
Final Answer: 1/3 + √15/6
2. [Maximum mark: 5]
A curve has equation y = 3tan(½x)cos2x.
Find the gradient of the curve at x = ⅓π.
▶️Answer/Explanation
Step 1: Differentiate using product rule:
dy/dx = 3[½sec²(½x)cos2x + tan(½x)(-2sin2x)]
Step 2: Simplify:
= 3/2 sec²(½x)cos2x – 6tan(½x)sin2x
Step 3: Evaluate at x=⅓π:
= -1 – 3 = -4
Final Answer: Gradient is -4
3. [Maximum mark: 6]
(a) [4 marks] Find ∫410 4/(2x-5) dx, giving answer as ln a.
(b) [2 marks] Find exact value of ∫410 e2x-5 dx.
▶️Answer/Explanation
Part (a):
1. Integral → 2ln|2x-5|
2. Apply limits: 2(ln15-ln3) = 2ln5 = ln25
Part (b):
1. Integral → ½e2x-5
2. Apply limits: ½(e15-e3)
Final Answers:
(a) ln25
(b) ½(e15-e3)
4. [Maximum mark: 7]
(a) [2 marks] Sketch y=|3x-5| and y=2x+7 on same diagram.
(b) [3 marks] Solve |3x-5| = 2x+7.
(c) [2 marks] Solve |3y+1-5| = 2×3y+7.
▶️Answer/Explanation
Part (a):
– V-graph vertex at (5/3,0)
– Line with gradient 2, y-intercept 7
Part (b):
1. Case 1: 3x-5=2x+7 → x=12
2. Case 2: 3x-5=-2x-7 → x=-2/5
Part (c):
1. Let k=3y → solve |3k-5|=2k+7
2. Get k=12 → y=log312 ≈ 2.26
Final Answers:
(b) x=12 or x=-2/5
(c) y=2.26 (3 s.f.)
5. [Maximum mark: 9]
(a) [5 marks] Given p(x)=6x³+ax²+bx-20 has factor (x+2) and remainder -11 when divided by (x+1). Find a and b.
(b) [4 marks] Factorise p(x) and find exact roots of p(3x)=0.
▶️Answer/Explanation
Part (a):
1. p(-2)=0 → 4a-2b=68
2. p(-1)=-11 → a-b=15
3. Solve: a=19, b=4
Part (b):
1. Factorise: (x+2)(2x+5)(3x-2)
2. Solve p(3x)=0 → x=-2/3, -5/6, 2/9
Final Answers:
(a) a=19, b=4
(b) Roots: -2/3, -5/6, 2/9
6. [Maximum mark: 9]
(a) [3 marks] Show cosecθ(3sin2θ+4sin³θ) ≡ 4+6cosθ-4cos²θ.
(b) [3 marks] Solve cosecθ(3sin2θ+4sin³θ)+3=0 for -π<θ<0.
(c) [3 marks] Find ∫cosecθ(3sin2θ+4sin³θ)dθ.
▶️Answer/Explanation
Part (a):
1. Rewrite cosecθ as 1/sinθ
2. Simplify using sin2θ identity
3. Convert sin²θ to cos²θ terms
Part (b):
1. Form quadratic: 4cos²θ-6cosθ-7=0
2. Solve: cosθ=(6-√148)/8 ≈ -0.770
3. θ ≈ -2.45 rad
Part (c):
1. Use identity from (a)
2. Rewrite cos²θ term
3. Integrate: 2θ+6sinθ-sin2θ+C
Final Answers:
(b) θ=-2.45 (3 s.f.)
(c) 2θ+6sinθ-sin2θ+C
7. [Maximum mark: 11]
(a) [4 marks] Curve e2x-18x+y³+y=11 has stationary point (p,q). Find exact p.
(b) [2 marks] Show q=∛(2+18ln3-q).
(c) [2 marks] Show q is between 2.5 and 3.0.
(d) [3 marks] Use an iterative formula, based on the equation in (b), to find the value of q correct to 4 significant
figures. Give the result of each iteration to 6 significant figures.
▶️Answer/Explanation
Part (a):
1. Differentiate implicitly
2. Set dy/dx=0 → e2p=9
3. p=ln3
Part (b):
1. Substitute x=p into original equation
2. Rearrange to show required form
Part (c):
1. Calculate f(2.5)≈-0.18
2. Calculate f(3.0)≈0.34
Part (d):
1. Iteration formula: qn+1=∛(2+18ln3-qn)
2. Converges to q≈2.673
Final Answers:
(a) p=ln3
(d) q=2.673 (4 s.f.)