Sec 3 WA3 Revision Time Trial 2023 Bartley

Secondary 3 Mathematics WA3 Revision Time Trial — Bartley (2023) Duration: 47 minutes Score: / 31 Topics tested: Graphs of Functions; Graphs in Practical Situations; Further Trigonometry; Applications of Trigonometry. Section A: Graphs of Functions [10 marks] 1. (a) Complete the table of values for y = −x3 + 3x − 3. [2] x −3 −2 −1 0 1 2 3 y −1 −5 −3 −5 −21 (b) On the grid on the next page, draw the graph of y = −x3 + 3x − 3 for −3 ≤ x ≤ 3. [3]

x y O−4 −3 −2 −1 1 2 3 5 10 15 20 25 −5 −10 −15 −20 −25 −30

(c) By drawing a tangent, find the gradient of the curve at x = −1.5. [2] Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (d) By drawing suitable lines on your graph, find the value(s) of (i) x when x3 = 3x − 3, [1] Answer x = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (ii) k for which −x3 + 3x − 3 = k has exactly two solutions. [2] Answer k =. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .or . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Section B: Graphs in Practical Situations [4 marks] 2. The diagram below shows a speed-time graph of a cyclist’s journey. (a) Find the speed of the cyclist at t = 8 seconds. [1] Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . m/s (b) Find the deceleration for the last 30 seconds of his journey. [1] Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . m/s2 (c) Area under the speed-time graph represents the distance travelled. Find the average speed of the cyclist’s whole journey. [2] Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . m/s

Section C: Further Trigonometry [12 marks] 3. In the diagram, AD is a straight line, AB = CD = 17 cm, and BE = DE = 8 cm. Leaving your answers in fractions, find (a) cos ∠ABC , [1] Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (b) sin ∠ABC . [2] Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4. In triangle ABC , angle ABC is 15◦, AB is 20 cm, and AC is 10 cm. (a) Find angle ACB . [2] Answer Angle ACB = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .◦ (b) Find angle BAC . [1] Answer Angle BAC = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .◦ (c) Find the area of triangle ABC . [2] Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cm2

(d) Find the shortest distance from C to AB. [2] Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cm (e) The point C is now moved to point D, such that BD is 25 cm. Show that angle BAD is approximately 108.2◦. [2] Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Section D: Applications of Trigonometry [5 marks] 5. Amir’s apartment is 75 m above ground. Betty’s apartment is in a neighbouring block that is 200 m away from Amir’s block. [Assumption: the thickness of the wall is negligible.] (a) Calculate the angle of depression from Amir’s apartment to the base of Betty’s block. [2] Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .◦ (b) The angle of elevation from Amir to Betty’s apartment is 10◦. Calculate the height of Betty’s apartment above the ground. [3] Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . m — END OF PAPER —

Answer Key Sec 3 WA3 Revision Time Trial — Bartley (2023) Section A: Graphs of Functions [10 marks] 1(a) At x = −3: y = −(−27) + (−9) − 3 = 15. At x = 1: y = −1 + 3 − 3 = −1. [2] 1(b) All seven points plotted and joined with a smooth cubic curve: falling from (−3, 15) through a local minimum at (−1, −5), rising to a local maximum at (1, −1), then falling to (3, −21). [3] 1(c) T angent drawn at x = −1.5. Exact gradient: dy dx = −3x2 + 3 = −3(2.25) + 3 = −3.75 (accept −4.5 to −2.5). [2] 1(d)(i) x3 = 3x − 3 ⇔ − x3 + 3x − 3 = 0 , i.e. where the curve cuts the x-axis (y = 0). From the graph: x ≈ −2.1 (accept −2.3 to −2.05). [1] 1(d)(ii) y = k (a horizontal line) meets the curve exactly twice when it passes through a turning point: local maximum (1, −1) or local minimum (−1, −5). ∴ k = −1 or k = −5. [2] Section B: Graphs in Practical Situations [4 marks] 2(a) The cyclist accelerates uniformly from 0 to 30 m/s over the first 10 s: v 8 = 30 10 ⇒ v = 24 m/s. [1] 2(b) Deceleration = 30 − 0 30 = 1 m/s2. [1] 2(c) Distance = area under graph = 1 2 (60 + 20)(30) = 1200 m. Average speed = 1200 60 = 20 m/s. [2]

Section C: Further Trigonometry [12 marks] 3(a) In right-angled triangle ABE : cos ∠ABE = BE AB = 8 17 . ∠ABC = 180 ◦ − ∠ABE (adjacent ∠s on straight line EBC ) ∴ cos ∠ABC = − cos ∠ABE = − 8 17 . [1] 3(b) AE = √ 172 − 82 = 15 cm (Pythagoras). sin ∠ABC = sin ∠ABE = AE AB = 15 17 . [2] 4(a) By the Sine Rule: sin ∠ACB 20 = sin 15◦ 10 sin ∠ACB = 2 sin 15◦ = 0.5176 . . . ⇒ ∠ACB = 31.2◦ (rejected) or 148.8◦ (from the diagram, ∠ACB is obtuse). [2] 4(b) ∠BAC = 180 ◦ − 15◦ − 148.8◦ = 16.2◦. [1] 4(c) Area = 1 2 (20)(10) sin 16.2◦ = 27.86 . . . = 27.9 cm2 (3 s.f.). [2] 4(d) T akingAB as base: 1 2 (20)h = 27.86 . . . h = 2.79 cm (3 s.f.). [2] 4(e) By the Cosine Rule in triangle ABD: cos ∠BAD = 202 + 102 − 252 2(20)(10) = −125 400 = −0.3125 ∠BAD = 108.21 . . .◦ ≈ 108.2◦ (shown). [2] Section D: Applications of Trigonometry [5 marks] 5(a) Angle of depression = angle of elevation of Amir from the base (alternate ∠s): tan θ = 75 200 ⇒ θ = 20.556 . . . = 20.6◦ (1 d.p.). [2] 5(b) Height of Betty’s apartment above Amir’s level: h = 200 tan 10◦ = 35.265 . . . m. Height above ground = 75 + 35 .265 . . . = 110.26 . . . = 110 m (3 s.f.). [3]