Sec 3 WA3 Revision Time Trial 2023 Broadrick

Secondary 3 Mathematics WA3 Revision Time Trial — Broadrick (2023) Duration: 36 minutes Score: / 24 Topics tested: Graphs of Functions; Further Trigonometry. Section A: Graphs of Functions [10 marks] 1. The variables x and y are connected by the equation y = −x3 + 4x2 − 3. Some corresponding values of x and y are given in the table below. x −2 −1.5 −1 0 1 2 2.5 3 4 y 21 a 2 −3 0 5 6.4 6 −3 (a) Find the value of a, correct to 1 decimal place. [1] Answer a = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (b) Using the grid on the next page, draw the graph of y = −x3 + 4x2 − 3 for −2 ≤ x ≤ 4. [2]

x y 0−2 −1 1 2 3 4 4 8 12 16 20 24 −4

(c) By drawing a tangent, find the gradient of the curve at the point where x = −1. [2] Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (d) The horizontal line y = k touches the curve y = −x3 + 4x2 − 3 at only one point. State one possible positive integer value of k. [1] Answer k = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (e) The equation −2x3 + 8x2 − 3x − 4 = 0 can be solved by drawing a suitable straight line on the grid. (i) Find the equation of this straight line. [1] Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (ii) By drawing the line in (e)(i), find the three solutions to the equation −2x3 + 8x2 − 3x − 4 = 0 . [3] Answer x = . . . . . . . . . . . . . . . . . . . . . . . . . .or . . . . . . . . . . . . . . . . . . . . . . . . . .or . . . . . . . . . . . . . . . . . . . . . . . . . .

Section B: Further Trigonometry [14 marks] 2. Given that 6 sin x = 4, find the two possible values for angle x, where 0◦ ≤ x ≤ 180◦. [2] Answer x = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .◦ or . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .◦ 3. The area of triangle ABC is 96 cm 2. Two of the sides are of length 24 cm and 16 cm respectively. Find the length of the third side. [4] Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cm

4. In trapezium P QRS, P Q = 25 cm, P S = 21 cm, P R = 29 cm and RS = 20 cm. QR is parallel to P S. (a) Show that triangle P RS is a right-angled triangle. [2] Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (b) (i) Write down the value of tan ∠P RS. [1] Answer tan ∠P RS = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(ii) Find angle P RQ. [2] Answer ∠P RQ = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .◦ (iii) Hence, or otherwise, find angle QP R, given that angle P QR is obtuse. [3] Answer ∠QP R = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .◦ — END OF PAPER —

Answer Key Sec 3 WA3 Revision Time Trial — Broadrick (2023) Section A: Graphs of Functions [10 marks] 1(a) At x = −1.5: y = −(−3.375) + 4(2.25) − 3 = 9 .375. ∴ a = 9.4 (1 d.p.). [1] 1(b) All nine points plotted and joined with a smooth curve: falling from (−2, 21) to a local minimum at (0, −3), rising to a local maximum ≈ (2.7, 6.5), then falling to (4, −3). [2] (Marking scheme graph shown, including the line y = 3 2 x − 1 for (e).) 1(c) T angent drawn at x = −1. Exact gradient: dy dx = −3x2 + 8x = −3 − 8 = −11 (accept −8 to −13.5). [2] 1(d) y = k meets the curve only once when k is above the local maximum ( ≈ 6.5): e.g. k = 8 (accept k = 7 or any larger integer within the grid). [1] 1(e)(i) −2x3 + 8x2 − 3x − 4 = 0 ⇒ − x3 + 4x2 − 3 2 x − 2 = 0 ⇒ − x3 + 4x2 − 3 = 3 2 x − 1. So draw y = 3 2 x − 1. [1] 1(e)(ii) Where the line cuts the curve: x ≈ 3.9, 1.1 and −0.5 (accept ±0.1 each). [3]

Section B: Further Trigonometry [14 marks] 2. sin x = 4 6 ⇒ x = 41.8◦ or 180◦ − 41.8◦ = 138.2◦ (1 d.p.). [2] 3. 1 2 (24)(16) sin ∠BAC = 96 ⇒ sin ∠BAC = 1 2 ⇒ ∠BAC = 30 ◦ (acute, from the diagram). By the Cosine Rule: BC 2 = 16 2 + 242 − 2(16)(24) cos 30◦ = 166.89 . . . BC = 12.9 cm (3 s.f.). [4] 4(a) P R2 = 29 2 = 841 and RS2 + P S2 = 20 2 + 212 = 841. Since P R2 = RS2 + P S2, ∠RSP is a right angle and triangle P RS is right-angled (by the Converse of Pythagoras’ Theorem ). [2] 4(b)(i) tan ∠P RS = P S RS = 21 20 . [1] 4(b)(ii) ∠P RS = tan−1 21 20 = 46.397 . . .◦ QR ∥ P S, so ∠QRS = 90 ◦ and ∠P RQ = 90 ◦ − 46.397 . . .◦ = 43.6◦ (1 d.p.). [2] 4(b)(iii) By the Sine Rule in triangle P QR: sin ∠P QR 29 = sin 43.602 . . .◦ 25 sin ∠P QR = 0.8000 . . . ⇒ ∠P QR = 53.1◦ or 126.9◦; given obtuse, ∠P QR = 126.869 . . .◦. ∠QP R = 180 ◦ − 126.869 . . .◦ − 43.602 . . .◦ = 9.5◦ (1 d.p.). [3]