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(a) x -3 -2 -1 0 1 2 3 4 \(y = 3x^{2} - 5x - 7\) 35 15 1 -7 -9 -5 5 21 (b) (c) \(y + 3x + 2 = 0 \implies y = -3x - 2\) when x = -2, y = 6 - 2 = 4. when x = 3, y = -9 - 2 = -11. (d) \(3x^{2} - 2x - 5 = 0 = 3x^{2} - 5x - 7 = -3x - 2\) The roots of the equation \(3x^{2} - 2x - 5 = 0\) is when the line \(y = -3x - 2\) cuts the equation i.e. x = 1.0 or 1.7.

(a) \(S = \frac{n}{2} [2a + (n - 1)d]\) \(2S = n[2a + (n - 1) d]\) \(\frac{2S}{n} = 2a + (n - 1)d\) \(\frac{2S}{n} - 2a = (n - 1) d\) \(d = \frac{\frac{2S}{n} - 2a}{(n - 1)}\) = \(\frac{2S - 2an}{n(n - 1)}\) (b) (i) Given: circle ABP To prove < AOB = 2 < APB Construction : Join PO produced to Q. Proof : /OA/ = /OB/ (radii) \(\therefore x_{1} = x_{2} \) (base angles of isosceles triangle) \(\therefore < AOQ = x_{1} + x_{2}\) (exterior angle of triangle AOP) \(\therefore < AOQ = 2x_{2} (x_{1} = x_{2})\) Similarly, < AOB = < BOQ - < AOQ. = \(2y_{2} - 2x_{2} = 2(y_{2} - x_{2}) = 2 \times < APB\) \(\therefore < AOB = 2 < APB.\) (ii) \(z = 2x \) (angle subtended at the centre) \(z = 2(126^\circ) = 252^\circ\) \(\therefore x = 360^\circ - 252^\circ = 108^\circ\) \(\therefore y = \frac{1}{2} \times 108^\circ = 54^\circ\) \(x = 108^\circ ; y = 54^\circ ; z = 252^\circ\)

(a) Difference in longitudes = 124^\circ + 56^\circ = 180^\circ Distance PQ = \(\frac{180^\circ}{360^\circ} \times 2\pi r\) where \(r = R \cos \theta\) \(/PQ/ = \frac{180}{360} \times 2 \times 3.142 \times R \cos 68.7^\circ\) \(/PQ/ = 3.142 \times 6400 \times 0.3633\) = \(7305.53 km\) \(\approxeq 7300 km\) (3 sig. figs) (b) Red balls = 4, white = 3 ; green = 5 Total = 4 + 3 + 5 = 12 balls. (i) P(green) = \(\frac{5}{12}\) P(not green) = \(1 - \frac{5}{12} = \frac{7}{12}\) (ii) P(picking red and white) = P( red then white) or P(white then red) = \(\frac{4}{12} \times \frac{3}{11} + \frac{3}{12} \times \frac{4}{11}\) = \(\frac{2}{11}\)

(a) By cosine rule, \(/PQ/^{2} = 20^{2} + 15^{2} - 2 \times 20 \times 15 \times \cos 145^\circ\) = \(400 + 225 - 600 \cos 145^\circ\) = \(625 + 600 \times 0.8192\) = \(491.52 + 625\) \(/PQ/^{2} = 1116.52 \implies /PQ/ = \sqrt{1116.52} = 33.41\) \(\approxeq 33km\) ( to the nearest whole number) (b) By sine rule, \(\frac{\sin < QPR}{15} = \frac{\sin 145}{33.41}\) \(\sin < QPR = \frac{15 \sin 145}{33.41}\) \(\sin < QPR = \frac{15 \times 0.5736}{33.41} = 0.2575\) \(< QPR = 14.92^\circ\) \(\therefore\) Bearing of Q from P = 25^\circ + 14.92^\circ = 39.92^\circ. \(\approxeq 40^\circ\) (to nearest whole number)

The sentence "Imona couldn't have attended lectures yesterday" suggests a situation where it is unlikely or impossible for Imona to have attended lectures on the specified day. The word "couldn't" implies a strong negation or impossibility. Option A, "It was impossible that Imona attended lectures yesterday," is the correct interpretation. It reflects the idea that attending lectures yesterday is not a possible or likely scenario for Imona.

No expanded working has been attached for this question yet. The verified answer is Option C.

M1U = (M1 + M2)V = 4.2 x 10 = (4.2 = 2.8)V ∴ V = 4.2 x 10/7.0 = 6m/s

T - ma = mg T = mg + ma = 2 x 10 + 2 x 2 = 24N

F = ma = MV/t ∴ t = MV/F = 20/50 = 0.40s

Micrometer screw gauge reading = Main scale reading + vernier scale reading = 3 + 0.22 = 3.22mm

No expanded working has been attached for this question yet. The verified answer is Option D.

Distance of A = 1/2 x (10 + 20) x 5 = 1/2 x 30 x 5 = 75m. Distance of B = 20 x 5 = 100m Distance of C = 1/2 x 5 x 20 = 50/225 ∴ Total distance is 225m

Impulse = Ft = mu - (-mu) = mu + mu = 2mu = 2 x 5 x 2 = 20kgms-1

V = √2gR V = √2 x 10 x 6.4 x 102 1.13 x 104ms

Efficiency = M.A/V.R x 100% = 80/100 = 120 x 10/E E = 250N