AISS Prelim AMath Paper 1 Marking Scheme

3 AISS PRELIM/4E/4049/01/2025 [Turn over 1 It is given that P, Q and R are the angles of a triangle. (a) Show that cos P = – cos (Q + R). [2] (b) Given that Q = 45 and R = 60, find cos P in the form 1 4 ( a – b ), where a and b are integers. [3] cos P = cos [180   (Q + R)] [M1 – for replacing] = cos 180  cos (Q + R) + sin 180 sin (Q + R) = –cos ( Q + R) + 0 [A1 – apply addition formula & = –cos (Q + R) [evaluate to arrive at result] cos P = –cos (45  + 60) = –cos 45  cos 60 + sin 45 sin 60 [M1 – correct use of formula expansion] = – 2 2 ( 1 2 ) + 2 2 ( 3 2 ) [M1 – correct special angles trigo ratios] = 1 4 ( 6 – 2 ) [A1]

4 AISS PRELIM/4E/4049/01/2025 2 Baking powder is poured onto a flat surface at a constant rate of 312 c m s  and formed a right circular cone. The radius of the cone is always 1 18 of its height. Find the rate of change of the radius of the cone after 3 seconds of pouring. [5]   2 2 3 11 2 1 1 3 1 183 6 d 18 [B1d 6 6 [M1 - finding c a orresponding ] Vol. of cone, After 3 sec 1[ A 1 ] dd d [M1 - T o connect o e 2 nd d s, 6 rate of change]dd d d d8 d 11 he rate f h 9 c 1 d rr r r r V r r r VV r tr t r t V t r r r V r r                    1 [A1 o.e.]nge required is = cm9 /s.

5 AISS PRELIM/4E/4049/01/2025 [Turn over 3 (a) Determine the set of values of m for which the equation 26242 2  mxmxx has real roots. [4] (b) Hence state what can be deduced about the curve 2)1(2  xy and the line .26  xy Justify your statement. [2] 22( 4 6 ) 2 2 0 2 40 2(4 6 ) 4(2)(2 2) 0 216 48 36 8(2 2) 0 236 64 0 (9 16) 0 160 9 xm x m ba c mm mm m mm mm mo r m                [M1 - correct Discriminant] [M1 – simplification in factors] [A2, minus 1 mark if inequality sign is wrong due to earlier wrong D sign] 224 2 6 2xx x  By comparing with (a), m = 1 [B1 – correct m value] When m = 1, it is not within the set of values of m for which there will be real roots, hence the curve will not meet the line/ the curve will not cut the line. [B1]

6 AISS PRELIM/4E/4049/01/2025 4 (a) Show that d ln cos tand x xx  . [2] (b) Differentiate tanx x with respect to x. [2] (c) Using the results from part (a) and (b), find 2sec dx xx and hence show that 24 0 1sec d ln 2 42xx x   . [4]  d1 dln cos cosdc o s d sin = tancos x xxx x x xx    [M1 – show working] [A1 – show fraction] 2d tan sec tand x xx x xx  [M1 – show product rule] [A1 – correct ans for both]   2From (b), sec tan d tanx xx x x x C  2sec d tan d tanx xx xx x xC   2sec d tan tan d = tan ln cos x xx x x xxC x xx C      2 44 0 0 1 2 sec d tan ln cos tan ln cos 0 ln144 4 1ln4 2 ln 24 1 ln 242 xx x xx x                   [M1 – use part (b), ‘C’ must be seen] [M1 – proper integration and final correct ans, ‘C’ must be seen] [minus 1 mark if “C” is not seen] [M1 – proper evaluation] [A1]

7 AISS PRELIM/4E/4049/01/2025 [Turn over 5 (a) In the expansion of 2 n x , where n is a positive integer, the coefficient of x2 is twice the coefficient of x. Find the value of n. [3] (b) Find the value of the term that is independent of x in the expansion of 15 4 12 4x x   . [4] [M1 – correct coeff] [M1 – simplify] [A1 – with rejection] Max 2 marks if able to gather 15 – 5r for exponent of ‘x’ and equate to zero with correct r value. [M1] [M1 - gather x and let power = 0] [M1 for r value] [A1]     211 22 2 22 02 1 12(4) 18 9 nnnn nn n n n        

8 AISS PRELIM/4E/4049/01/2025 6 The diagram shows a circle passing through the points P, Q and R. The point Q lies on the line RB. AB is a tangent to the circle at P. The points S and T lie on PR and PQ respectively. Given that AB is parallel to ST, prove that (a) triangle PST is similar to triangle PQR, [3] (b) PQ × PT = PR × PS, [2] SPT = QPR (common angle) [M1] PST = SPA (alt. s, AB // ST) OR PTS = TPB (alt. s, AB // ST) = PQR (Alt. Segment Thm) = PRQ (Alt. Segment Thm) [M1] triangle PST is similar to triangle PQR. [A1] (2 pairs of corresponding angles are equal) From above result, PQ PS = PR PT [M1]  PQ  PT = PR  PS [A1]

9 AISS PRELIM/4E/4049/01/2025 [Turn over (c) Determine if STQR is a cyclic quadrilateral. [4] QTS = 180  PTS (adj. s on a st. line) = 180  QRS (from (a) result) [M1] So QTS + QRS = 180 [M1] RST + RQT = 360  (QTS + QRS) (  sum of a quadrilateral) = 360   180 = 180 [M1] By converse of angles in opposite segment, STQR is a cyclic quadrilateral and all four vertices lie on the circumference of a circle. [A1 – with correct reason, accept even if no mention of four vertices]

10 AISS PRELIM/4E/4049/01/2025 7 The diagram shows an isosceles triangle ABC in which  3, 0A  , 1 ,32B  and 5, 2C . M is the foot of perpendicular from B to AC. (a) Find the coordinates of M. [1] (b) Find the equation of the perpendicular bisector of AC. [2] y O M x M = 35 02,22    = (1, 1) [B1] Gradient of AC = 20 1 5( 3 ) 4   Gradient of perpendicular BM = – 4 [M1 – use 1 mm   ] o r 31 41 12 BMm    Equation of perpendicular bisector: 14 1 45 yx yx      [ A 1 ]

11 AISS PRELIM/4E/4049/01/2025 [Turn over (c) Given that ABCD is a kite with 2 7BM BD , find the coordinates of D. [3] (d) Find the area of the kite ABCD. [2] Correct ratio: 2 : 5 11:124 M1 Coordinates of D = 11, 1 54  M1 = 12, 44   A1 Area of ABCD = 1132 5 31 422 04 2 3 0   M1 (ecf 1 – correct method) = 1 31.5 282  (anti-clockwise) = 29.75 units 2 (or 119 4 ) A1 (accept 29.7 with 3 s.f. if using other method such as Pythagoras’ hf i d h h i h f h i l ) Alternative: using vectors  2 7 0.5 2 [M1 - position vectors]2 7 0.5 0.5 2.253.5 [M1]23 4 BM BD OD OB OD                       D= 12, 44   [A1]

12 AISS PRELIM/4E/4049/01/2025 8 A prototype consists of a cyli ndrical container of height h cm and radius r cm inscribed in a hollow sphere with centre O. The sphere has a surface area of 6400  cm 2 and both the sphere and container have negligible thickness. (a) Show that the volume of the cylinder container, V cm 3, is given by 222 1600Vr r  . [3] 246 4 0 0 40 (radius R of sphere) R R    [M1 – find radius R] By Pythagoras’ Theorem, 2 22 402 hr  [ M 1 ]  22 4 1600hr  241 6 0 0hr 22 1600hr 2 22 2 1600 Vr h rr    