ASRJC Prelims 2025 H2 Math P2 Solutions

Paper 2 Solution Section A: Pure Mathematics [40 marks] 1 Relative to the origin O , the position vectors of two points A and B are a and b respectively. The length of a is k units and b is a unit vector. The angle between a and b is 6 π radians. It is g iven that the point M lies on the line segment AB such that AM :AB is 3:4, find exact area of OAM∆ , giving your answer in terms of k. [4] Solution 3 4OM += ba  Area of 1 2OAM OA OM∆=×     13 24 += ×  baa 1 38= × +×a baa 3 8= ×ab Since ×=aa 0 3 sin86 π= ab n where is a unit vector 31 82k =  3 16 k= units2

2 (a) Without the use of a graphing calculator, solve the inequality 42 12 xx x ++ ≥+ . [3] (b) Hence solve 3 e21 e1 2 x x ++≥ + . [2] Solution (a) 42 12 xx x ++ ≥+ 24 021 xx x ++ −≤ + ( 2)( 1) 2( 4) 02( 1) xx x x + +− + ≤+ ( ) 2 6 021 xx x +− ≤+ ( )( ) ( ) 23 01 xx x −+ ≤+ Multiplying both sides by 2( 1)x + : ( )( )( )2 3 10xxx− + +≤ , 1x ≠− 3x ≤− or 12 x−< ≤ (b) 3 e21 e1 2 x x ++≥ + e4 e2 e1 2 xx x ++ ≥+ Replace x with ex e3x ≤− (NA) or 1e 2 x−< ≤ (No solution e0x > ) 0e 2 x<≤ ln 2x ≤ –3 –1 2 y = 2 y = –1 ln 2

3 (a) Given that y = 1 1tan 5 x− , show that ( ) 2 2 2 dd25 2 0 dd yyxx xx+ += . [2] (b) By further differentiation of this result, find the Maclaurin’s series of y up to and including the term in x3. [3] [Solution] Method 1 Differentiating wrt x: 2 d 11 d5 11 5 y x x = +   21 d11 25 d 5 yx x  += Method 2 y = 1 1tan 5 x− ⇒ tan y = 1 5 x Differentiating wrt x: sec2y d d y x = 1 5 ( ) 2 d11 tan d5 yy x+= 21 d11 25 d 5 yx x  += ( ) 2 d25 5 d yx x+= (Shown) Differentiating wrt x: ( ) 2 2 2 dd25 2 0 dd yyxx xx+ += Differentiating wrt x: ( ) ( ) 3 22 2 3 22 d d dd25 2 2 2 0d d dd y y yyx xx x x xx+ + + += When x = 0, y = 0, d1 d5 y x = , 2 2 d 0d y x = , 2 2 d2 d 125 y x = − Hence the Maclaurin expansion required is y = 0 + 3 2 1 1250 ...5 3!xx − ++ + = 311 ...5 375xx−+

4 The curve C is defined by the parametric equations x = cos t + 1 2 cos 5t, y = sin t + 1 2 sin 5t for 0 ≤ t ≤ π 2 . (a) Find the coordinates of the point on the curve C corresponding to t = 0. [1] Another curve D has the following parametric equations x = 2 + cosk θ , y = 3 sin2 θ for 0 ≤ θ ≤ 2π and k > 0. (b) Find the cartesian equation for curve D. [2] (c) Sketch the curves C and D on the same diagram. [2] (d) Hence determine the range of values of k such that the equation ( ) ( ) 22 cos 0.5cos5 2 4 sin 0.5sin 5 19 t t tt k +− + += has no real solutions. [1] [Solution] (a) When t = 0, x = 1 + 1 2 = 3 2 , y = 0. Hence the coordinates of the points on the curve corresponding to t = 0 are 3 ,02   (b) x = 2 + cosk θ ⇒ cos θ = ( )1 2x k − y = 3 sin2 θ ⇒ sin θ = 2 3 y Since cos2θ + sin2θ = 1, ( ) 2 2 12 21 3xy k  −+ =   ( ) 2 22 4 19 x y k − +=

(c) (d) For ( ) ( ) 22 cos 0.5cos5 2 4 sin 0.5sin 5 19 t t tt k +− + += to have no solutions, there must not be any intersections b/w the 2 curves. Hence 2 k− > 3 2 ⇒ k < 1 2 Since k is positive, we have 0 < k < 1 4 2 – √k y x C 0 3/2 3/2 D 2 + √k – 3/2 2

5 (a) Solve the following integral. (i) 2 12 d 32 x x xx − +− ⌠ ⌡ . [2] (ii) ( ) 2ln 2 dx xx −∫ , where 22 x− << . [3] Solution (a)(i) 2 12 d 32 x x xx − = +− ⌠ ⌡ ( ) 22 2 22 1 d 32 21 x x xx x − − +− −− ⌠ ⌡ 21 12 3 2 sin 2 xxx c− −= +−− +  (a)(ii) ( ) 2ln 2 dx xx −=∫ ( ) 2 2 2 2 2ln 2 d2 22 x xx xx x  −−−   − ⌠ ⌡ ( ) 23 2 2ln 2 d22 xx xx x= −+ − ⌠⌡ ( ) 2 2ln 22 x x= −+ 2 2 d2 xxx x−+ − ⌠⌡ ( ) 2 2 2 2ln 2 d22 xx xx x x −= − + −− − ⌠⌡ ( ) ( ) 22 22ln 2 ln 222 xx x xc= −− − −+ OR ( ) 22 21 ln 222 xx xc − − −+ OR ( )( ) ( ) 2 22ln 2 1 ln 22 x x xc−− − −+ For integration by parts: ( ) 2 2 2 dln 2 d d2 d2 2 uux x x ux x vxx = −= −= =− Integration by parts: d du v uv v u= −∫∫ For long division: ( ) 23 3 2 2 2 x xx x x x − + − − − Reverse Chain Rule: f( ) d ln f ( )f( ) x x xcx ′ = +⌠⌡ Reverse Chain Rule: [ ] [ ] 1 f( )f ()f () d 1 n n xxx x c n + ′ = + +∫ Integrating 1 Quadratic : #1 Complete the square #2 Choose correct formula from MF 27: 1 22 1 d sin xxc ax a − = + −   ⌠ ⌡

(b) A function f is defined by f( ) e 2 xx = − . (i) Sketch the graph of f( )yx= , indicating clearly the equation(s) of asymptote(s), if any. [2] (ii) Hence find ln 3 0 f( ) dxx∫ , giving your answer in exact form. [3] Solution (b)(i) For e 2xy = − : (b)(ii) ln 3 0 e 2dx x−=∫ ln 2 ln 3 0 ln 2 e 2d e 2dxx xx−+ + −∫∫ ln 2 ln 3 0 ln 2 e2 e2xx xx  = −+ + −   (2 ln 2 2) (0 1) (3 2 ln 3) (2 2 ln 2)= −−− +− −− 4 ln 2 2 ln 3= − Alternative answers: 16ln 9 ; 42 ln3 y x y = 2 (0, 1) (ln 2, 0) When 0y = , e 20x −= ln 2x = When 0x = , 0e2 1y = −= − y x y = 2 (0, 1) (ln 2, 0) ln 3 y = ex – 2 y = –ex + 2 From (i), ( ) e 2, ln 2f e 2, ln 2 x x xyx x  −>= = −+ ≤

6 The points P and Q are represented by the complex numbers p and q respectively, where 2 2ip = + , 2arg( ) 3q π= and 2q = . (a) Find p and arg( )p in exact form. [2] (b) Sketch the points P and Q on an Argand diagram. [2] (c) Use the Argand diagram to deduce Re( q) and Im(q), giving your answers in exact form. [2] (d) The point R is represented by the complex number pq+ . What can you deduce about the shape of quadrilateral OPRQ, where O is the origin? [1] (e) By considering arg( )pq+ or otherwise, show that 11t a n 632 224 π =+++ . [3] [Solution] (a) ( ) ( ) 22 || 2 2 2p = += 1 2arg( ) tan 42 p π− = =  (b) Diagram:

(c) Method 1 Let F be the foot of perpendicular from Q to the y-axis Re(q) = –QF = – 2| | sin 3q π = –1 Im(q) = OF = 2| | cos 3q π = 3 Method 2 222 cos i sin33q ππ= + 132 i 22 = −+ 1 3i= −+ Re(q) = –1 Im(q) = 3 (d) Quadri lateral OPRQ is a rhombus. (e) First, 12arg( ) 24 3pq ππ+= +  11 24 π= Also, 2 2i 1 3ipq+ = + −+ ( ) ( )2 1 3 2i= −+ + Therefore, 11 3 2tan 24 21 π + = − 3 2 21 21 21 ++= × −+ 632 2=+++

Section B: Probability and Statistics [60 marks] 7 Anthony plays a game using a four-sided fair die with the numbers 1, 2, 3 and 4 printed on its sides. Anthony rolls the die and should the outcome be a prime number, he wins an amount equivalent to that number in dollars. Conversely, if the outcome is a non- prime number, Anthony will lose an amount equivalent to that number in dollars. T he random variable X represents the amount of money Anthony wins in a roll of the die. (a) Tabulate the probability distribution of X and fine the value of E(X). [2] (b) Find the value of the variance of X. [1] (c) The random variables X1 and X2 are two independent observations of X. By drawing a table of outcomes for 12XX+ or otherwise, find E( )12XX+ . [2] Assume now that a biased six-sided die with numbers 2, 3, 4, 5, 6 and 7 printed on its sides, is used instead for the same game. It is known that for this biased die, the probability of obtaining a “3” is p, the probability of obtaining a “5” is 2p and the remaining four numbers each have an equal probability of occurring. (d) Find the exact value of p for the game to be fair using the biased die. [3] [Solution] (a) x –4 –1 2 3 P(X = x) 1 4 1