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 soluti