SOTA 2023 Prelim MAA HL Paper 1 Solutions

1 2023 Prelim MAA HL Paper 1 Student Solutions Section A No Solutions 1(a) 1(b) So a multiple of 4. 2(a)(i) 2(a)(ii) 2(b) 3(a) 3(b) Let b2−4ac=(−m)2−4(2)(m−2)=0

2 4 5(a) P(X = 2) > 0 5(b) Hence, 5(c) 2sin2θ+sinθ−1=0(2sinθ−1)(sinθ+1)=02sinθ−1=0orsinθ+1=0sinθ=12orsinθ=−1(rej)

3 6(i) Recognise that 6(ii) Either Or 7(a) 7(b) Attempt to show reflection about y-axis 7(c) 8(a) Eliminate one variable and obtain 2 equations e.g. Eliminate another variable 8(b) 8(c) Attempt to let one of the variable be the parameter e.g. let v=dsdt3s2dsdt+dsdt−2=0v=dsdt=21+3s2a=dvdt=dvdsdsdtdvds=−12s(1+3s2)2a=−12s(1+3s2)2dsdt6sdsdt⎛⎝⎜⎞⎠⎟2+3s2d2sdt2+d2sdt2=0d2sdt2=−6sdsdt⎛⎝⎜⎞⎠⎟21+3s2a=−24s(1+3s2)3 p=1x∈!,x≠03y+6z=10−bandy+(1−a)z=1a=−1b=7z=t,t∈!y=1−2tx=3+3t

4 9(a) 9(b) Recognise that Amanda can win in her first shot, second, third, e.g. Recognise that it is a geometric series First term is and common ratio is Correct substitution into the sum to infinity formula i.e. 9(c) Use of conditional probability, e.g. 13⎛⎝⎜⎞⎠⎟12⎛⎝⎜⎞⎠⎟23⎛⎝⎜⎞⎠⎟=1923+13⎛⎝⎜⎞⎠⎟12⎛⎝⎜⎞⎠⎟23⎛⎝⎜⎞⎠⎟+13⎛⎝⎜⎞⎠⎟12⎛⎝⎜⎞⎠⎟13⎛⎝⎜⎞⎠⎟12⎛⎝⎜⎞⎠⎟23⎛⎝⎜⎞⎠⎟+!!2316231−16451945536

5 Section B No Solutions 10(a)(i) Amplitude = 0.5 10(a)(ii) Minimum value = 1 10(a)(iii) 10(b)(i) Period = 10(b)(ii) 10(b)(iii) 10(c)(i) 10(c)(ii) 11(a) 11(b) −122⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟×−134⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟=22−1⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟12AB!"!!×AC!"!!=1222+22+(−1)2=32

6 11(c) 11(d) 11(e) (i) Substitute the line into the plane Substitute into the line (ii) Attempt to find the distance between the point of intersection and D e.g. 9 11(f) Valid approach e.g. 11(g) 9 12(a) Attempt to expand Change to polar form 12(b) xyz⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟⋅22−1⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟=31−3⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟⋅22−1⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟2x+2y−z=11r=−3−34⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟+λ22−1⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟,λ∈!λ=3(3,3,1)−3−34⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟−331⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟=(−6)2+(−6)2+32−3+x2,−3+y2,4+z2⎛⎝⎜⎞⎠⎟(9,9,−2)2×13×basearea×heighte0=1z2−z(cosθ−isinθ+cosθ+isinθ)+1z2−2zcosθ+1−1=cisπ

7 12(c) 12(d)(i) Attempt to solve , , 12(d)(ii) z=cisπ+2kπ4⎛⎝⎜⎞⎠⎟,k∈!eiπ4,ei3π4ei(−π4),ei(−3π4)z4+1=(z−eiπ4)(z−ei(−π4))(z−ei3π4)(z−ei(−3π4))(z2−2zcos(π4)+1)(z2−2zcos(3π4)+1)z8=−1eiπ8,ei3π8ei5π8,ei7π8e−iπ8,e−i3π8e−i5π8,e−i7π8z8+1=(z2−2zcos(π8)+1)(z2−2zcos(3π8)+1)(z2−2zcos(5π8)+1)(z2−2zcos(7π8)+1)i8+1=(−2icos(π8))(−2icos(3π8))(−2icos(5π8))(−2icos(7π8))2=16cos(π8)cos(3π8)cos(5π8)cos(7π8)cos(π8)cos(3π8)cos(5π8)cos(7π8)=18