2010 Y6 ACS Mathematics HL Mock Paper 1 (Solutions)

Anglo-Chinese School (Independent)/ Mathematics Department / IBDP Mathematics HL / Mock Exam / Paper 1 Solutions Page 1 of 8 2010 Y6 Mathematics HL Mock Paper 1 (Solutions) Qn Solution Marks 1 [6 marks] (a) det 𝐴 = −2 − 6𝑘, det 𝐵 = 7ℎ + 9 Since det 𝐴 = det 𝐵 and det 𝐴𝐵 = 256ℎ 𝐝𝐞𝐭 𝑨𝑩 = 𝐝𝐞𝐭 𝑨 ⋅ 𝐝𝐞𝐭 𝑩 = (det 𝐵)2 = (7ℎ + 9)2, i.e., (7ℎ + 9)2 = 256ℎ. Expand and simplify, we have 49ℎ2 − 130ℎ + 81 = 0. (b) By solving the equation in part (a), (ℎ − 1)(49ℎ − 81) = 0, ℎ = 1 or ℎ = 81/49 (N.A., since ℎ is an integer) Note that det 𝐴 = det 𝐵, so −2 − 6𝑘 = 7ℎ + 9, which implies 𝑘 = −3. REMARK: Don’t waste your time to calculate the matrix 𝑨𝑩. Qn Solution Marks 2 [6 marks] The intersections of the two curves are given by 𝑦 = 2 − 3𝑥 + 𝑥2 = 2 + 𝑥 − 𝑥2, i.e., 2𝑥2 − 4𝑥 = 0. Thus, 𝑥 = 0 or 2. Area = ∫ (2 + 𝑥 − 𝑥2 − 2 + 3𝑥 − 𝑥2)𝑑𝑥 2 0 = ∫ (4𝑥 − 2𝑥2)𝑑𝑥 = 2 0 [2𝑥2 − 2 3 𝑥3] 0 2 = 8 3 REMARK: To find out which curve is higher, substitute x = 1 to see which y-coordinate is greater. Qn Solution Marks 3 [6 marks] (a) ∫ 𝑓(𝑡)𝑑𝑡 𝑎 −𝑎 = 1 ⇒ ∫ 𝜋 4 cos (𝜋𝑡 2 ) 𝑑𝑡 𝑎 −𝑎 = [1 2 sin (𝜋𝑡 2 )] −𝑎 𝑎 = 1 2 sin (𝑎𝜋 2 ) − 1 2 sin (− 𝑎𝜋 2 ) = sin (𝑎𝜋 2 ) = 1 Thus, sin (𝑎𝜋 2 ) = 1 ⇒ 𝑎𝜋 2 = 𝜋 2 Thus, = 1 . (b) 𝐹(𝑥) = ∫ 𝑓(𝑡)𝑑𝑡 𝑥 −1 = ∫ 𝜋 4 cos (𝜋𝑡 2 ) 𝑑𝑡 𝑥 −1 = [1 2 sin (𝜋𝑡 2 )] −1 𝑥 = 1 2 sin (𝑥𝜋 2 ) − 1 2 sin (− 𝜋 2) = 1 2 (sin (𝑥𝜋 2 ) + 1) REMARK: You can also find c.d.f. in the following way: 𝐹(𝑥) = ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝜋 4 cos (𝜋𝑥 2 ) 𝑑𝑥 = 1 2 sin (𝑥𝜋 2 ) + 𝐶, then use 𝐹(−1) = 0 𝑜𝑟 𝐹(1) = 1 to find C. (c) 𝐹(𝑥) = 1 2 (sin (𝑥𝜋 2 ) + 1) So,

Anglo-Chinese School (Independent)/ Mathematics Department / IBDP Mathematics HL / Mock Exam / Paper 1 Solutions Page 2 of 8 𝐹(𝑥) = 1 4 ⇒ 1 2 (sin (𝑥𝜋 2 ) + 1) = 1 4 ⇒ sin (𝑥𝜋 2 ) = − 1 2 ⇒ 𝑥𝜋 2 = − 𝜋 6 ⇒ 𝑥 = − 1 3 𝐹(𝑥) = 3 4 ⇒ 1 2 (sin (𝑥𝜋 2 ) + 1) = 3 4 ⇒ sin (𝑥𝜋 2 ) = 1 2 ⇒ 𝑥𝜋 2 = 𝜋 6 ⇒ 𝑥 = 1 3 Thus, the interquartile range is =    −− 3 1 3 1 2 3. Qn Solution Marks 4 [6 marks] Let 𝑋 be the marks, and 𝑋~𝑁(𝜇, 𝜎2). We are given 𝑃(𝑋 ≥ 80) = 0.1 and 𝑃(𝑋 ≥ 65) = 0.3 Then by inverse normal distribution table (NOT GDC) 80 − 𝜇 𝜎 = 𝑖𝑛𝑣𝑛𝑜𝑟𝑚(0.9) = 1.2816 65 − 𝜇 𝜎 = 𝑖𝑛𝑣𝑛𝑜𝑟𝑚(0.7) = 0.5244 i.e., {𝜇 − 1.2816𝜎 = 80 𝜇 − 0.5244𝜎 = 65 Solve the simultaneous equations (2 s.f.): {𝜎 = −19.81 ≈ −20 𝜇 = 54.61 ≈ 55 REMARK: You also need to know how to use table to find other values, for example invnorm(0.1). Qn Solution Marks 5 [6 marks] • The range of arcsin function is [− 𝜋 2 , 𝜋 2]: 4 > 𝜋, so − 𝜋 2 ≤ 𝜋 − 4 ≤ 𝜋 2, then arcsin(sin 4) 4 = arcsin(sin(𝜋 − 4)) 4 = 𝜋 − 4 4 = 𝜋 4 − 1 • The range of arccos function is [0, 𝜋] : 0 < 3 < 𝜋, then arccos(cos 3) 3 = 3 3 = 1 • The range of arctan function is ]− 𝜋 2 , 𝜋 2[ (open interval): 2 > 𝜋 2, so − 𝜋 2 ≤ 2 − 𝜋 ≤ 𝜋 2, then arctan(tan 2) 2 =