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 = arctan(tan(2 − 𝜋)) 2 = 2 − 𝜋 2 = 1 − 𝜋 2 • The range of arccot function is ]0, 𝜋[ (open interval): 0 < 1 < 𝜋, then arccot(cot 1) 1 = 1 1 = 1 Thus, arcsin(sin 4) 4 + arccos(cos 3) 3 + arctan(tan 2) 2 + arccot(cot 1) 1 = 2 − 𝜋 4. 4 is out of the range ]2,2[ − 4 is in the 3rd quadrant so )4sin()4sin(4sin −=−−=  3 is within the range ],0[  2 is out of the range    − 2,2  2 is in the 2nd quadrant so ( ) −=−−= 2tan)2tan(2tan 1 is within the range  ,0 REMARK: Graph of inverse trigo functions:

Anglo-Chinese School (Independent)/ Mathematics Department / IBDP Mathematics HL / Mock Exam / Paper 1 Solutions Page 3 of 8 𝑦 = arcsin 𝑥 and 𝑦 = arccos 𝑥 𝑦 = arctan 𝑥 and 𝑦 = arccot 𝑥 Qn Solution Marks 6 [6 marks] Given that 𝑏 is real and positive, 𝑏 + 𝑖 is in the first quadrant ( ) 2arg0 + ib . Thus, 0 < 𝑎𝑟𝑔(𝑏 + 𝑖)2 < 𝜋, so 𝑎𝑟𝑔(𝑏 + 𝑖)2 = arg 𝑧 = 𝜋 3. arg(𝑏 + 𝑖) = 𝜋 6 implies 𝑏 = 1 × cot 𝜋 6 = √3 since 1 𝑏 = 𝑡𝑎𝑛 𝜋 6. Qn Solution Marks 7 [6 marks] 𝐸 [(𝑋 − 𝐸(𝑋)) 2 ] = 𝐸 [𝑋2 − 2𝑋𝐸(𝑋) + (𝐸(𝑋)) 2 ] = 𝐸[𝑋2] + 𝐸[−2𝑋𝐸(𝑋)] + 𝐸 [(𝐸(𝑋)) 2 ] = 𝐸(𝑋2) − 2𝐸(𝑋)𝐸(𝑋) + (𝐸(𝑋)) 2 = 𝐸(𝑋2) − (𝐸(𝑋)) 2 Note that (𝑋 − 𝐸(𝑋)) 2 ≥ 0 for any random variable 𝑋, so we have 𝐸 [(𝑋 − 𝐸(𝑋)) 2 ] ≥ 0. Thus, 𝐸(𝑋2) − (𝐸(𝑋)) 2 ≥ 0, i.e., 𝐸(𝑋2) ≥ (𝐸(𝑋)) 2 Qn Solution Marks 8 [6 marks] First, change the equations of lines to the standard form: 𝑥 − 1 1 = 𝑦 − 1 −2 = 𝑧 − 2 1 and 𝑥 + 1 3 = 𝑦 − 2 −3 = 𝑧 + 2 5 . The normal of the plane is given by ( 1 −2 1 ) × ( 3 −3 5 ) = ( −7 −2 3 ) So the equation of the plane is of the form −7𝑥 − 2𝑦 + 3𝑧 = 𝑑 Substitute the point (1,1,2), we have 𝑑 = −3, i.e., the equation of the plane is −7𝑥 − 2𝑦 + 3𝑧 = −3.

Anglo-Chinese School (Independent)/ Mathematics Department / IBDP Mathematics HL / Mock Exam / Paper 1 Solutions Page 4 of 8 OR The equation of the plane is given by 𝒓 = ( 1 1 2 ) + 𝜇 ( 1 −2 1 ) + 𝜆 ( 3 −3 5 ). REMARK: The second solution is much easier than the first one. But when the question is in section B, and the later parts need the Cartesian equation, you still need to use the first method. Qn Solution Marks 9 [6 marks] (a) From the graph the amplitude is 3, and max is 2, i.e., 𝑎 = 3, 𝑐 = −1. You can consider the transformation from 𝑦 = sin 𝑥 to 𝑦 = 𝑎 sin(𝑥 + 𝑏) + 𝑐 first. So the transformation is given by • Translate the graph along negative 𝑦-axis by 𝑐; • Scale the graph along 𝑦-axis by 1 𝑎 • Translate the graph along positive 𝑥-axis by 𝑏; (b) From the graph the amplitude is 3, and max is 2, i.e., 𝑎 = 3, 𝑐 = −1. The point ( 3𝜋 4 , 2) will be shifted to the left to become      2 ,2  So 44 3 2  −=−=b 14sin3 −     −= xy becomes      −= 4sin3 xy , then      −= 4sin xy and then xy sin= Qn Solution Marks 10 [6 marks] (1 − 2𝑥)5(1 + 3𝑥)4 = (1 − 10𝑥 + 40𝑥2 − ⋯ )(1 + 12𝑥 + 54𝑥2 + ⋯ ) = 1 + 2𝑥 − 26𝑥2 So, 𝑎 = 1, 𝑏 = 2, 𝑐 = −26. Qn Solution Marks 11 [6 marks] (a) 𝑓(𝑔(𝑥)) = (𝑎𝑥 + 𝑏 + 2)2 − 3 = 𝑎2𝑥2 + 𝑎𝑏𝑥 + 2𝑎𝑥 + 𝑎𝑏𝑥 + 𝑏2 + 2𝑏 + 2𝑎𝑥 + 2𝑏 + 4 − 3 = 𝑎2𝑥2 + 𝑥(2𝑎𝑏 + 4𝑎) + (𝑏2 + 4𝑏 + 1) (Given that) = 4𝑥2 + 6𝑥 − 3 4 Equating coefficients of 𝑥2 gives 𝑎2 = 4. Thus, 𝑎 = 2, since 𝑎 > 0. Equating coefficients of 𝑥 gives 2𝑎𝑏 + 4𝑎 = 6, so 𝑏 = − 1 2. (b) ℎ(𝑘(𝑥)) = 5(𝑐𝑥2 − 𝑥 + 2) + 2 = 0 5𝑐𝑥2 − 5𝑥 + 12 = 0 Condition for real equal roots is 𝑏2 − 4𝑎𝑐 = 0, thus, 25 − 240𝑐 = 0, i.e., 𝑐 = 5 48.

Anglo-Chinese School (Independent)/ Mathematics Department / IBDP Mathematics HL / Mock Exam / Paper 1 Solutions Page 5 of 8 true. 1 isP Qn Solution Marks 12 [14 marks] (a) (b) By the graph of part (a), there is only one intersection between the two graphs. For the intersection, it satisfies 𝑥 + 1 𝑥 − 1 = 3𝑥 − 5 ⇒ 𝑥 = 9 + √33 6 or 9 − √33 6 (N. A. , since 𝑥 < 5 3) Therefore, the solution to the inequality is given by 𝑥 < 1 𝑜𝑟 𝑥 > 9 + √33 6 . Qn Solution Marks 13 [16 marks] (a) Let Pn be the statement: 𝑑𝑛 𝑑𝑥𝑛 (cos 𝑥) = cos (𝑥 + 𝑛𝜋 2 ) for +n . P1: 𝑑 𝑑𝑥 (cos 𝑥) = −sinx = cos (𝑥 + 𝜋 2) Assume Pk is true i.e. 𝑑𝑘 𝑑𝑥𝑘 (cos 𝑥) = cos (𝑥 + 𝑘𝜋 2 ) Then for 𝑛 = 𝑘 + 1 𝑑𝑘+1 𝑑𝑥𝑘+1 (cos 𝑥) = 𝑑 𝑑𝑥 [ 𝑑𝑘 𝑑𝑥𝑘 (cos 𝑥)] = 𝑑 𝑑𝑥 [cos (𝑥 + 𝑘𝜋 2 )]

Anglo-Chinese School (Independent)/ Mathematics Department / IBDP Mathematics HL / Mock Exam / Paper 1 Solutions Page 6 of 8 = − sin (𝑥 + 𝑘𝜋 2 ) = sin (−𝑥 − 𝑘𝜋 2 ) = cos [𝜋 2 − (−𝑥 − 𝑘𝜋 2 )] = cos (𝑥 + (𝑘 + 1)𝜋 2 ) Pk+1 is true whenever P1 and Pk are true. Hence Pn is true for all +n by Mathematical Induction. (b) We know that sin 4𝑥 = cos ( 𝜋 2 − 4𝑥) , cos(𝑥) = cos(𝑥