ACSI 2022 Y6 Prelim P1 Student solution with comments

Student solution and comments 2022 Prelim Y6 HL Math P1 Section A 1. Comments (a)(i) sin 𝐴𝐴 = 1 √3 tan 𝐴𝐴 = 1 √2 Some students forgot to square root the side 2 (ii) tan 𝐴𝐴 = 2 tan 𝐴𝐴 2 1 − tan2 𝐴𝐴 2 Let 𝑥𝑥 = tan 𝐴𝐴 2 : ⇒ 1 √2 = 2𝑥𝑥 1 − 𝑥𝑥2 1 − 𝑥𝑥2 = 2√2𝑥𝑥 𝑥𝑥2 + 2√2𝑥𝑥 − 1 = 0 𝑥𝑥 = −2√2 ± ��2√2� 2 − 4(−1) 2 𝑥𝑥 = −2√2 ± √12 2 = −2√2 ± 2√3 2 𝑥𝑥 = −√2 + √3 ⇒ tan 𝐴𝐴 2 = √3 − √2 Students did not realize that the quadratic equations can be solved using formula method, instead tried to factorize.

(b) 2 sin 𝑥𝑥 = tan 𝑥𝑥 2 sin 𝑥𝑥 = sin 𝑥𝑥 cos 𝑥𝑥 2 sin 𝑥𝑥 cos 𝑥𝑥 − sin 𝑥𝑥 = 0 sin 𝑥𝑥 (2 cos 𝑥𝑥 − 1) = 0 sin 𝑥𝑥 = 0 ⇒ 𝑥𝑥 = 0 2 cos 𝑥𝑥 − 1 = 0 ⇒ 𝑥𝑥 = 1 2 ⇒ 𝑥𝑥 = 𝜋𝜋 3 or 𝑥𝑥 = − 𝜋𝜋 3 sin x was deleted on both side, thus missing a root which is sin 0x= ; do not cancel out on both sides of the equation, instead shift everything to LHS and factorise. 2. Comments (a) 𝑛𝑛 = 10 Generally well done; remember number of terms in binomial expansion = n+1 (b) 𝑟𝑟𝑡𝑡ℎ𝑡𝑡𝑡𝑡𝑟𝑟𝑡𝑡 = �10 𝑟𝑟 � (𝑥𝑥2)10−𝑟𝑟 �𝑘𝑘 𝑥𝑥� 𝑟𝑟 = �10 𝑟𝑟 � (𝑥𝑥)20−3𝑟𝑟𝑘𝑘𝑟𝑟 20 − 3𝑟𝑟 = 14 ⇒ 𝑟𝑟 = 2 ∴ �10 2 � 𝑘𝑘2 = 180 where �10 2 � = 10! 8! 2! = 9×10 2 = 45 45𝑘𝑘2 = 180 𝑘𝑘2 = 4 ⇒ 𝑘𝑘 = 2 Generally well done 3. Comments (a) 𝑐𝑐(−2) + 6 = 0 ⇒ 𝑐𝑐 = 3 Generally well done (b) 𝑦𝑦 = 1 3 Generally well done

(c) x-intersection at origin (d) 𝑥𝑥 3𝑥𝑥 + 6 = 2 ⇒ 𝑥𝑥 = − 12 5 Hence 0 < 𝑥𝑥 3𝑥𝑥+6 < 2 : 𝑥𝑥 < − 12 5 or 𝑥𝑥 > 0 Since “hence” is used, attempt to find intersection must be seen. 4. Comments (a) 63 32 uu uu= Generally well done 15 12 12 1 dd dd ++ =++ 2 22 (1 2 ) (1 5 )(1 ) 14 4 1 5 d dd d d d dd += ++ + + =++ + 2 20dd+= ( 2) 0dd += 0d = (rej.) 2d =− (b) 1 ( 1)( 2) 17nun= +−−= − Generally well done 2 20 10 n n −= − = 10 1 1 ( 1) ( 3) ... ( 17)r r u = =+− +− + +−∑ Generally well done; some

( ) 10 1 10 1 ( 17)2 r r u = = +−∑ students did not read the question carefully and tried to find sum of GP 10 1 80r r u = =−∑ 5. Comments (a) Let nP be the statement that 61n − is divisible by 5 for 2n≥ and .n +∈  Generally well done OR Let nP be the statement that 6 15, ,n KK +−= ∈  for 2n≥ and .n +∈  2 :P 26 1 35 5 7−= =× 2P∴ is true. Some students accidentally wrote P1 instead Assume kP is true for some integer k 6 1 5,k QQ +−= ∈  ( ) 1 1 : 6 1 66 1 k kk P+ + −= − ( )65 1 1 30 5 5(6 ) 5 5(6 1) 5, Q Q Q Q RR + = +− = + = + = + = ∈  Some students assume LHS = 5A and manipulate from the result. Always start from one side and show to the other. Conventionally start from LHS. Since 2P is true, kP is true 1kP+⇒ is true, by MI, nP is true for 2n≥ and .n +∈  * Make sure initial step is true ( 2P is true); inductive step ( kP is true 1kP+⇒ is true) is true, these two are a pair with “if.. then…” relationship. Some students used ambiguous phrasing. 6. Comments (a) Assume 2a b ab+< for ,ab +∈  Some students did not write down the correct negation. Conditions should be kept the same (a, b are positive integers and not equal), while the therefore statement is

negated. ( you can seen < as the complement set of )≥ 2( )4a b ab+< 22 22 24 20 a ab b ab a ab b + +< − +< 2( )0ab−< However, 2( )0ab−≥ and since ab≠ So 2( )0ab−> (contradiction) By contradiction, 2a b ab+≥ The contradiction should not allow some cases to be true. Eg. Some students assumed 22 ( )0a b ab a b+≤ ⇒ − ≤ and contradiction is 2( )0ab−≥ , allowing 2( )0ab−= to be true. 7. Comments (a) Point of inflexion occurs when () 0fa′′ = and ()fx′′ changes sign at .xa= Minimal point of ()fx′ satisfy the condition, therefore Well done 3x= is the point of inflexion. (b) 1 0 ( ) (1) (0)f x dx f f′ = −∫ Well done (1) 1 1f= −= (1) 2f⇒= 4 1 ( ) (4) (1) 5f x dx f f′ =−= −∫ Quite a number of students forgot the integral result is the signed area under graph. (4) (1) 5ff = − (4) 2 5 3f =−= − 8. Comments (a) Method 1: using geometry formula rectangle area + trapezium area =1 Well done 1 ( 2 )(1) 12k kk++ = 3 12 5 12 kk k += =

2 5k = Method 2: using integration 34 23 ( 2) 1kdx k x dx+ −=∫∫ [ ] 42 3 2 3 212 93 2 88 6 1 2 3 12 5 12 xkx k x k kk kk k +−=  − + −− − =   += = 2 5k = (b) Area of rectangle = 3 2 21 52kdx= <∫ ( ) 3 22 1 255 2 m x dx+ −=∫ Some students forgot to add the area under graph from x=2 to 3 to the 0.5 probability 2 3 21 25 2 10 m x x −= 2 91262 24 m m −−−=  22 8 61mm− += 22 8 50mm− += 8 64 40 4m ±−= Some students forgot to reject one answer 82 4 46 42m ±±= = 46 2m += since 3m> 9. Comments (a) 332 3xxxx e dx x e x e dx= −∫∫ Well done. A few students did not show any working for second and third integration by parts. Do be reminded that all steps should be 32 3xxx e x e dx= − ∫ 32 32x xxx e x e xe dx= −−  ∫ 32 36xx xx e x e xe dx= −+ ∫

shown for show question. 32 32 36 3 66 x x xx x x xx x e x e xe e dx x e x e xe e C = −+−  = − + −+ ∫ 32( 3 6 6)xex x x C= − +−+ (b) 2x t xt=⇒= 2xdx dt= 22 2 23 00 0 22tx xte dt x e xdx x e dx= =∫∫ ∫ Quite a number of students forgot to substitute the limits while t is replaced by x 32 2 02[ ( 3 6 6)]xex x x= − +− 22[ (2 2 6 6 2 6) ( 6)]e= − + − −− Quite a number of students did not realize that the last 6 does not have a factor of 2e . 22 (8 2 12) 12e= −+ Section B 11. Comments (a) ℎ(𝑥𝑥) = (sin(2𝑥𝑥) + cos(2𝑥𝑥))2 = (sin(2𝑥𝑥))2 + 2 sin(2𝑥𝑥) cos(2𝑥𝑥) + (cos (2𝑥𝑥))2 = 1 + sin(4𝑥𝑥) Well done (b)(i) sin 𝑥𝑥 → sin(4𝑥𝑥): stretch the graph of 𝑦𝑦 = sin 𝑥𝑥 by factor of ¼ units parallel to the x-axis. sin(4𝑥𝑥) → sin(4𝑥𝑥) + 1: translate the graph of 𝑦𝑦 = sin(4𝑥𝑥) + 1 by 1 unit in the positive y-axis. Well done

(ii) Well done, a few students draw outside the given domain. Do note that only graph within given domain should be shown, else one mark will be deducted in IB. (b)(i) Period = 𝜋𝜋 2 Well done (ii) Range = [0, 2] Well done (c) Area bounded by curve, x-axis and y-axis = � (sin(4𝑥𝑥) + 1 3𝜋𝜋 8 0 ) 𝑑𝑑𝑥𝑥 = �− cos(4𝑥𝑥) 4 + 𝑥𝑥� 0 3𝜋𝜋 8 = � − cos(3𝜋𝜋 2 ) 4 + 3𝜋𝜋 8 − �− cos(0) 4 �� = 3𝜋𝜋 8 + 1 4 A lot of students did not identify the correct region under graph. Note that if y-axis is specified, it must line one of the sides of the area. All the sides specified must appear in the region you identify. 11. Comments (a) 2 2 ()1( ) arcsin ()1 xfx x −−−= −+ 2 2 1arcsin ( )1 x fxx  −= = + ( ) ()f x fx−=

()fx is an even function OR graph is symmetrical about the y-axis as shown in the diagram, So ( ) ()f x fx−= , function must be even. Generally well done. (b) 2 2 2 2 111lim lim 1 11 1 xx x x x x →∞ →∞ −− = =+ + Or apply L’hopital’s rule 2 2 1 22lim lim lim 11 22 HH x xx xx xx→∞ →∞ →∞ −∞= = = =+∞ arcsin(1) 2 π= So horizontal asymptote 2y π= Students missed crucial steps or wrote mathematically incorrect notation like 1 ∞ (c) (i) 22 2222 2 1 ( 1)(2 ) ( 1)(2 )() ( 1)11 1 x xx xfx xx x + −−′ = + −− + Take out 22 1 ( 1)x + from inside the square root in the denominator 222 22 14() ( 1)( 1) ( 1) xfx xxx ′ = ++−− 22 14() ( 1)4 xfx xx ′ = + 222 22() | | ( 1)( 1) xxfx xxxx ′ = = −− Students who were unable to do the question did not apply chain rule correctly. (ii) 2 ||2xx = = 42'( 2) 2(4 1) 5f −−= = − + Since () 0fx′ < , f is decreasing at 2.x=− Generally well done (d) 01(0) arcsin arcsin( 1)01 2f π−= = −= − +

Generally well done (e) (i)let ()y gx= , try to express x as the su