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