2022 H2 Math RI Prelim Paper 2 Solns

RAFFLES INSTITUTION 2022 Year 6 H2 Mathematics Preliminary Examination Paper 2 Questions and Solutions with comments Page 1 of 24 1 (i) Show that 2 d 1 sin d cos cos  = n kx xx x , where the values of the constants k and n are to be determined. [2] (ii) Hence use integration by parts to evaluate 234 0 sin sec dx xx π ∫ , leaving your answer in the form lnab c+ , where a, b and c are exact constants. [5] (i) 2 d1 d cos  xx ( ) 2d cosd − = xx 32(cos ) ( sin )−= −− xx 3 2sin cos= x x (i.e. k = 2 and n = 3) Generally, majority of students were able to do this part well. A very common mistake that a number of students made was the missing of the negative sign after differentiating cosine. (ii) 2 2344 3 00 sinsin sec d dcos xx xx x x ππ =∫∫ 4 3 0 2sin sin . dcos 2 π =∫ xx xx 4 4 22 0 0 1 sin 1 cos . . dcos 2 cos 2 π π = − ∫ xx xxx 4 4 2 0 0 1 sin 1 sec d2 cos 2 π π = − ∫ x xxx [ ]4 02 sin11 4 0 ln(sec tan )22cos 4 π π = −− + π  xx 1 11 2 ln(sec tan ) ln(1 0)12 2 44 2   ππ= − + −+   11 ln( 2 1)22 = −+ This part posed some difficulty to some as they could not identify which should be “u” in the by parts process. Students are reminded to pay attention how to apply the “Hence” method for this type of by parts question. Some common errors were as follows: 1. “1 2 ” was missing. 2. Did not recognize 44 00 1 d sec dcos x xxx ππ =∫∫ and proceeded to lengthy and tedious working which often were wrong. 3. 4 4 00 1 d ln coscos xxx π π = ∫ 4. 2 221 22 ++=

2 Do not use a calculator in answering this question. (a) Let f( )z be a polynomial in z of degree 4 with real coefficients. The equation f( ) 0z = has four roots, namely , , and αβγ δ such that they satisfy the following two conditions: 0αβγδ < and 2 222 0αβγδ+++< . Based on the two conditions, a student concludes that the equation f( ) 0z = has one positive real root, one negative real root and a pair of complex conjugate roots. State, with reasons, whether the student’s claim is true. [3] (b) It is given that 43 2g( ) 2 4 24.zzz z z=+− +− Verify that 2iz= is a root of the equation g( ) 0.z = Hence find the other roots of the equation. [5] (a) [3] The student’s claim is true. Reasons: 1. From 2 222 0αβγδ+++< , it shows that at least one of the roots is complex. 2. As the coefficients of ( )f z are real, we know that complex roots exist in conjugate pairs, so there is at least one pair of complex conjugate roots. 3. If there are 2 pairs of complex conjugate roots, then 0αβγδ > . However, given that 0αβγδ < , then there is only a pair of complex conjugate roots and 2 real roots of opposite signs. So with 1, 2 and 3, we can conclude that the equation ( )f0 z = has one positive real root, one negative real root and a pair of complex conjugate roots. This part posed some difficulty to a number of students as they took the wrong approach of using the claim to check on the conditions instead of the other way as intended by the questions. These students often could not explain why with the claim they began with could satisfy the second condition. Students are also reminded to read the question carefully and answer the question completely as some did not state whether the claim is true or not and just proceeded to give reasons to justify. (b) [5] ( ) ( ) ( ) ( ) ( ) 43 2 Since f 2i 2i 2i 2 2i 4 2i 24 16 8i 8 8i 24 0, =+− +− = − ++ − = so 2iz= is a root of the equation ( )f0 z = (verified). As complex roots occur in conjugate pair, so 2iz=− is the other complex root. Now ( )( ) 22i 2i 4zz z− += + . Most students were able to handle this part well though they are reminded to improve in their method to use the most efficient way to solve as suggested here and not continue to use their very tedious and long method.

Hence ( )( ) 43 2 2 22 4 24 4 6z z z z z z az+− +−= + +− . Comparing the coefficient of 3z : 1a= . Thus ( )( ) ( )( )( ) 43 2 2 2 2 2 4 24 4 6 432 zz z z z zz z zz + − + − = + +− = ++− So the other 3 roots of the equation ( )f0 z = are 2i, 2 and 3−− . Some students missed out verifying that 2iz= is a root of the equation and a number of students did not know that 2iz= is a root while 2iz− is a factor.

3 The line 1L has equation 11 , 2 2 zyx −−= = , and meets the the xy -plane at point P . The point A has position vector 3 1 2  − with reference to the origin O. (i) Find a vector equation of the line 2L which passes through O and P. [3] (ii) Find an equation of the plane π containing both 1L and 2L , in the scalar product form. [2] (iii) The points A and C are on different sides of π such that AC is perpendicular to π. The distance of C from π is t times the distance of A from π . Find, in terms of t, the position vector of C. [5] (iv) Find the value of t such that the line OC is parallel to the plane with equation 2 02 1 .   = r . [2] (i) [3] Method 1 The z-coordinate of P is 0, since P lies on the xy-plane. Thus, putting z = 0 into 1L , 01 31 22yy −−= ⇒ = The position vector of P is 2 3 2 0 OP   =    . Hence, equation of 2L is 4 3 0 λ  =  r , λ∈  Method 2 Vector equation of line 1L : 20 11 12 λ      = +−       r , µ∈  Equation of xy-plane : 0 00 1   = r To find P, consider 2 00 1 1 00 1 21 µ      +− =        This part is well done. Quite a significant had careless mistake, missing negative sign or placed at the wrong component. The vector equation of a line is of the form ,λλ= +∈ra b  . Missing “ =r ” (i.e. writing just λ+ab ) is considered incomplete. This is analogous to writing the Cartesian equation of a 2D straight line as mx c+ (i.e. without y = ).

12 0µ+= 1 2µ =− Hence, 2 04 111 13 221 20 OP        = +− − =            . Hence, equation of 2L is 4 3 0 λ  =  r , λ∈  (ii) [2] 40 6 3 3 1 8 24 02 4 2 −         ×− =− = −         −     Hence, equation of π is 3 40 2 −  = r This part is also well done. (iii) [5] Method 1 (Find the foot of perpendicular first) Let F be the foot of perpendicular from A to π . Equation of the line AF is 33 14 22 λ −   = −+    r . Since F lies on the line AF, then 33 14 22 OF λ −   = −+      , for some λ∈  . Since F is also on π, then 3 33 1 4 40 2 22 λ  −−     −+ =        ( )9 4 4 (9 16 4) 0λ−−+ + + + = 9 29λ = Hence, 3 3 60 9114 729 292 2 76 OF −     = −+ =         . Quite a few wrote that 33 14 22 AF λ −   = −+      . Recall that “ r ” in the equation of a line represents the position vector of a point on the line (i.e. a vector from the origin to a point on the line). You need get the concept of the line correct. See Chap 4C page 2.

Method 1.1 (Find C using ratio theorem) By Ratio Theorem, ( 1) tOA OCOF t += +      60 3 60 27 ( 1) 1( 1) 7 1 7 3629 2976 2 76 18 t tOC t OF tOA t t t −  +   =+ − = −−= +      +         Method 1.2 (Find C without using ratio theorem) 60 3 27 11 7 1 3629 2976 2 18 AF OF OA −         = − = −− =                    Since FC t AF=     , then 60 27 1 7 3629 2976 18 OC OF FC OF t AF t =+=+ −     = +                  Method 2 ( Use projection vector) A few applied