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 th