2022 H2 Math RI Prelim Paper 1 Solns

RAFFLES INSTITUTION 2022 Year 6 H2 Mathematics Preliminary Examination Paper 1 Questions and Solutions with comments Page 1 of 18 1 (i) On the same axes, sketch the graphs of xy xa= − and 2 1,y xa= −+ where a is a positive constant. [3] (ii) Hence solve the inequality 2 1,x xaxa > −+− leaving your answer in terms of a. [3] (i) Note that the asymptotes intersect at (a, 1), and the vertex of the modulus graph is at the same point. Also note the curve passes through the origin. (ii) Let the 2 graphs intersect at the point T. To solve for T, consider 2 2 2( ) 1 2( ) () 2 2 x xaxa x xa xa axa axa = −+− = − +− −= = ± Since xa> at the point T, 2 2 axa= + . Thus the solution is 2 .2 aaxa<<+ The graphs are in (i), so it is easier to solve the equation to find the x- coordinate of the intersection point, then refer to the graphs in (i) to write down the solution to the inequality. 0

Page 2 of 18 2 A function is defined as 2 2 12f ( ) , , 0, 2.2( 2 ) xxx xx xx +−= ∈≠ −  (i) Show that f( )x can be written in the form 2 2 2( ) ( )1 xpq xp  −+  +− , where p and q are constants to be found. [2] (ii) Hence describe a sequence of transformations that will transform the graph of 2 2 2 1 xy x −= − onto the graph of f( ).yx= [2] (iii) Determine, with the help of a sketch or otherwise, the set of values of k for which the equation 2 2 2 1 x kx − =− has no real roots. [2] (i) 2 2 2 2 2 2 12 2( 2 ) 1 2 ( 2 1) 2 ( 2 1) 1 1 2 ( 1) 2 ( 1) 1 xx xx xx xx x x +− −  −−+=  − +−  −−=  −− Thus 11 and . 2pq= −= Do show enough working for a ‘show’ question. Note that a ‘show’ question is different from a ‘verify’ question, so here the given answer should not be expanded to compare with the expression for f. (ii) The 2 transformations (in either order) are 1. A translation of 1 unit in the positive x-direction 2. A scaling parallel to the y-axis by a factor of 1 .2 Remember to use standard mathematical terms and descriptions. Unacceptable ones include ‘along the x-axis’, ‘in the x-axis’, ‘on the x-axis’. (iii) The equation 2 2 2 1 x kx − =− has no real roots for values of k in the set ( ]2, 1−− . It is usually easier to use the method as suggested in the question. Here the question suggested a sketch, and the correct answer was mainly obtained by those who referred to the graph instead of the discriminant method. 1y=− 1x=− 1x= 2 2 2 1 xy x −= −

Page 3 of 18 Alternative Solution: 2 2 22 2 2 1 2 ( 1) ( 2) 0 x kx x kx k kxk − =− −= − + −+= If 1k =− , 2( 1) ( 2) 0kxk+ −+= becomes 10−= , so there is no solution. If 1k ≠− , the quadratic equation will have no real roots if [ ]0 4 ( 2)( 1) 0 ( 2)( 1) 0 21 kk kk k −−+ + < + +< − < <− So 21 k− < ≤− and the required solution set is ( ]2, 1−− . Note that the discriminant is only used for a quadratic expression. If k = -1, then the equation in the 3rd line is no longer quadratic. So, the case 1k =− should be considered separately.

Page 4 of 18 3 A curve has parametric equations 2 111 , x a y at t t   = += −     , where a is a constant and . (i) Find the equations of the tangent and the normal to the curve at the point P where 1 2t =− . [5] (ii) The tangent at P meets the y-axis at Q and the normal at P meets the y-axis at R. Show that the area of triangle PQR is 2241 120 a . [2] (i) [5] 2 d d xa t t =− ; 3 d2 1d y at t = + 33 2 21d2 d ayt t axt t + += =− − When 1 d 15 9, , , 2d 4 2 yt x ay ax= − = = −= − Equation of tangent at P: 9 15 ()24y a xa+= + 15 3 44y xa= − Gradient of normal at P 4 15=− Equation of normal at P: 94 ()2 15y a xa+= − + 4 143 15 30y xa= −− A generally well done question apart from some careless mistakes. Some candidates did not substitute the t-values and had to go through a whole lot of calculation for the next part of the question. (ii) [2] At point Q, 30, 4xy a= =− At point R, 1430, 30xy a= =− Area of triangle PQR 2 2 1 143 3 2 30 4 1 241 2 60 241 (shown)120 a aa a a = − =   = Most mistakes in this part were carried forward from the previous part. Note that the value of a could be positive or negative. Thus, candidates are reminded to be careful with the positive negative signs and ensure that the final expression is obtained correctly. 0t ≠ P Q R

Page 5 of 18 4 (a) The point R has position vector r. Given that r = 1 24 35 a × , where a is a real number, describe geometrically the set of all possible positions of the point R, as a varies. [2] (b) (i) The points P and Q have position vectors p and q respectively. Show that the point F, the foot of perpendicular from the origin O to the line passing through P and Q, has position vector ( )1 λλ−+ pq , where 2 2 ..λ −= − p pq qp [4] (ii) Write down an inequality satisfied by λ for F to lie within the line segment PQ. [1] (a) [2] r = 1 2 20 2 4 53 3 5 3 5 42 2 4 a aa a −−         × = −+= +−         −−     , a∈  R lies on the line passing through the point with coordinates ( 2, 3, 2)−− , and the line is parallel to the vector 54− j+ k . Many candidates described the points as “moving 5 units in the negative j direction, 4 units in the positive k direction” and so on without addressing what would the “set of points” be geometrically. The line should be correct described, or its equation correctly stated. (b) [4] Since F lies on the line PQ, then ()OF λ= +−p qp   , for some λ∈  . Since OF   is perpendicular to the line PQ, then ( ) () () 0λ+ − −=p qp qp  2 () 0 λ−+ − =p qp qp 22 0λ−+ −=pq p q p 2 2λ −= − p pq qp  Substitute value of λ into (1) : ( ) 2 2 22 22 ( 1 1 OF λλ −= +− −  −−= − + −− = −+ p pqp q p) qp p pq p pqpq qp qp pq     Candidates who used dot product were generally able to get this part correct. Candidates are reminded that all workings should be showed clearly since this is a “show” question.

Page 6 of 18 Alternative Method: Consider PF   as the projection vector of PO   onto PQ   ,  ( )  ( ) ( ) ( ) 2 2 PF PO PQ PQ= ⋅ −⋅ − −= × −− −⋅= − − pqp qp qp qp p pq qp qp         Hence we have ( ) ( ) 2 2 22 221 1 OF OP PF λλ = + −⋅= +− −    −⋅ −⋅= −  +    −−   = −+ p pqp qp qp p pq p pq pq qp qp pq       where 2 2λ −⋅= − p pq qp (shown) Candidates who used the projection vector method, with careful consideration of the directions of the vectors, were also able to conclude correctly. It is important to consider the direction of the vectors when using this method that involves projection vector. Thus, Both FP    and PF   are parallel to PQ   , but the correct direction is determined by PO   . Thus,  ( ) FP PO PQ PQ≠⋅         . Candidates who used the length of projection method are reminded to be careful when dealing with the modulus sign. Using merely length of projection would have omitted the direction. (b)(ii) [1] For F to lie within the line segment PQ, 01 λ≤≤ Generally well done for those who attempted this part. P Q F

Page 7 of 18 5 Do not use a calculator in answering this question. The complex numbers z and w are given by sin i cos66z ππ = +    and 2 sin icos63w ππ  = +    . (i) Find z and ( )arg z