EJC H2 MATH P1

2018 JC2 H2 Mathematics Preliminary Examination [Turn over EUNOIA JUNIOR COLLEGE JC2 Preliminary Examination 2018 General Certificate of Education Advanced Level Higher 2 MATHEMATICS Paper 1 [100 marks] 9758/01 12 September 2018 3 hours Additional Materials: Answer Paper List of Formulae (MF26) READ THESE INSTRUCTIONS FIRST Write your name, civics group and question number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use an HB pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all questions. Give non-exact numerical answers correct to 3 signifi cant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use an approved graphing calculator. Unsupported answers from a graphing calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphing calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 7 printed pages.

2 2018 JC2 H2 Mathematics Preliminary Examination 1 The shaded region R bounded by the curve yx=− , the line yx=− and the x-axis is rotated about the x-axis through O360 . Find the volume of the solid fo rmed, leaving your answer to 2 decimal places. [4] Suggested Solution yx=− 15yx=± − 15yx=+ − intersect yx=− at (4 , 4 )− 15yx=− − cuts the x-axis at (4,0) Volume required () () () 2255 2 44 115 d 15 d 4 ( 4 ) 3 201.06 (to 2 d.p.) xx xxππ π − =+ − − − − − =  2 (i) Solve the inequality 2xa x a axa −− ≥− , where a is a positive real constant, leaving your answer in terms of a. [4] (ii) Hence, by using a suitable value for a, solve the inequality 24e e 1 1 4e 1 4 xx x −− ≥− leaving your answer in exact form. [3] y x O (5,1) R

3 2018 JC2 H2 Mathematics Preliminary Examination Suggested Solution (i) 2xa x a xa −− − ≥ a ( xa≠ ) () () 2 22 0 2 0 xa x aa x a xa xa x a a xa −−− − ≥− −+ − ≥− Consider () 22 2xa x a a−+ − = 0 x = () () () 2 222 4 ( 1 ) 2 aa a a−− ± − − − = 24 2 aa± = aa± Method 1 (test critical points) 22 2 0xa x a a xa −+ − ≥− can be rewritten as ()() ()() 0 xa a xa a xa −+ −− ≥− Using sign test, we can check whether each factor is positive or negative for the different range of values of x Thus, we have or a a xa xa a−≤ < ≥ + Method 2 () () 2xa x aa x a xa −−− − − ≥ 0 () () 22 2xa x a x a a−− + − ≥ 0 ( xa≠ )

4 2018 JC2 H2 Mathematics Preliminary Examination or a a xa xa a−≤ < ≥ + (ii) Given 24e e 1 1 4e 1 4 xx x −− ≥− . () () () 2 11ee 144 1 4e 4 xx x −− ≥ − . Replace x with ex and let 1 4a = , 11 e 44 x−≤ < or 3e 4 x ≥ 1 0 e 4 x<< or 3e 4 x ≥ 1 ln 4x <  or 3 ln 4x ≥  3 The parametric equations of a curve C are 3,xa t ya t== , where a is a positive constant. (i) The point P on the curve has parameter p and the tangent to the curve at point P cuts the y-axis at S and the x-axis at T. The point M is the midpoint of ST. Find a Cartesian equation of the curve traced by M as p varies. [5] (ii) Find the exact area bounded by the curve C, the line 0x = , 3x = and the x-axis, giving your answer in terms of a. [3] Suggested Solution Solutions: (i) 3,xa t ya t== 2dd y ,3dt dt x aa t==

5 2018 JC2 H2 Mathematics Preliminary Examination 2 2 d3 d 3 ya t xa t = = Therefore, gradient of the tangent at P 23p= Equation of tangent at P: () 32=3ya p pxa p−− 32 3 23 =3 3 32 ya p p x a p yp xa p −− =− For point S, 30, 2xy a p== − . () 3 is 0, 2Sa p∴− For point T, 3 2 220, 33 apyx a p p== = 2 is ,03Ta p ∴  Midpoint M = 31 ,3 ap ap − . 31 ,3xa p y a p== − 3 3 2 3 , 32 7 x pa xxya aa = =− =− (ii) Required area 3 0 dyx=  () 3 3 0 dta at a = () 3 3 0 3 4 2 2 0 dt 81=a 44 a a at a t a  =  = Alternatively, find the cartesian of the given curve and use it to find the required area.

6 2018 JC2 H2 Mathematics Preliminary Examination 3 3 3 2 ,xa t ya t xxya aa == ∴= =  Required area 33 34 4 22 2 2 0 0 11 3 8 1d0 44 4 xx xaa a a      == = − =     

7 2018 JC2 H2 Mathematics Preliminary Examination 4 It is given that 11sin cosyx x −−= , where 11 x−≤ ≤ . (i) Show that 21 1d1c o s s i nd yxx xx −−−= − . [1] (ii) Show that () 2 2 2 dd12 dd yyxx xx−− = − [2] (iii) Hence find the exact value of A, B and C if y can be expressed as 23Ax Bx Cx++ , up to (and including) the term in 3x . [4] (iv) A student used (iii) to estimate that () () 11 2 3sin 0.8 cos 0.8 0.8 0.8 0.8 ABC−− ≈+ + . Explain, with working, if his estimate is a good one. [1] Suggested Solution (i) 11 11 22 dc o s s i nsin cos d 11 yx xyx x x xx −− −− −=  =+ −−  21 1d1c o s s i nd yxx xx −− −= − ---(1) [shown] (ii) Diff (1) wrt x, () () 1 2 22 2 2 22 1d d 1 112 12d d 11 yyxx x xx xx − −−−+ − = − −− () () () 22 22 22 dd d d11 1 2 1 2dd d d yy y yxx x xxx x x −+ − = − − = −  −− = − --(2)[shown] (iii) Diff (2) wrt x, () 23 2 2 23 2 dd d d21 0dd d d yy y yxx xxx x x −+ − − + =  When 23 23 dd d0, 0, , 2, d2 d d 2 yyyxy xx x ππ== = = − = . Therefore, ()11 2 3 2 3 2 2sin cos 22 ! 3 !2 1 2xx x x xx x x π ππ π−− − ≈+ + = − + . (iv) The estimate is not good as () () 11sin 0.8 cos 0.8 0.597−− = (to 3 sf)

8 2018 JC2 H2 Mathematics Preliminary Examination But () () () 23 0.8 0.8 0.8 0.75121 2 ππ −+ = (to 3 sf) Alternative Explanation The graphs illustrated that at 0.8x = , the two graphs are quite different from each other. 23 21 2yx x xππ=− + 11sin cosyx x −−=

9 2018 JC2 H2 Mathematics Preliminary Examination 5 (a) Referred to the origin O, the points A and B have position vectors a and b . Point C is on the line which contains A and is parallel to b. It is given that the vectors a and b are both of magnitude 2 units and are at an angle of () 1sin 1/ 6− to each other. If the area of triangle OAC is 3 units 2, use vector product to find the possible position vectors of C in terms of a and b . [5] (b) Referred to the origin O, the points P and Q have position vectors p and q where p and q are non-parallel, non-zero vectors. Point R is on PQ produced such that :1 :PQ QR λ= . Point M is the mid-point of OR. (i) Find the position vector of R in terms of λ ,p and q . [1] F is a point on OQ such that F, P and M are collinear. (ii) Find the ratio OF:FQ, in terms of λ . [4] Suggested Solution (a) for some .OC λλ=+ ∈ab uuur ¡ Area of triangle OAC () () () () () 1 2 1 2 1 2 1 2 1 sin2 11 1 2226 3 OA OC λ λ λ λθ λλ =× =× + =× + × =× = == aa b aa ab ab ab uuuru u ur Since area of triangle OAC= 3, 1 33 9o r 9 λ λ = =− 9OC =±ab uuur (b)

10 2018 JC2 H2 Mathematics Preliminary Examination By the ratio theorem, () 1 1 OR OPOQ OR λ λ λλ += + =+ − qp uuuru u u ruuur uuur Since the point F lies on line OQ, OF t= q uuur , for some .t ∈ ¡ () () 1 2 1 22 1 122 PM OM OP OR λ λ λ λ =− =− +=− − + =− +  p qp p qp uuuur uuuur uuu r uuur Since the point F also lies on line PM, () () ,for some . 1 122 11 22 OF sPM s s ss s λ λ λλ =+ ∈ + =+ − +   +=− − + p pq p pq uuuru u u u r ¡ Since p and q are non-parallel & non-zero vector s, comparing coefficients of p and q against OF t= q uuur , we have 10 2 112 12 212 s s s s λ λ λ