JPJC 9758 2023 Prelim P2 Solution

Jurong Pioneer Junior College H2 Mathematics JC2-2023 Prelim Paper 2 Solution 1 (i) Method 1 ( ) 2 2 2 2 22 2 22 2 ln 1 e ln 1 1 2 ln 2 2 ln 2 1 24 ln2 ln 1 24 24ln2 24 2 ln2 24 8 1ln2 28 x xx xx xx xx xx xx xx x x x −  + ≈ + −+  = −+ = −+  = + +− +  −+ = +− + − + = −+ − +  = −+ +    Method 2 ( ) ( )f ln 1 e xx −= + ( ) ( )11f ' e1e e 1 x xxx − −= −= −++ ( ) ( ) 2 f '' e 1 e xxx − = + When 0x= , ( )f 0 ln 2= , ( ) 1f ' 0 2=− , ( ) 1f '' 0 4= ( ) 211f ln2 ......2 4 2! xxx =−+ +  ( ) 21f ln2 ...... 28 xxx= −++ (ii)

2 ( )d e1ln 1 ed 1e 1e x x xxx − − − −+= = − ++ Using the series from part (i), 21d 1 ln 21e d 2 8x x xx ≈− − ++  = 12 11 28 24xx−− + = − . 2 d 9.8d v Rt = − ( ) d 9.8 (Since , where 0)d 1 d 1 d 9.8 1 d 1 d 9.8 1 ln 9.8 ln 9.8 (9.8 0) 9.8 e , where e When 0, 0 (helicopter is stati kt kC v kv R kv kt vtkv k vtk kv kv t Ck kv kt kC kv kv A A tv −− = −= > =− −−= − − −= + − = −− − > −= = = = ∫∫ ∫∫ 0 onary) 9.8 (0) e 9.8 9.8 9.8e 9.8(1 e ) kt kt kAA kv v k − − − = ⇒= ∴−= −= 1The terminal velocity of the para 9.8As , e chuti 0, st is 9.8 ms kttv k k − − →∞ → → ∴ 3 (a) Method 1 eee 2xxxyy y= →= −→= −+ Reflect about the x-axis. Translate by 2 units in the positive y-direction. Method 2 ( )e e2 e2 = e2xx x xyy y= →=−→= − − −+ Translate by 2 units in the negative y-direction. Reflect about the x-axis. (b)

3 After A : 36yx= + After B : ( )3 2 63yx x= − += After C : ( )1 33y xx= = (c) (i) Asymptotes : xa= , 1y d= Axial intercepts : ( ),0c , 10, b   (ii) Asymptotes : xd= , yc= Axial intercepts : ( )0,a , ( ),0b 4 (i) h= 2 3 ( ) 6 1 f (2 f(2 ) f (2 2 ) f (2 3 ) f (2 4 ) f (2 5 ) f (2 6 ) n n h h hh hh hh hh hh hh = + =+++ ++ ++ ++ ++∑ represents the sum of areas of six rectangles drawn and the rectangles are above the curve, so it is greater than area of A. (ii) ( ) 5 0 f (2 ) n nh h = +∑ (6, f(6)) y 2 O x 6 2+h 2+2h 2+3h 2+4h 2+5h

4 (iii) Upper bound of area of A = ( ) 266 2 3 11 22f(2 2 e 291433 n nn nh h n  + = = += + ≈  ∑∑ Lower bound of area of A = ( ) 255 2 3 00 22f(2 2 e 131133 n nn nh h n  + = = += + ≈  ∑∑ (iv) Exact area of A 6 2 66 2 2 662 2 e d e ed 6e 2e e x xx x xx xx= − =−−  ∫ ∫ = 625e e− units2 (v) 5 (a) (i) Given = ••rq pq . R represents any point on the plane that is perpendicular to q and containing the point P. (ii) Given = ××rq pq ×−×=rqpq 0 ( )− ×=rp q 0 ( ) ( ) // , , kk kk − −= ∈ ∴=+ ∈ rp q rp q rp q   R represents any point on the line containing the point P and parallel to q . Let ux= and d ed xv x = . Then d 1d u x = and exv= (6, g(6)) y = g(x) 2+h 2+2h 2+3h 2+4h 2+5h