VJC 2023 JC1 Promos 9649 (Solutions)

1(a) 3 3 2 2 1 10 (1 ) ( 1 ) 0 Since is non-real, it satisfies 1 0. w w ww w ww w        2 1.ww   - GC is not allowed and given 2 marks, the intent is not to solve for w. (b)    22 22 2 2 2 32 (1 2 3 )(1 2 3 ) 3 LHS: 1 2( ) 1 2( ) 12 ( 1 ) 12 ( 1 ) 11 () 1 1( 1 )1 3 ww w w ww w w w w ww ww ww w                     Alternative  22 3 4 2 3 23 4 23 4 2 LHS = (1 2 3 ) (2 4 6 ) (3 6 9 ) 1 5 11 13 6 1 5 11 13 6 1 14 11 11 14 11( 1) 3 RHS ww w w w ww w ww ww ww w w w w ww                       - Make good use of result in (a) to simplify. - Note the efficient way to evaluate 4.w 2(a) 22 22 22 2 22 2 2 distance from ( , ) to (0,2 ) = ( 2 ) distance from ( , ) to line = (2 ) (2 )( ) (2 )( ) () () (2 ) (2 ) (2 ) ( s h o w n ) xy d x y d xy x d xy d x d xy d x d yd x d x xd x xd x dxd yd d dx                  From definition, d > x. |x – d| = d – x

(b)  3 3 23 33 2 32 3 22 3343 22 Volume = 2 d 2 d (2 )2 d 22 d( 2 ) d () ( 2 ) ( 2 ) d2 (2 ) (2 4 2+ )922 2 3 4 d d d d d d dd dd d d dd dd dd xy y xy y dy d yyd dy y y y dy d yd y yd yd y d dy d y ddd d d dd                                     34 4 3 3 3 2 43 8 cu. units3 dd ddd d      Alternative (Disc Method) 2( 2) ( 2) 2 ( 2)yd d dx yd d dx      Required volume = volume generated by upper curve Uy – volume generated by lower curve Ly  22 22 00 222 0 2 0 2 0 3/2 2 0 3 3 dd 2( 2 ) 2( 2 ) d 8( 2 ) d 182 ( 2 ) d2 (2 )4 3/2 8 03 8 cu. units3 dd uL d d d d yx yx dd d x dd d xx dd d x x dd d d x xd dd x d d                             Alternative: Expand 2 (2 )yy d ; the method shown is simpler to evaluate limits. By shell method, 2 2 2 2 (2 ) (2 ) (2 )2 (2 ) 22 Ax y y yd d dx yddx d dy dx d       

3(a)            22 22 3 2 2 2 122 2 21 23 21 23 4646 2 21 23 2 0 (2 1) & (2 3) 0 for 021 23 kk k kkk kk k k kkkk kk kk kkk               22 2 21 23 kk kk   (b) Let Pn be the proposition that 2 , .21 n n nun n    . Consider P1: LHS 1 2 given3u RHS  2(1) 2 21 1 3  Hence 1 2 3u  1P i s t r u e Assume that Pk is true for some positive integer k, 2 21 k k ku k   . To show Pk+1 is true, 1 22 23 k k ku k    LHS of Pk+1 1 1 22 23 22 2 23 21 22 22 2 22 since 23 23 21 23 22 23 k k k k k u k uk kk kk kk k k kk k k k k                     1 1 22 23 k k ku k       Pk is true 1Pk is true Since P1 is true and Pk is true 1Pk is true, by mathematical induction, P is true for all positive integers .n n implies tends to maps to    4(a) 1 1 1 1 f( ) f '( ) f( )By Newton-Raphson method, f' ( ) kk n nn n k n nn k n xxa xk x xxx x xaxx kx          

1 11 1 1 1 ( 1) (shown) k n nn kk nn n n k n n k n x axx kx kx x ax kk x akxkx           A1 (b) 33Since 25 27 3, 3 is a reasonable initial estimate. B1 (c) 1 2 3 4 5 3 Using 3, from GC, 2.92592... 2.924018... 2.924017... 2.924017 23 2.924 (3 dp) x x x x x        M1 A1 (d) Question stated to sketch suitable graphs (hence more than 1), so the graphs of yx and 2 12 523yx x   are expected 5 21 422nn nvvv   , 1 = 2,v 2 = 0v     21 21 21 11 2 2 21 4 becomes 22 4 22 4 For this equation to be homogeneous, 4. 42 , 44 . 22 0 22 nn n n nn n nn n nn n nn n vwkvk w kw kw kw www k k wv wv www vvv                      Auxiliary equation is 2 22 0  24 8 1i2     1 2 12 2c o s s i n 44 22 44 2 ( 0 ) 2 0 22 s i n 4 42 s i n 4 n n n n n n nnwA B wA B vB B A nw nv                It is totally unthinking to write auxiliary equation as: 2 22 4 . 

(b)  42 12 42 22 22 22 (4 2)42 s i n 4 42 s i n 2 1 2 4 2 ( 1), is odd 4 2 (1), is even 4(1 4 ), is odd 4(1 4 ), is even n n n n n n n nv n n n n n                 6(a) x 3 4   5 4  2y 0 1 4 0 5 4 2 3 4 53 1144 d 0 0 4 0.262 (3sf)64 1 2yx             (b) 5 4 2 3 4 1Area of the inner loop = d2 1 0.131 (3sf)21 2 2 4 r       7(a) Locus is an ellipse. (b) 22 22 22 2 2 22 2 2 4 4 cos 2 (use cosine rule to find ) 44 c o s2 44 c o s 44 41 c o s 4 ( ) (shown) 1c o s OG GF a rr c r c a G F rcr c a r rcr c aa r r car a c a ca ar c a                   2 and .ccpa q aa  A1 p A1 q 

(c) q represents the eccentricity of the conic section. (d) 2 2 0 2 2222 40 dArc length = d d sin d 1c o s 1c o s rr ccc aa aaa c c a a                      For small values of c, the ellipse approaches the circle with centre at O and radius a. arc length . a Alternative: 2 2222 22 40 sin 0lim 111c o s 1c o s c ccc aa aa aaa ac c a a                 0 arc length d aa      8(a) Number of ways 47900160012! (b) WWWWW MMMMM MMMMMMM WWWWWMM Number of ways = 7 55! 7! P 5! 2! 3 2419200   (c) Consider complementary cases involving two particular girls are adjacent to each other. Case 1: Both seated in front row 10 32! 4! 7! 29030400C   Case 2: Both seated in back row 10 52! 6! 5! 43545600C   Number of ways 479001600 29030400 43545600 406425600    (d) Number of ways = 12 8 4 444 57753! CCC  9 (a) 66 40 01 66 4 2 40 6 / / 3 0 1 24 12 26 2 equation of plane is 3 0 3 12 12 0 12 Cartesian equ AB AC n ABC r                                                            ation is 2 3 12 12.xy z 

(b) Consider line OD: 2 0, 3 r     When OD intersects plane ABC, 2(2 ) 3(0) 12(3 ) 12.   0.3  point of intersection is (0.6, 0, 0.9). Answer in coordinates form, not position vector. (c) Shortest distance = 02 11 0 8ˆ 03 4 9 144 157 91 2 AD n               units. (d) Area of plane ABC = 11 628 15722AB AC      Volume of tetrahedron = 1 108157 36 cu. units.3 157  10(a) 2 22 2 ,2 , dd 2, 2dd d1 d d1Gradient of d 1tan gradient of (shown) 202Gradient of 1 2tan 1 P x at y at a t a xy at att y xt yTS xt TS t at tQP at a t t t                2 2 2 2 2 tantan 1 1 tan 2t a n tan (1 tan ) 2ta n 1t a n tan 2                Since 0, , 2 ( s h o w n )     When tan A = tan B, there are many possible relationships between A and B. A good answer should include justifying why A = B. (b)  2 (corresponding angles) 2 (sum of angles on a straight line) (shown) QPR TPQ TPQ             (c) The inner reflective surface of the car headlight is to have a parab