CJC 8865 2022 Prelim Solutions vetted

1 CATHOLIC JUNIOR COLLEGE H1 MATHEMATICS 2022 JC2 PRELIM EXAMINATION SOLUTION Q1 Solution ( ) 2 10kx k x k+ − +  for all real values of x So, graph has no real roots and is U-shaped Discriminant < 0 2 40b ac− and 0k  ( ) ( )( ) 2 1 4 0k k k− −  23 2 1 0kk− − +  23 2 1 0kk+ −  ( )( )3 1 1 0kk− +  Alternatively, If 23 2 1 0kk+ − = , 22 2 4(3)( 1) 2(3) 11 or 3 k −  − −= =− 1 3k  1 (rej 0)kk−  or Set of values of k is 1: 3kk   Q2 Solution Let the polynomial be 32y ax bx cx d= + + + 2d 32d y ax bx cx = + + At ( )1, 13− , 13 --- (1)a b c d+ + + =− 3 2 0 -------- (2)a b c+ + = At ( )1,7− , 7 --- (3)a b c d− + − + = 3 2 12 --- (4)a b c− + =− Solving (1), (2), (3) and (4), 2a= , 3b= , 12c=− , 6d =− 1− 1 3 k

2 Hence the polynomial is 322 3 12 6x x x+ − − Q3 Solution (a) (i) 2d lnd2 xe xx   + ( )( ) 2d ln ln( 2)d xexx= − + ( )d 2 ln( 2)d xxx= − + 12 2x=− + Alternative Method (Quotient rule) 2d lnd2 xe xx   + ( ) ( ) 22 22 222 2 xx x x e ex e x  +−+=  + 12 2x=− + (a) (ii) d2 1d n xx + 1 2 221 n n xx −    = + −       1 2 22 1 n n xx − =− +  (b) 1 1 2 2 2 2 xx x x x x x x +−= − − − − + − ( ) ( ) 22 2 2 xx xx +−= −− 2 2 xx+−= 11 222 dx x x dx xx = + − −− ( ) 3 3 2 221 332 22 xx C   −= + +           , where C is an arbitrary constant ( ) 3 3 2 2 1 23 x x C= + − + 

3 Q4 Solution (i) ( )ln 2 4 2 6y x x= − − + 1 2d 2 1 2d 2 4 2 y xxx −=− −  11 2x x =− − For stationary points, set d 0d y x = So, 11 02x x −=− 11 2x x =− 2xx=− Method 1 Squaring both sides, ( ) ( ) 2 2 2xx =− 2 44x x x= − + 2 5 4 0xx− + = ( )( )1 4 0xx− − = 1x= or 4x= Since 2x , so 4x= Method 2 20xx− + = Let yx= , then 2 20yy− + = ( )( ) 2 20 1 2 0 yy yy − − = + − = So, 2yx== or 1yx= =− (no solution) Hence 4x= 11 02x x −=− 11 2x x =− When 4x= , ( )ln 2 4 4 2 6 ln 4 4 6 2 ln 4yx=  − − + = − + = + So turning point is ( )4, 2 ln 4+

4 (ii) (iii) At 9x= , d 1 1 1 1 1 1 3 7 4 d 2 9 2 7 3 21 21 9 y xx x −= − = − = − = =−−− and ( )ln 2 9 4 2 9 6 ln14 6 6 ln14y=  − − + = − + = Method 1 Substituting point and gradient into y mx c=+ ( )4ln14 9 21 c= − + ( )4 12ln14 9 ln1421 7c = + = + Hence equation of tangent at 9x= is 4 12 ln1421 7yx=− + + Method 2 Hence equation of tangent at 9x= is ( )4ln14 9 21yx− =− − 44 9 ln1421 21 4 12 ln1421 7 yx yx =− +  + =− + + x y O V.H. ( )ln 2 4 2 6y x x= − − +

5 (iv) Area = area under tangent – area under curve ( ) 99 0 2.0212859 4 12 ln14 d ln 2 4 2 6 d21 7x x x x x= − + + − − − + 10.2 (to 3 s.f.)= x y O V.H.

6 Q5 Solution (i) Since N is in hundreds, So, N = 481 ( ) ( ) 5 5 481 500 140e 19e 140 1 19ln5 140 0.399 0.4 (1 d.p.) k k k k =− = =  =− =− (ii) 0.4d 56ed tN t −= d d N t represents the rate of change of recorded influenza cases with respect to time. Or d d N t represents the rate of increase of recorded influenza cases with respect to time. Or d d N t implies that for every increase of 1 day, the number of recorded influenza cases increase by 0.456e− hundreds. (iii) (iv) As t becomes large, 0.4e0 t− → , 500N → So, number of recorded influenza cases will tend to 50 000. No, it is not realistic as it is not possible to have the number of cases stay at a constant as sick people should get well eventually. (or any other reasonable reason.) N t O (0, 360) N = 500 500 140e ktN =−

7 (v) 24 33d 250 7 d 3 3 Q ttt − =− For max/min values, d 0d Q t = 24 33250 7 033 tt − −= 24 33250 7 33 tt − = 24 33250 7tt − = 2250 7 t= 250 7t = (since 0t ) t 5.97 250 7 5.98 d d Q t 0.05203 0 -0.03266 Hence, 250 7t = days gives max value of Q. (vi) (vii) x-coordinates of intersection: 2.3655512, 7.874977 2.37 7.87t   days N t O (17.218, 0) (0, 120) 500 140e ktN =− 17 33250 120Q t t= − +

8 Q6 Solution (i) Consider AB as one group. Number of ways 7! 2!= 10080= (ii) A J S J S J S J or J S J S J S J A Number of ways ( )4! 3! 2=   288= (iii) Probability nbr of ways to select 1 Junior and 4 Seniors nbr of ways to select 5 members= 44 14 8 5 CC C = 1 14= Alternative Method Probability 4 4 3 2 1 58 7 6 5 4=      1 14= Arrange 7 objects Adam and Bernice can swap places Arrange 4 Juniors Arrange remaining 3 Seniors Adam can be on the left or right of row.

9 Q7 Solution (i) (i) Let X be the random variable denoting number of passengers rejected by the facial recognition scanner out of 50 passengers. Then ( )~ B 50,Xp Given ( )Var 3.68X = ( ) ( ) 2 2 1 3.68 50 1 3.68 0.0736 0.0736 0 np p pp pp pp  − =  − =  − =  − + = Method 1 (Using GC to find the roots) Since 0.5p , 0.08p= Method 2 (factorize to find the roots) ( )( ) 2 2 2 0.0736 0 46 0625 625 625 46 0 25 2 25 23 0 pp pp pp pp − + = − + = − + = − − = So, 2 25p= or 23 25p= (reject since 0.5p ) Hence, 0.08p= Method 3 (using GC graph to find the roots) Using GC, since 0.5p , 0.08p=

10 (ii) Let Y be the random variable denoting number of passengers rejected by the iris scanner out of 100 passengers. Then ( )~ B 100,0.07Y P(plane departs on time) ( )P 13Y= ( )P 12 0.9775924796 0.978 (to 3s.f.)Y=  =  (iii) Let W be the random variable denoting number of planes delayed out of 75 planes. Then ( )( )~ B 75,P 13WY  ( )~ B 75,1 0.9775924796W − ( )~ B 75,0.0224075204W P(more than 5 planes delayed) ( ) ( )P 5 1 P 5WW=  = −  0.0068237351 0.00682 (to 3s.f.)= (iv) ( )~ B 75,0.0224075204W w ( )P Ww= 0 0.1827 1 0.3141 2 0.2664 Hence the most likely number of planes delayed is 1.