JPJC_8865_2022_Prelim_Solutions_vetted

2022 JPJC J2 H1 Maths Preliminary Examinations Solutions: 1 22 4 1 0kx x x k+ − + −  ( ) ( ) 21 4 1 0k x x k+ − + −  Discriminant < 0 and 10k+ ( ) ( )( ) 2 4 4 1 1 0kk− − + −  1k − ( ) ( )( ) 2 2 2 16 4 1 0 20 4 0 50 5 5 0 5 or 5 k k k kk kk − −  − − + −  −  Since 1k − , therefore 5k  { : 5}xx 2(i) 2 d 3 1 4 x x− ( ) ( ) ( ) ( ) ( ) 1 2 1 2 1 2 1 2 2 1 4 d3 142 ,13 42 142 32 1 143 1 14 3 where C is an arbitrary constant xx x C x C xC xC −=−   −=+  −  −=+ −  =− − + =− − +  k

2 2(ii) 2 24 24 5 253e 9e 30ee xx xx  − = − + 449e 30 25exx −= − + ( ) ( ) 2 2 2 44 44 d5 3ede 9 4e 25 4e 36e 100e x x xx xx x − −  − = + − =− Hence, 36, 100pq= =− 3(i) Let the original prices of Set Meals A, B and C be a, b and c respectively 11 51.80 13 2 73.4022 a b c abc a b c + − = + + = + + = Using GC, 14.50, 16.90, 20.40abc= = = Therefore, the original prices of Set Meals A, B and C are $14.50, $16.90 and $20.40 respectively. 3(ii) Total amount they will spend if they use the membership discount to pay ( )0.80 14.50 2 16.90 2 20.40 71.28 = +  +  = As $71.28 < $73.40, the total amount that they need to pay using membership discount is lesser than using the sales promotion. Hence, they should pay for the food that they intend to order using the membership discount. 4(i) 4(ii) ( )5 ln 2yx= + − d1 d2 y xx=− − x y 0 When 0x= , 5 ln 2y=+ When 0y= , ( ) ( ) 5 5 5 ln 2 0 ln 2 5 2e 2e x x x x − − + − = − =− −= =− Alternative method: ( ) ( )( ) ( ) 2 2 2 222 2 2 2 2 44 44 d5 3ede d 3e 5ed 2 3e 5e 6e 10e 2 18e 30 30 50e 36e 100e x x xx x x x x xx xx x x − −− − −  − =− = − + = + − − =− Hence, 36, 100pq= =−

3 When 1x=− , ( ) d 1 1 d 2 1 3 y x =− =−−− , ( )5 ln 2 5 ln 3yx= + − = + Equation of tangent: 1 3y x c=− + , where C is an arbitrary constant Substitute ( )1,5 ln 3−+ : ( )15 ln 3 1 3 14 ln 33 c c + =− − + =+ Therefore, equation of tangent is 1 14 ln 333yx=− + + , where 1 3m=− and 14 ln 33c=+ 4(iii) Area = ( ) 0 3 5 ln 2 d xx − +− = 18.7 unit2 5(i) ( )1 e2 btPa=+ When 0t= , 2P= : ( ) 012 e 32 aa= +  = When 1t= , 1.75P= : ( )11.75 3 e2 b=+ 1e 2 1ln ln 22 b b = = =− 5(ii) Using GC, at 5t= , d 0.0108d P t =− The value means that the population of the endangered birds in the 5th year is decreasing at 0.0108 thousand per year. 5(iii) 5(iv) As t→ , ( )ln 2 e0 t− → , 1.5P→ As observed from the equation and graph, the population decreases to (or approach es) 1.5 thousand in the long run. t P 0 2000

4 5(v) 32 1 5 5 1 1 1110 10 2 2Q t t t t t = + + = + + 3 d 1 1 d 10 Q tt =− At minimum point, 3 3 3 d 1 1 0d 10 11 10 10 2.1544 2.15 Q tt t t t = − = = = = ( ) ( ) 2 1 1 12.1544 0.8231710 2 2 2.1544 Q= + + = t 2 2