HCI 9758 2024 Promo Solutions

1 An athlete on a special diet would like to have battered fish fillet, coleslaw and fries for lunch. He must ensure that his intake (in grams) of protein, carbohydrates and fat per meal is 55g, 130g, and 70g respectively. The table below shows the nutritional breakdown for one serving of each item. Calculate the number of servings of battered fish fillet, coleslaw and fries that the athlete should take for his lunch. [3] Suggested Solutions Let x, y and z be the number of servings of battered fish fillet, coleslaw and fries respectively. 20 4 3 55x y z+ + = 20 15 45 130x y z+ + = 15 8 16 70x y z+ + = By GC, x = 2, y = 3, z = 1 Therefore, the runner should take 2 servings of battered fish fillet, 3 servings of coleslaw and 1 serving of fries for his lunch. Protein (in grams) Carbohydrates (in grams) Fat (in grams) Battered fish fillet 20 20 15 Coleslaw 4 15 8 Fries 3 45 16

2 © HCI 2024 2 (a) Given that ( )( ) 2 1 1 2 16 n r nr n n = = + + , find 1 11 23 rn r rr =    −+        , in terms of n. [4] Suggested Solutions (a) ( )( ) ( ) ( )( ) ( ) 1 2 1 1 1 2 11 23 11 23 11122 1 2 1 11 661 2 11 1 2 1 126 111 23 rn r rn n n r r r n n n rr rr nnn n n n nn n n = = = =    −+       = − −  − = − + + − + − = − − + + + = − − +     (b) Explain why 1 11 123 rn r rr =    − +         for all positive integers n. [1] Suggested Solutions (b) Since n + , ( ) 21 0 and 1 023 n n n  +  . Thus 1 11 123 rn r rr =    − +         . 3 The diagram below shows a curve with equation 1yx=− . Let A be the area bounded by the curve and the axes. A total of n rectangles, each of equal width, are constructed to approximate the value of A. 1 0 ⋯

3 © HCI 2024 [Turn over (a) Show that the total area of n rectangles is given by 1 0 1 n r nr nn − = − . [2] Suggested Solutions (a) Area of n rectangles ( ) 1 0 1 0 1 0 1 1 1 11 1 ... 1 11 0 1 ... 1 1 n r n r n n n n n n n nnnn n n n n nr nn nr nn − = − =       −= − + − + + −                 −−−−= + + +  −= −  

4 © HCI 2024 (b) Without using a calculator, evaluate 1 0 1lim n n r nr nn − → = − . [3] Suggested Solutions (b) 1 0 1 0 1 0 1lim Area of infinitely many rectangles Area under the curve from 0 to 1 1 d 1 d n n r nr nn xx xx xx − → = −= = = = =− =− − −    ( ) 1 3 0 2 1 3 2 3 x= − − = 4 (a) Sketch the curve with equation 2 1 xqy x −= − , where 2q  , stating the equations of the asymptotes and the coordinates of the points where the curve meets the axes. [3] Suggested Solutions (a) 2 1 xqy x −= −

5 © HCI 2024 [Turn over (b) Hence, by adding a suitable line on the same diagram in part (a), solve the inequality 2 2 0,1 xq xx − −− giving your answer in terms of q. [3] Suggested Solutions (b) 2 21 xq xx − − To find the x-coordinate of point of intersection: 2 2 21 2 2 2 xq xx x q x x −−= − − + = − 22xq= 2 qx= Since 0,x 2 qx = For 2 2 0,1 xq xx − −− 1x or 1 2 qx

6 © HCI 2024 5 Show that the distance L between any point ( ),xy on the curve 2e3xy=− and the fixed point ( )2, 1A − satisfies the equation ( ) ( ) 2222 2 e 2 xLx= − + − . [1] Using differentiation, find the x-coordinate of the point P on the curve that is closest to the point A, leaving your answer in 4 decimal places. You do not need to show that P is closest to A. [3] A variable point Q moves along the curve 2e3xy=− such that its x-coordinate is increasing at a rate of 0.5 units per second. Find the exact rate of change in L when Q is on the y-axis. [3] Suggested Solutions ( ) ( )( ) ( ) ( ) 22 2 2222 2 e 3 1 2 e 2 (Shown) x x Lx Lx = − + − − − = − + − Method 1 ( ) ( ) ( ) ( )( ) ( ) ( )( ) 2222 22 22 42 42 2 e 2 d2 2 2 2 e 2 2ed d 2 e 2 2ed 2 2e 4e 2e 4e 2 x xx xx xx xx Lx LLx x LLx x x x = − + − = − + − = − + − = − + − = − + − Method 2 ( ) ( ) 22 22 e 2 xLx= − + − ( ) ( )( ) ( ) ( ) 22 22 2 2 2 2 e 2 2ed d 2 2 e 2 xx x xL x x − + − = − + − ( ) ( ) 24 22 2 d 2e 4e 2 d 2 e 2 xx x Lx x x − + −= − + − For shortest distance, d 0d L x =

7 © HCI 2024 [Turn over 422e 4e 2 0xx x− + − = By GC, 0.42445 0.4245x= (to 4 dp) Method 3 To minimise L, we can minimise 2L ----- (*) ( ) ( )( ) 2 22d 2 2 2 e 2 2ed xxL xx = − + − ( ) ( )( ) 220 2 2 2 e 2 2e xxx= − + − By GC, 0.42445 0.4245x= (to 4 dp) Method 1 When x = 0, ( ) ( ) 22 2 1 2 5L= − + − = ( ) ( ) ( )d5 2 1 4 1 2d d4 d 5 L x L x = − − =− d d d d d d 4 0.5 5 2 5 L L x t x t= =−  =− Method 2 When x = 0, ( ) ( ) ( ) ( ) 22 2 1 4 1 2d4 d 52 1 2 L x −−= =− − + − d d d d d d 4 0.5 5 2 5 L L x t x t= =−  =−

8 © HCI 2024 Method 3 ( ) ( ) ( ) ( )( ) ( ) ( )( ) 2222 22 22 2 e 2 d d d2 2 2 2 e 2 2ed d d dd 2 e 2 2edd x xx xx Lx L x xLx t t t LxLx tt = − + − = − + − = − + − When x = 0, ( ) ( ) 22 2 1 2 5L= − + − = Substitute 0x= , 5L= and d 0.5d x t = : ( ) ( ) ( )( ) ( ) 00d5 2 e 2 2e 0.5d d2 d 5 L t L t = − + − =− L is decreasing at a rate of 2 5 unit per second. 6 The curve C has equation 22ln y x y= , where , 1 0xy −   . It is given that C has only one turning point. (a) Show that ( ) 22d 22d y x y xyx −= . [3] Suggested Solutions (a) ( ) 2 2 2 22 1 d d22 dd d2 2d d 22d yyy xy xy x x y x xyxy y x y xyx  = +  −= −=

9 © HCI 2024 [Turn over (b) Find the coordinates of the turning point. [2] Suggested Solutions (b) For turning point, d 0d y x = . 220xy= Since 0y , 0x= . When 0x= , 2ln 0 1 y y =  = Since 10 y−   , the turning point is ( )0, 1− . (c) Calculate the value of 2 2 d d y x at the turning point and determine whether the turning point is a maximum or a minimum. [3] Suggested Solutions (c) ( ) ( ) 22 2 2 2 2 2 d 22d d d d d2 2 4 2d d d d y x y xyx y y y yx y x xy xy yx x x x −= − + − − = + Substitute 0x= , 1y=− and d 0d y x = : ( ) ( ) 2 2 2 2 2 d 2 2 1d d 10d y x y x =− = Therefore ( )0, 1− is a minimum point. 7 (a) Describe a sequence of two transformations that will transform the graph of ( )fyx= to the graph of ( )fyx =+ , where  and  are positive constants. [2] Suggested Solution (a) Translate  units in the negative x-direction. Scale parallel to x-axis by a scale factor of 1  . OR Scale parallel to x-axis by a scale factor of 1  .

10 © HCI 2024 Translate   units in the negative x-direction. Diagram 1 and Diagram 2 show the graphs of ( )fyx= and ( )fyx =+ respectively. The turning points with coordinates (0,0) and (2,4) on ( )fyx= correspond to the points with coordinates ( 2,0)− and (2,4) respectively on ( )fyx =+ . The asymptotes 1x= and yx =+ on ( )fyx= correspond to the asymptotes 0x= and yx   = + + respectively on ( )fyx =+ . (b) Find the value of  and of . [2] Suggested Solution (b) Method 1 By observation, 1 ,12== Method 2 ( )0 2 2 ----- I  − =−  = ( )2 2 2 2 ----- II  − =  − = Solving (I) and (II), 1 ,12== . Diagram 1 Diagram 2