RVHS 8865 2022 Prelim Solutions vetted

1 ©RIVER VALLEY HIGH SCHOOL 8865/01/2022 Section A: Pure Mathematics [40 marks] 1 Solution [4] Inequalities For ( ) 21 24 0k x x k− − −  , Discriminant < 0 ( ) ( )( ) 2 2 2 1 0 & 24 4 1 0 1 & 24 4 4 0 1 & 6 0 1 & ( 3)( 2) 0 1 & 2 or 3 k k k k k k k k k k k k k k k −  − − − −   + −   − −   − +   −  taking intersection, 3k  2 Solution [7] Differentiation & Integration Techniques (i) Let ( ) 3 212yx=− ( )( ) ( ) 11 22 d3 2 1 2 3 1 2d2 y xxx = − − =− − (ii) ( ) ( ) ( ) 1 2 1 2 1 d 1 2 d 12 12 1 22 12 x x x x x c xc − =− − −=+ − =− − +  (iii) ( )6 5 3 1 2 2xx− =− − − ( ) ( ) ( ) ( ) 3 2 3 2 3 1 2 265 dd 1 2 1 2 23 1 2 d 12 13 1 2 d 2 d 12 1 2 2 1 2 1 2 2 1 2 xx xx xx xx x x x x x x x c x x c − − −− = −− = − − − − = − − − − = − − − − + = − + − +   

2 ©RIVER VALLEY HIGH SCHOOL 8865/01/2022 3 Solution [8] Solving Equations using GC (i) 22 2 2 11e 8 e4 4 2 for stationary points, let 0 11 e042 1e 2 1ln( )22 1ln or ln 44 xx x x x dyy dx dy dx x x = − +  = − = −= = = =−  (ii) (iii) 2 2 e ln( )4 e 8 8 ln( )4 Sketch =8 ln( ) from GC, 8.51 or 2.48 x x x x x x yx x − =− − − + = − − −− =− − 4 Solution [8] Graphing + Appln of Differentiation & Integration (i)

3 ©RIVER VALLEY HIGH SCHOOL 8865/01/2022 (ii) ( ) 12 1 2 1 2 3 2e d 2 2 e 4ed x xx y y x − −− =+ = − =− When 2x= , ( ) ( ) 1 2 2 3 1 2 2 3 3 2e 3 2e d 4e 4ed y y x − − − − = + = + =− =− Equation of tangent at x = 2: ( ) ( ) ( ) 33 3 3 3 33 3 2e 4e 2 4e 8e 3 2e 4e 3 10e yx yx x −− − − − −− − + =− − =− + + + =− + + (iii) Required area ( ) ( ) 2 1 2 3 3 0 2 3 2e 4e 3 10e d 2.07105 2.07 units x xx− − −= + − − + + =   5 Solution [13] Graphing + Application of Differentiation (i) 32 2 30 585 1980 12000 d 90 1170 1980d x m m m x mmm = − + + = − + At stationary point, ( ) ( )( ) ( ) ( ) 2 2 d 0d 90 1170 1980 0 1170 1170 4 90 1980 2 90 1170 810 2 90 2 or 11 x m mm m = − + = −= = = When 2m= , 13860x= When 11m= , 2925x= 2 2 2 d 90 1170 1980d d 180 1170d x mmm x mm = − + =−

4 ©RIVER VALLEY HIGH SCHOOL 8865/01/2022 When 2m= , ( ) 2 2 d 810 0d x m =−  ( )2,13860 is a maximum point When 11m= , ( ) 2 2 d 810 0d x m = ( )11,2925 is a minimum point (ii) (iii) Required area ( ) 12 32 0 30 585 1980 12000 d 105120 m m m m= − + + =  The total production of fishballs by Todo Fishball Company for the fiscal year 2021 is 105120 kg. (iv) When 0d = , 40 5 35y= − = Therefore, an employee produces 35 kg of fishball immediately after the training programme. (v) The model is suitable as the employee shows gradual improvement in the efficiency over time after the training and the improvement tapers off which is realistic. d y 35 40

5 ©RIVER VALLEY HIGH SCHOOL 8865/01/2022 Section B: Statistics [60 marks] 6 Solution [6] Probability (i) (ii) P( hit bull’s-eye in his 2nd throw) = 0.8 x 0.9 + 0.2 x 0.8 = 0.88 (iii) P(hit bull’s-eye on 1st throw | hits bull’s-eye on 2nd throw) 0.8 0.9 0.88 90.818 (or )11 = = 7 Solution [6] Permutations and Combinations (i) Required number of 7-letter code-words = 75 = 78125 _ _ _ _ _ _ _ (each of blank _ can be filled by any of the 5 letters AUXYZ) (ii) Required number of 7-letter code-words = 46 = 4096 E.g. AUXZAUX or AUXZUXY _ _ _ Z _ _ _ (each of blank can be filled by any 4 of the letters A, U, X, Y) (iii) [Note: The first 4 letters A, U, X, Y are fixed. There is only one way to do/fix that.] For *** , choose 1 letter from 5 to be identical: 5 1   , then from remaining 4 choose 1 to be the different letter: 4 1   (e.g. AAU, UUX, UXX etc)  No. of ways = 54 2011    =       OR 52 2021    =      

6 ©RIVER VALLEY HIGH SCHOOL 8865/01/2022 (i.e. choose 2 letters from 5, then from the 2 chosen decide which 1 to be the identical.) (iv) Case 1: all 3* identical letters ( AAA, UUU, XXX, YYY or ZZZ) No. of ways = 5 Case 2: 2* identical, 1 different letter No. of ways = 20 (from part (iii)) Case 3: all 3* different (AUX, UXZ etc) No. of ways = 5 3   = 10 Required number of codewords = 5+20+10 = 35 8 Solution [9] Binomial distribution (i) Each student has the same probability of 0.08 of being left - handed. OR The event that a student is left-handed is independent of another student. (ii) Expected number = 30 x 50 x 0.92 = 1380 students (iii) Let X be the number of left-handed students in a class of 30. ~ B(30,0.08)X Probability ( ) ( 4) 1 ( 3) 1 0.784206 0.215794 0.216 3 s.f. PX PX = = −  =− = = (iv) Let W be the number of left-handed students in a lecture theatre of 250 students, and c the number of chairs in the lecture theatre for left-handed students. Then ~ B(250,0.08)W Want c such that ( ) 0.9P X c . Using GC, ( 25) 0.8971 ( 26) 0.9306 ( 27) 0.9547 PX PX PX = = = Hence, a minimum number of 26 such chairs are needed in each lecture theatre in order to be 90% certain of meeting the needs of the left-handed students.

7 ©RIVER VALLEY HIGH SCHOOL 8865/01/2022 9 Solution [7] Probability A and B not mutually exclusive 0xy +  ( ) ( ) ( )( ) ( )( ) 22 | 5 15 25 55 5 55 15 25 275 60 5 375 40 25 5 5 P A S P A x x y x x y x x y x y x x x y xy x x y xy xy yx = + + +=+ + + + + + = + + + + + + + = + + + + =+ =− Alternatively ( ) ( ) ( ) ( )( ) ( )( ) 22 5 15 25 55 55 55 5 55 15 25 275 60 5 375 40 25 5 5 P A S P A P S x x y x x y x y x y x x y x y x x x y xy x x y xy xy yx = + + + +=+ + + + + + + + + = + + + + + + + = + + + + =+ =− (i) ( ) 3| 8 35 8 25 7 2 P B S x x x y = += + = = (ii) ( )( ) 30 37P' 55 64 xS A B xy +  = = ++ 10 Solution [10] Correlation and Linear Regression (i)

8 ©RIVER VALLEY HIGH SCHOOL 8865/01/2022 (ii) Using the GC, 0.988r=− Since r is close to -1, there is a strong negative linear correlation between the age of a bicycle and its price. (iii) 1.23 119.59yx=− + (to 2 d.p.) 119.59c= means that the resale value of a brand new bicycle of that particular model is $119.59. (iv) When 72x= , ( )1.23216 72 119.590 30.87y=− + = (OR ( )1.23 72 119.590 31.03y=− + = ) The cost of the 72 month old bicycle will be around $30.87. This estimate is unreliable as 72x= is outside the data range of the values of x. 11 Solution [10] Hypothesis Testing (i) Unbiased estimate of the population mean, 539 15.435x == Unbiased estimate of the population variance,

9 ©RIVER VALLEY HIGH SCHOOL 8865/01/2022 ( )2 2 2 21 34 35 1 5398647.434 35 10.2 x sx   − =−  = =  (ii) Let µ be the population mean travel time from Town A to Park B. To test H0:  = 14.5 against H1:  > 14.5 at 5% significance level Test statistic: Under H0, 2 N 14.5, 35 sX   approximately by Central Limit Theorm since 3035n=  is large, ( )14.5 ~ N 0,1 / 35 XZ s −= approximately. p-value = 0.0477 (3 s.f.) As p-value < 0.05, we reject H0. There is sufficient evidence at 5% significance level that the mean travel time from Town A to Park B is greater than 14.5 minutes. (iii) At 5% level of significance means there is a probability of 0.05 that the test will conclude that the mean travel time from Town A to Park B is more than 14.5 minutes when in fact it is 14.5 minutes. OR At 5% level of significance means there is a probability of 0.05 that the test will indicate to reject the claim that the mean travel time from Town A to P