2013 H1 Mathematics 8864 (SAJC)

1 2013 H1 Mathematics 8864/01 Section A 1. Find the set of values of k for which the equation     2 2 2 1 0x k x k     has no real roots. [4] 2 2 2 2 ( 2) (2 1) 0 0 ( 2) 4(1)(2 1) 0 4 4 8 4 0 12 0 ( 12) 0 x k x k D kk k k k kk kk                  { : 0 12}kk   2. (i) Differentiate 2ln (1 2 )x . [2] (ii) Use a non-calculator method to find the exact value of   0 41 1 d 13 x x  . [4] (i)   2ln 1 2d xdx  2 2 14 12 4 12 x x x x     (ii)         0 41 0 4 1 0 4 1 03 1 3 1 13 13 1 ( 3) 1 33 131 33 11194 7 (0.109375)64 dx x x dx x dx x                      

2 3. A piece of card has the shape of a trapezium ABCE. The point D on CE is such that ABCD is a rectangle. It is given that AB = y cm, BC = 4x cm and DE = 3x cm (see diagram). The area of the card is S cm2. Given that the perimeter of the card is 20 cm, (i) find an expression for S in terms of x, [3] (ii) find the maximum value of S, justifying that this value is a maximum. [3] (i) By Pythagoras Theorem, length of EA =     22 3 4 5x x x 20 3 2 4 5 12 2 10 6 x y x x xy yx            2 2 2 1 3 4 42 64 6 4 10 6 40 18 S x x xy x xy x x x xx       d 40 36d S xx  When S is maximum, d 0d S x  Therefore, 40 36 0 10 9 x x   Using second derivative test, A B E D C y 4x 3x

3 2 2 d 36 0d S x   for all x. When 10 9x , 2 2 d 36 0d S x   Therefore, S is max when 10 9x . 2 2 10 10Max value of 40 18 99 200 2 22 cm99 S             4. The curve C has equation 32 36y x ax x    , where a is a constant. (i) Find, in terms of a, the gradient of the normal to C at the point P where x = 1. [3] [This part is not in syllabus] The normal at P passes through the point ( -5, 3 ). (ii) Show that a satisfies the equation 2 10 24 0aa   and hence find two possible values of a. [5] (iii) For the smaller value of a, find the coordinates of the point of intersection of the normal at P and the line y = x. [2] i) 32 36y x ax x    2d 3 2 3d y x axx    At P, 1x d 3 2 3 6 2d y aax      Therefore, the gradient of the normal at 1 62P a  ii) When 1, 1 3 6 10x y a a       At P, 5x , 3y

4 The normal passes through (1, 10 – a) and (-5, 3). 1 (10 ) 3 6 2 1 ( 5) 17 6 2 6 a a a a          2 2 7 6 2 6 2 20 48 0 10 24 0 (shown) aa aa aa              2 2 10 24 0 10 10 4 1 24 2 4 or 6 aa a aa      iii) Smaller value of 4a , Eqn of normal is : Equation of the normal at P :     13 ( ( 5))62 13 ( ( 5))6 2 4 1 11 (1)22 (2) yx a yx yx yx              Solving (1) and (2), x = 11 , y = 11 The point of intersection of the normal at P and the line y = x is (11, 11) 5 (i) By taking logarithms, find the exact root of the equation 2 2 2 2xxee  [3] (ii) Use differentiation to show that the curve C with equation 22 2xxy e e  has a stationary point at ( 2, 2e ). [3] (iii) Sketch C, stating the exact value of the x-coordinate of its point of intersection with the x-axis.[2]

5 (iv) Use your calculator to find the area of the region bounded by C, the x-axis and the lines x = 0 and x = 1. [1] i) 22 2xxee      22ln ln 2 2 2 ln 2 ln 2 2 ln 2 2 ln 2 xx x ee xe xx x           ii) 22 2xxy e e  22d 22d xxy eex   For a stationary point, d 0d y x    22 22 22 2 2 2(2) 2 2 2 2 2 0 20 0 (no solution) or 0 2 2 Stat point is (2, ).(Shown) xx xx xx x ee e e e e e e ee x y e e e e                       iii) Intercepts : When x = 0, y = e2 – 2 When y = 0, 2 ln 2x (from (i)) Asymptotes: As 2 2 2, 0 0 0 xx exy ee      Stat point: 2(2, ) e (from (ii) iv)

6 Area of the region bounded by C =   1 2 2 2 0 2 1.93 units (from GC)xxe e dx  Section B [This part is not in syllabus] 6. Suky is organising a pop concert. She sells 5000 tickets at $X each, 10000 tickets at $Y each and 15000 tickets at $Z each. Suky wants to find out whether those who bought the tickets thought that the price they paid was good value for money. She decides to do this by choosing a stratified random sample of size 150. (i) Describe how Suky might choose her sample. [3] (ii) State one reason for using stratified random sampling in this context. [1] i) Divide the 30000 ticket holders into three strata based on the price of the tickets – for $X each , $Y each and $Z each. For $X ticket holders, 5000 150 2530000 For $Y ticket holders , 10000 150 5030000 For $Z ticket holders, 15000 150 7530000 Within each stratum, use simple random sampling to choose her sample. For example, for the stratum $X ticket holders, list them from 1 to 5000. Use a computer to generate 25 random numbers and select the ticket holders corresponding to the numbers generated. Repeat the process for the other two strata. ii) Stratified sampling will ensure that the sample of ticket holders would be a good representative of the different groups of ticket holders with different ticket values. Suky can then analyse whether each group of ticket holders (for $X each, for $Y each and $Z each) thought that the price they paid was good value for money. 7. A particular type of electronic device is being tested to determine for how long information stored in it is retained after power has been switched off. A random sample of 250 such devices is chosen and the time, T hours, for which information is retained is measured for each one. The results obtained are summarised as follows:  75 305t   2 75 29555t Find unbiased estimates of the population mean and variance. [3]

7 This type of device has previously been considered capable of retaining information for 75 hours, on average, after power is switched off, but the manufacturers now claim that information is retained for longer than this. Test at the 12%2 significance level whether the claim is justified. [4] 8. A shop sells batteries in packs of 10. An advertiser claims that individual batteries each have a lifetime of at least 100 hours. The probability that an individual battery has a lifetime less than 100 hours is 0.2, independently of all other batteries. (i) Find the probability that, in a randomly chosen pack of 10 batteries, each of the batteries satisfies the advertiser’s claim. [1] Customers are satisfied if at least 8 of the batteries in a pack have a lifetime of at least 100 hours. (ii) Find the probability that a randomly chosen pack will satisfy customers. [3] [This part is not in syllabus] A customer buys a batch of 80 packs of these batteries. (iii) Using a suitable approximation, estimate the probability that at least 75% of packs in the batch will satisfy the customer. State the mean and variance of the distribution that you use. [4] ( 75) 75250 305 75250 76.22 tt     2 2 1 (305)29555250 1 250 117.2004016 117 s     Let X be the r.v. “retained time for the particular type of electronic device after which the device has been switched off” and  be the population mean. 0 1 : 75 : 75 H H     Under 0H , since 250 50n is large, 117.2004~ (75, ) approximately by CLT250XN p-value = 0.0374 > 0.025 Therefore, we do not reject 0H and conclude that at 2.5% significance level, there is insufficient evidence that the retained time for the particular type of electronic device after which the device has been switched off is longer than 75 hours. Thus, the manufacturer’s claim is not valid.

8 i) Let X be the r.v. “no of batteries each have a lifetime of less than 100 hours out of 10 batteries”. ~ B(10,0.8)X P( 10) 0.107 (3 sig. fig.)X  ii) P(a pack of 10