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