2015 H1 A Level Mathematics Solution_for 8865 (SAJC)

1 2015 H1 A Level Mathematics Solution Q1. For   22 4 2x k x k    to be always negative, Discriminant < 0 and coefficient of 2 20x        2 2 2 4 4 2 2 0 ( 8 16) 16 0 8 16 0 kk k k k kk            2( 4) 0k But 2( 4) 0k for all real values of k. Therefore there is no solution for 2( 4) 0k and there are no values of k for which   22 4 2x k x k    is always negative. Graph of 2( 4)yx Q2. (i)         5 45 d 3 24 3 4 2 2 1d 2 1 2 1 xx xx       (ii) 2 1 1 2 2x dxx   1 2 1 2 2 1 22 1 2 13 1 2 44 44 443 76 24 x dx x x x dx x x x             

2 Q3. (i) 2 2 12 8 d 12 16d x x y x e y ex     For stationary point, 2 2 2 d 0d 12 16 0 16 12 12 3 16 4 x x x y x e e e        2 3ln ln 4 3 1 32 ln ln 4 2 4 xe xx                 (ii) Area of Region bounded by C =     2 0 22 0 22 2 2 2 12 8 64 6 4 0 4 6 4e + 4 units a x ax a a x e dx xe ae a             Q4. y 2 y 2 2 2 By Pythagoras' Theorem, 3 24 yyy  2 x x 2 2 2 By Pythagoras' Theorem, 3 24 xxx  F A D P D B A B

3 Perimeter of the remaining shape PQRSTU = 30 cm  3 2 3 30 10 10 y x x yx yx           Method 1: Area of the remaining shape PQRST, A = Area of triangle FDE – 3 ( Area of triangle PDQ)         22 22 22 2 2 22 2 11 sin 60 3 sin 6022 1 3 1 3 32 2 2 2 3 34 3 10 34 3 100 20 34 3 50 102 ooyx yx yx xx x x x xx                Method 2: Area of the remaining shape PQRSTU, A = Area of triangle FDE – 3 ( Area of triangle PDQ)             22 2 22 2 1 3 1 3 32 4 2 4 1 3 3 3102 2 2 2 3 1 3 100 202 2 2 1 3 50 10 (shown)2 yxyx x x x x x x xx                                            

4     21 3 50 102 d1 3 10 2d2 A x x A xx     For stationary value of A, d 0d 10 2 0 5 A x x x    2 2 d 20d A x    A is maximum when x = 5. Maximum value of   221 753 50 10(5) (5) 3 cm22A              Q5. (i) At x axis intercept (y = 0) At y axis intercept (x = 0)         0.5 ln 1 0 0.5 ln 1 Using GC, 0.786 x x x x x          0 0.5 ln 1 1y   The coordinates of points of intersection with the axes are ( 0 , 1 ) and (0.786, 0 ). Asymptote : x = -1 (ii) When x = 0.5, y = 0.30164 0.5 d 1.1568 1.157d x y x    ( 3 dec. places) (iii) Equation of normal is not in syllabus. But we can change the question to the following: y x O x = – 1  0.786,0

5 The tangent to C at P meet the x-axis at A and y-axis at B. Find the length AB. [5] Equation of tangent to C at P    0.301