2016 H1 A Level Mathematics Solution edited 2020 (SAJC)

1 2016 H1 A Level Mathematics Solution Section A Q1. (i)         2 22 d 2 122ln 3 4 6d 3 4 3 4 xxxx xx     (i)         3 23 d 1 1 3 2 3 1 3d2 2 1 3 1 3 xx xx        Q2. Substituting xue into inequality 22 9 3xxee  22 9 3xxee  becomes 22 9 3uu    22 3 9 0 3 2 3 0 uu uu       33 or 2 33 or e 2 xx uu e     Since 0 xe  for x , 3 xe  is rejected. Hence, 3 2 3ln 2 xe x    Q3. (i) y x (0,1) (0.703,0) -3 3/2

2 (ii) From GC, gradient of C when x = 0.5 is -1.61 ( 3 sig. fig. ) (iii) OUT of SYLLABUS when x = 0.5, gradient of the normal to C at the point = 1 0.6224591994 0.622 (3 sig.fig.)1.606531    At point (0.5, 0.3565307) Equation of the normal C when 0.5x  0.35653 0.62246 0.5 0.622 0.0453 (3 s.f) yx yx     (iv)   2 0 k xe x dx  3 0 3 0 3 3 0 33 1 3 k x k k xe kee ke                  Q4. (i) 23 2 1 6 3 4 d 6 6 12d y x x x y xxx        For stationary point, 2 d 0d 6 6 12 0 1 or 0.5 y x xx xx       When 1 , 4xy  When 0.5 , 2.75xy Coordinates of stationary points on the curve are ( -1, -4) and (0.5, 2.75). (ii) 2 2 d 6 24d y xx   For 1x

3   2 2 d 6 24 1 18d y x     ( > 0) (-1, -4) is a minimum point. For 0.5x   2 2 d 6 24 0.5 18d y x    ( < 0 ) (0.5, 2.75) is a maximum point. (iii) The coordinates of the points where the curve crosses the x-axis are ( -1.59, 0 ) , ( -0.157, 0) and (1, 0). (iv)

4 Numerical value of the area under C between x = 0.5 and x =1 1 2 3 2 0.5 1 6 3 4 d 0.9375 unitx x x x     Q5. Since DEF and is an equilateral triangle, 60oEFD Area of the remaining shape ABEDFCA = Area of triangle ABC – Area of triangle EDF       22 22 112 sin 60 sin 60 properties of equilateral triang les22 33 4 xy xy     Alternative Height of Triangle ABC = 22(2 ) 3x x x [Pythagoras Theorem] Height of Triangle EDF= 2 2 3() 22 yyy  Area of the remaining shape ABEDFCA = Area of triangle ABC – Area of triangle EDF   22 1 1 3(2 )( 3 ) ( )2 2 2 33 4 yx x y xy   Given that Area of the remaining shape ABEDFCA = 23 cm2. cos 60 / 2yy  2 cos60xx  Area of triangle = ½ absinC Area of triangle = ½ (base)(height)

5   22 22 22 33 2 3 4 1 24 4 8 shown [Equation 1] xy xy xy    (ii) Given that perimeter of the shape ABEDFCA = 10 cm    2 2 2 2 10 4 2 2 10 6 10 x y x y x y x y xy          Substitute 10 6yx into Equation 1     22 22 2 2 4 10 6 8 4 100 120 36 8 32 120 108 0 32 120 108 0 1.5 or 2.25 xx x x x xx xx xx                For 1.5 , 1xy For 2.25 , 3.5 (reject as length cannot be less than 0 )x y y  The