2017 H1 A Level Mathematics Solution (SAJC)

1 2017 H1 A Level Mathematics Solution Section A Q1. For     2 4 7 0x k x k     for all real values of x, Discriminant :         2 2 2 2 40 4 4 7 0 ( 8 16) 4 28 0 4 12 0 2 6 0 b ac kk k k k kk kk                  The set of values of k is  : , 2 6k k k    . Q2. (i)         13 22 3 2 d 1 d 1 5 5 2 5 2 5d d 2 52 2 5 2 xxxx x x             (ii)       22 4 2 22 22 2 1 4 4 1 dd 4 4 d x x x xxxx x x x           341 43 x x c x    where c is an arbitrary constant y y = 2 4 12kk (-2, 0) (6, 0) O x

2 Q3. (i) Area of the sign = Area of 2 half circles with diameters AB and DC + Area of rectangle ABCD =   2 2 ADxx  To find the length of AD, Perimeter of the sign =  2 2 ADx  = 10 (Given) AD =  1 10 2 52 xx   Area of the sign =     2 2 2 2 5 10 10 cmx x x x x x x         (Shown) (ii) For a non-calculator method, 210 d 10 2d y x x y xx     For stationary points, d 0d 10 2 0 10 5 2 y x x x      For 5x  2 2 d 2d y x  ( < 0) Hence the stationary value at 5x  gives a maximum area.

3 Required Area 2 2 2 10 5510 50 25 25 units xx                   The maximum value of this area is 225 units . Q4. (i) At x axis intercept (y = 0)   0 ln 4 5 0 4 5 e 4 5 1 3 2 x x x x     The coordinates of point of intersection with the axes are 3 ,02   . Asymptote : x = 5 4 .[Found by setting 4 5 0x ] x y O x = 5 4 3 ,02    ln 4 5yx

4 (ii) When x = 2.5, y = ln 5.  ln 4 5 d1 4d 4 5 yx y xx    [where  d 4 5 4d xx  ] 2.5 d 4 4 d 4(2.5) 5 5x y x    Equation of Tangent to the curve at the point x = 2.5,   45ln 5 52 4 2 ln 55 5 4 10 5ln 5 4 5 5ln 5 10 yx yx yx xy                 where 4, 5, 5ln 5 10a b c    (iii) When y = 0, 10 5ln 5 4x  , point A is 10 5ln 5 ,04   or  0.48820,0 When x = 0, 5ln 5 10 ln 5 25y    , point B is  0,ln 5 2 or  0, 0.39056 P ( 0.48820 , 0 ) Q ( 0 , - 0.39056 ) O ( 0 , 0 )

5 By Pythagoras’ Theorem, Length of PQ   2 2( 0.3908 0 0.3905 8 0.62520 0.625(3 sig. fig) 6 0.48820 0)     Q5. (i) Let x, y, z be the manufacturing cost of a pair of Supers, Runners and Walkers respectively. 2 ...............(1) 10 7 50 ...............(2) 2 6 4 481 ...............(3) xz yx x y z      Rearranging 2 0 ...............(1) 7 10 50 ...............(2) 2 6 4 481 ...............(3) xz xy x y z        Using GC, 55,x 43.5,y