2017 A Level H2 Maths Paper 1 Suggested Solutions CWY

1 2017 H2 Mathematics Paper 1 (9758/01) – Suggested Solutions 1 By using standard expansions of ex and  ln 1 x ,             2 2 3 2 3 2 2 3 3 2 2 3 3 2 3 2 2 3 e ln 1 2 2 1 2 ... ... 2! 3! 2 3 2 2 ...2 3 2 2 ...2 3 x ax x x ax ax x ax a x a x ax ax a x ax a a ax a x a a x                                                The expansion for   2e ln 1 x ax  up to the term in 3x is 2 3 2 2 3 2 2 2 3 a a ax a x a a x                  Given that there is no term in 2x , 2 2 0 2 a a   24 0 or 0 4 Since , 4 a a a a a a         2(i) y (0 , ab ) ( a , 0) 0

2 (ii) Point of intersection occurs when 1 b x a x a    . From graph, point of intersection occurs at x a  .     2 1 1 1 1 1Since Point of intersection occurs at x a x a b x a x a b x a x a x a b b x a x a b                       Hence, using the graph in (i), 1 b x a x a    when x a  or 1x a b   . 3(i)       2 2 2 2 2 5 10 0 Differentiating with respect to , d d 2 2 2 10 0 d d d 2 2 2 10 (1) d dAt the stationary points of , 0 d 2 10 0 5 Substitute 5 into the equation of : 5 2 5 5 10 0 y xy x x y y y x y x x x y y x y x x yC x y x y x y x C x x x x                          2 2 20 10 1 1 2 2 1 1 -coordinates of stationary points of are and 2 2 x x x x C x x         (ii) For stationary point where x > 0, 1 2 x  From (1),

3     d 2 2 2 10 d d 5 (2) d y y x y x x y y x y x x         Differentiating (2) with respect to x,      2 2 2 2 2 2 2 2 d d d d 1 5 d d d d 1 5 d At , and 0. d2 2 Substituting these values into the above equation: d 5 1 0 1 0 5 d 2 2 d 4 5d 2 d 5 2 0d 4 1Hence the stationary point at is 2 y y y y y x x x x x yx y x y x y x y x x                                     a maximum. 4(i)   2 4 9 1 42 2 Differentiating with respect to , d 1 d 2 xy x x x y x x         C is defined for all values of x , except for 2x       2 0 22 \xx      ℝ    2 d 1 d 2 \ 2xy x x    ℝ Gradient is negative for all points of .C (ii) 14 , where 4, 1.2y a b x     Equations of asymptotes are 4y  and 2x   . (iii) The graph of C undergoes the following transformations: 1) Translation of 4 units in the negative direction of the y-axis, 2) Translation of 2 units in the positive direction of the x-axis. Note that the order mentioned above can be interchangeable.

4 5(i)             When 1, 1 8 7 1 When 2, 8 4 2 12 4 2 4 2 When 3, 27 9 3 25 9 3 2 3 Using the GC to solve equations 1 , 2 , 3 : 3 3 , , 7 2 2 x a b c a b c x a b c a b c x a b c a b c a b c                                    (ii)     3 2 2 2 2 3 3 f 7, 2 2 3f' 3 3 2 13 2 1 3 3 2 4 x x x x x x x x x x x                           ℝ 2 2 1Since 3 2 1 3 f'( 0 , 3 0) 2 4 .4 x x x x x                       ℝ ℝ The gradient of the curve is always positive for all real values of x, Hence the function f is a strictly increasing function. This implies that f is one-to-one and therefore the line 0y  will cut the graph exactly once, which implies that  f 0 x  has exactly one real root. Using the GC, the root is at 1.33 x  (3 s.f.). (iii) When tangent is parallel to the line 2 3 y x   , the tangent has gradient 2.

5   2 2 f' 2 1 3 3 2 2 4 1 5 2 12 1 5 2 12 1 5 2 2 3 1 1 or 2 6 2 6 15 15 x x x x                            x-coordinates of the required points are 1 1 or 2 6 2 15 5 6 1  . 6(i) The equation describes the set of points with position vector r, where the points lie on a line parallel to the vector b, passing through a fixed point with position vector a. (ii) The equation describes the set of points with position vector r, where the points lie on a plane with a normal vector n, at a distance d units away from the origin. (iii) Since 0b n iɶ ɶ , the line is not parallel to the plane, hence there is a point of intersection between the line and the plane.   , Substitute into : Since 0, t d t d t d t d t d dt                           r a b r n r a b r n a b n a n b n b n b n a n a n b n dt             a n r a b a b b n r is the position vector of the point of intersection between the line with direction vector b, passing through the point with position vector a, and the plane with normal n, at a distance d units from the origin.

6 7(i)                                       sin2 sin2 1 2sin2 sin2 By factor formula 2 1 cos 2 cos 2 2 sin2 sin2 d 1 cos 2 cos 2 d 2 sin 2 sin 2 1 2 2 2 sin 2 sin 2 , where is an arb 4 4 mx nx mx nx x m n x m n mx nx x x m n x m n x x m n x m n Cm n m n x m n x m n C C m n m n                                       itrary constant (ii)                      2 0 2 0 2 2 0 2 2 0 0 2 0 f d sin2 sin2 d sin 2 sin 2 2sin2 sin2 d sin 2 sin 2 d 2sin2 sin2 d 1 cos4 1 cos4 d Using cos2 1 2sin 2 2 sin 2 sin 2 2 4 4 x x mx nx x mx nx mx nx x mx nx x mx nx x mx nx x A A x m n x m n m n m n                                                  0 Using part i         Note that when 0 or x x   ,  sin 2 0 kx  . Hence           0 sin 2 sin 2 2 0 4 4 x m n x m n m n m n               

7                 0 0 0 0 0 sin 2 sin 2 1 cos4 1 cos4 d 2 2 2 4 4 1 cos4 1 cos4 d 2 2 11 cos4 cos4 d 2 1 sin4 sin4 2 4 4 1 1 0 0 0 0 0 2 2 x m n x m n mx nx x m n m n mx nx x mx nx x mx nx x m n                                                                    8(a)                 2 2 2 1 2 5 5 0 Using the quadratic formula: 2 2 4 1 5 5 2 1 2 4 20 1 1 2 2 2 2 1 5 1 2 2 1 1 10 1 1 9 1 1 3 1 z i z i i i z i i i i i i i i i i                                        2 2 2 2 1 3 1 1 3 2 4 1 2 1 1 2 or 1 3 1 1 3 4 2 21 1 2 i i i i z i i i i i i i z i i i                         Hence the roots are 1 2 and 2 z i z i     

8 (b)(i)         22 2 2 3 2 4 2 1 1 2 1 1 2 2 2 1 2 2 2 2 2 2 1 2 2 2 2 2 2 4 i i i i i i i i i i i i i i i                                               4 3 2 39 58 0 Substituting the above information, 4 2 2 39 2 1 58 0 4 2 2 78 58 0 54 2 78 2 0 Comparing real and imaginary coefficients: 54 2 0 2 54 2 78 2 0 2 78 p q p i i q i p pi i q qi p q i p q p q p q p q p q                                                   54 2 2 78 54 4 24 6 78 2 6 66 p q p q p q p p q                    6, 66 p q     (ii) 4 3 2 6 39 66 58 0          1 is a root of the above equation. Sinc e the equation has only real coefficient s, hence, the complex conjugate, 1 is also a root. i i        1 1 i i        is a factor of 4 3 2 6 39 66 58         .           2 2 2 2 1 1 1 1 1 2 2 i i i i i                     2 2 2     is a quadratic factor.          4 3 2 2 2 4 3 2 6 39 66 58 2 2 , , , real values 2 2