SAJC H2 Promo 2023 (Qn)

2023 SAJC H2 Maths Promo Paper Duration: 3 hrs Marks: 100 Attempt all questions. 1 The sum of the first n terms of a sequence, nS , is given by 32 n bcadSn nn  , where a, b, c and d are constants . It is given that 12 3 45, 20, 57 and 128SS S S   . Find nS .             [4] 2 (a) Differentiate  1 2 sin 2 1, for 1 12 24 xx x  , with respect to x, simplifying your answer as much as possible as a single fraction. [3] (b) It is given that 24 3e 4xy  . By repeated differentiation, show that 22 2 2 dd 8dd yyy kxx y   , where k is a constant to be determined. [4] 3 (a) Given that    1 2 1 12 16 n r rn n n   , find  1 61 n r rr   , in terms of n. [2] (b) (i) Express   1 12rr r in partial fractions.  [1] (ii) Hence find NS in terms of N, where  1 1 12 N N r S rr r   , and deduce that 1 4 NS  . [6] 4 An arithmetic sequence 123,,,,, nuuu u  is known to have a common difference of ln 3. The rth term, rw , of another sequence 123,,,www , is given by e ru rw  . (i) Show that the sequence 123,,,,, nwww w  , is a geometric progression with common ratio 1 3 . [2] (ii) Explain why 1 r r w    converges. [1] (iii) Given that 1 ln 3u  , find the smallest possible value of n such that the sum of the first n terms of the geometric progression given by 123,,,www  is within 0.5% of the sum to infinity of the geometric series. [3]

5 The position vectors of the points A and B relative to the origin O are a and b respectively. (i) Point M on AB is such that :1 : 2AM MB  . Find OM  in terms of a and b. If the area of triangle OBM is 4 units2, find || ab . [5] (ii) Another point P has position vector p and p0 . Given () pa ba 0 , what can you deduce about the relationship between the vectors AP  and AB  ? Hence find the vector equation of line l that passes through points A and P in terms of b and a. [2] (iii) Given instead that 4aa b b , describe in words, the relation between a and b and find |b|. [3] 6 (a)                     O y x x x

The diagram above shows the graph of . The curve crosses the x-axis at  3.5, 0 and has a minimum point at  2, 0 . The lines 3x , 0x and 3y are asymptotes of the curve. On separate diagrams, sketch the graphs in (i) and (ii), stating the equations of any asymptotes and the coordinates of the points where the curve crosses the axes and of any turning points in exact form, if possible. (i)  fyx , and [2] (ii) f '( )yx . [4] (b) The transformations A, B and C are given as follows: A: Reflection about the x-axis; B: Translation of 4 units in the positive x-direction; C: Scaling parallel to the x-axis by a factor of 2. A curve undergoes in succession, the transformations A, B and C, and the equation of the resulting curve is  21 13yx . Determine the original equation of the curve. [3] 7 (i) Sketch, on the same diagram, the graphs 21 3 xy x   and  ln 1yx , stating clearly the coordinates of axial intercept(s) and equation of asymptote(s). Hence solve 21 ln 13 x xx   . [5] (ii) Solve 21 ln 13 x xx   . [1] (iii) Hence, solve 21 ln2 x xx   . [3] fyx

8 The function f is defined by 3f : , for , 1 xxx x x k    . (i) State the value of k and explain why this value has to be excluded from the domain of f. [2] (ii) Find  1f x . Hence find   2f x . [3] The function g is defined by g: 1, , 1xx x x   . (iii) Find the range of fg . [2] 9 A curve C is defined by the parametric equations 23,x ty t . (i) Prove that the equation of the tangent at the point   23,tt on the curve is 323 0 .yt x t  [3] (ii) This tangent passes through a fixed point ( a,b). Explain why there cannot be more than 3 tangents through ( a,b). [1] (iii) The tangent at the point P when t = 2 meets the curve again at the point Q where tk . Find the value of k. [3] (iv) Find the equation of the normal at the point P. [2] (v) Sketch the curve C, indicating the intersection point(s) with the axes. [2] (vi) By sketching the tangent and normal at the point P on the sketch in (v), explain why 11 1tan (3) tan . 32    [2]

10 The line 1l and the plane 1 have equations 13 32 , 32            r  and 2 54 p   r  respectively, where p is a real constant. (i) Given that the line 1l and the plane 1 intersect at the point A 5, 7, 7 , show that p = 3 . [2] (ii) Find the acute angle between the line 1l and the plane 1 . [2] (iii) B is the point on 1l where 1  . Find the position vector of the foot of perpendicular, F, from the point B to 1 . [4] (iv) Find the equation of the line of reflection of 1l in the plane 1 . [3] Another line 2l has the following properties. • 2l passes through point B , • 2l is perpendicular to 1l , and • 2l is parallel to 1 . (v) Find, in vector form, the equation of 2l . [2]

11 [It is given that the area of a sector of a circle with radius r and angle  is given by 21 2 r  .] A florist owns a greenhouse in the shape of a circular sector OPQ, with center O, and intends to enclose a rectangular area to grow roses. The garden is represented in the diagram below by a fixed sector OPQ where 10 mOP OQ and π 3POQ radians. The rectangular area WXYZ is inscribed in sector OPQ such that ZOP  radians. (i) By considering triangles OWX and OZY, or otherwise, show that the area of rectangle WXYZ, 2250 sin 2 sin 3 3A   . [3] (ii) Using differentiation, find the maximum value of A in exact form. [6] The florist subsequently decides to use remaining regions in the sector to grow marigolds. The cost of maintaining the plants each month, $C, can be broken down into $5/m2 for roses and $4/m2 for marigolds. (iii) Sketch the graph of C as  varies, indicating any turning point(s) and end point(s). [3] (iv) Hence determine the range of values of  for the florist to have a maintenance cost of less than $220. [1]