presbyterian high school 2023 A math paper 1 answer key

Name: Index No.: Class: PRESBYTERIAN HIGH SCHOOL ADDITIONAL MATHEMATICS 4049/01 Paper 1 18 August 2023 Friday 2 hours 15 min PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL 2023 SECONDARY FOUR EXPRESS / FIVE NORMAL (ACADEMIC) PRELIMINARY EXAMINATIONS MARK SCHEME Cynthia – Q1 to 10 Sabrina – Q11 to 13

2 1 The line 2 15yx=+ intersects the curve 2 63y x x= + + at points A and B. Find the value of p for which the distance AB can be expressed as 5.p [5] 2 A curve is such that 2 2 2 d 12e e . d xxy x −=+ The curve intersects the y-axis at (0, 5)P and the tangent to the curve at P is parallel to 4 3.yx=+ Find the equation of the curve. [6] 2 6 3 2 15x x x+ + = + M1 (equate curve to line) 2 4 12 0xx+ − = ( )( )2 6 0xx− + = M1 (factorise) 26x or x= =− 19 3y or y== M1 (find y) ( ) ( ) 22 2 ( 6) 19 3AB= − − + − M1 (apply distance formula) 320AB= 85AB= p = 8 A1 ( ) 22 1 d 12e e d 6e ed x x x xy xcx −−= + = − + M1 (any 2 correct terms) At ( )0,5 , d 4d y x = M1 (seen gradient at P = 4) 2(0) (0) 16e e 4 c−− + = M1 (sub. gradient at x = 0, attempt to find c1) 1 1c =− ( ) 226e e 1 d 3e ex x x xy x x c −−= − − = + − + M1 (any 2 correct terms) At ( )0,5 , 2(0) (0)3e e 0 5 c−+ − + = M1 (sub. x = 0 & y = 5, attempt to find c) 1c= 23e e 1xxyx − = + − + A1

3 3 A function is defined by 2f ( ) 2 2 3x x kx k= + + + for all real values of x, where k is a constant. (a) Find the discriminant of f ( )x in terms of k. [2] (b) Show that the discriminant of f ( )x in part (a) can be expressed in the form ( ) 2 4, k a b−− where a and b are integers. [2] (c) Find the range of values of k for which f ( ) 0x = has no real roots. [3] For 2f ( ) 2 2 3x x kx k= + + + , 22 4 (2 ) 4(1)(2 3)b ac k k− = − + M1 (apply discriminant) 24 8 12kk= − − A1 2 2 2 24 8 12 4 2 1 1 12k k k k − − = − + − − ( ) 2 4 1 1 12k= − − − M1 (completing the square) ( ) 2 4 1 16k= − − A1 2 40b ac− ( ) 2 4 1 16 0k− −  M1 (apply discriminant < 0) ( ) 2 21 2 0k− −  ( 1 2)( 1 2) 0kk− + − −  ( 1)( 3) 0kk+ −  M1 (factorise) 13 k−   A1 k 3 −1

4 4 It is given that 32f ( ) 2 5 4 12x x x x= − − + . (a) Show that 23x+ is a factor of f ( ).x [2] (b) Factorise f ( )x completely. [2] (c) Hence find the roots of the equation ( ) ( ) ( ) 322 2 5 2 4 2 12 0y y y− − + = . [3] 32 3 3 3 3f 2 5 4 122 2 2 2        − = − − − − − +               M1 (apply factor theorem) 27 45 6 1244 0 =− − + + = By the Factor Theorem, (2 3)x+ is a factor of )(f x . (shown) AG1 32f ( ) 2 5 4 12x x x x= − − + ( ) 2(2 3) 4 4x x x= + − + M1 (long division or comparing coefficients) 2(2 3)( 2)xx= + − A1 ( ) ( ) ( ) 32 2 2 5 2 4 2 12 0y y y− − + = Let 2yx= , ( ) ( ) ( ) 32 2 2 5 2 4 2 12 0y y y− − + = ( ) ( ) 2 2 2 3 2 2 0yy   + − =   ( ) ( )2 2 3 0 2 2 0yy or+ = − = 32 2 2 2 yy or=− = M1 (seen either one) (rejected) 1y= A1 M1 (attempt to let 2yx= and solve)

5 5 (a) Using long division, show that 32 2 2 5 10 2 5 x x x x x − + − =− + . [2] (b) Hence, by first expressing the denominator as a product of two factors, express 2 32 21 2 5 10 x x x x + − + − in partial fractions. [5] 2 3 2 3 2 2 2 5 2 5 10 ( 5 ) 2 10 ( 2 10) 0 x x x x x xx x x − + − + − −+ −− − − − M1 (attempt to use long division) 32 2 2 5 10 2 (shown) 5 x x x x x − + − = − + A1 22 3 2 2 2 22 22 2 1 2 1 2 5 10 ( 2)( 5) 21 2( 2)( 5) 5 2 1 ( 5) ( )( 2) xx x x x x x x A Bx C xx x x x A x Bx C x ++ = − + − − + ++ =+ −− + + + = + + + − M1 (seen both partial fractions) Sub. 2x= , 99 1 A A = = M1 (seen substitution or comparing coefficients) Comparing constant term, 1 5 2 1 5 2 2 AC C C =− =− = Comparing x2 term, 2 21 1 AB B B =+ =+ = A2 (any 2 correct) 2 3 2 2 2 1 1 2 22 5 10 5 xx xx x x x ++ = + −− + − + A1

6 6 (a) Find the first 3 terms, in ascending powers of x, of the binomial expansion of 8 2, 4 ax + where a is a non-zero constant. Give each term in its simplest form. [2] (b) Given that the coefficient of 2x is –320 in the expansion of ( ) 8 2 3 2 , 4 axx −+  find the possible value(s) of a. [4] ( ) ( ) 82 768 882 2 2 2 ... 124 4 4 ax ax ax         + = + + +                   M1 (apply Binomial theorem) 8 222 256 256 112 ...4 ax ax a x + = + + + A1 ( ) ( ) 8 2 2 2 23 2 9 6 256 256 112 ...4 axx x x ax a x − + = − + + + +  M1 (expansion) ( ) ( )2(9) 112 ( 6) 256 (1)(256) 320aa + − + =− M1 (comparing) 21008 1536 576 0aa− + = 221 32 12 0aa− + = ( )( )3 2 7 6 0aa− − = 26 37a or a== A1, A1

7 7 The diagram shows a quadrilateral PQRS whose vertices lie on the circumference of a circle. The diagonals PR and QS intersect at U. The tangent at R meets PS produced at T. If QR = RS, prove that (a) // ,QS RT [3] (b) triangle PQR is similar to triangle QUR. [3] S Q T U P R RQS RSQ = (base s of isos. Δ) B1 RQS TRS = (alt. segment theorem) B1 Since RSQ TRS = , //QR RT (alt. s are equal) AG1 RPQ RSQ = ( s in the same segment) B1 RPQ RSQ RQS = = (from part (a)) PRQ QRU = (common ) B1 Triangle PQR is similar to triangle QUR. (AA similarity) AG1

8 8 (a) The equation of a curve is ( ) 3ln . xy xe −= The normal to the curve at the point P has a gradient of 1 2 . Find the coordinates of P. [4] (b) The normal to the curve at P meets the x-axis at Q. Find the area of triangle OQP, where O is the origin. [3] ( ) 3ln ln 3 xy xe x x −= = − d1 3d y xx=− M1 1Gradient at point 1 2 2P=−  =− M1 123 11 1 x x x − = − = = M1 (equate 2dy dx =− & attempt to solve for x) ( ) 3ln 3ye −= =− Coordinates of P = (1, –3) A1 ( ) ( )131 2yx− − = − 17 22yx=− M1 (find equation of normal) At Q, 17 022x−= 7x= M1 (find x-intercept) Area of triangle OQP = 21 7 3 10.5 units2  = A1

9 9 Atmospheric pressure is a measure of the force exerted by the mass of air on an object. The atmospheric pressure, P millibars, exerted at the altitude h kilometres is related by the equation e,bhPA= where A and b are constants. The following table shows the mean atmospheric pressure at various altitudes. (a) Plot ln P against h and dr