ACJC EJC NJC RVHS FM 2024 Prelim P2

2024 H2 FM 9649 Prelim Paper 2 (ACJC_EJC_NJC_RVHS) Section A: Pure Mathematics [50 marks] P2 Q1 Prove by induction that ( ) ( )2 7 3 5 5nn +− is divisible by 24 for all positive integers n. [5] P2 Q2 Show that the substitution sinxt= , where ππ 22 t−   , reduces the differential equation ( ) 2 2 2 dd1 4 0 dd yyx x y xx− − + = to 2 2 d 4 0.d y yt += [4] Hence find the general solution for y in terms of x, giving your answer in non-trigonometric form. [4] P2 Q3 (a) Determine the eigenvalues and corresponding eigenvectors of the matrix 11 31 = − M . [4] (b) By expressing M in the diagonalised form 1−QDQ , find nM for positive integers n. [4] (c) Hence, or otherwise, show that nM is a diagonal matrix for all even values of n. [1]

2024 H2 FM 9649 Prelim Paper 2 (ACJC_EJC_NJC_RVHS) P2 Q4 A Gregorian telescope consists of a parabolic mirror 1C and an elliptical mirror 2C , both of which are concave. The major axis of the ellipse coincides with the axis of the parabola . In the diagram below, this common axis is the x-axis. The vertex of the parabola is at the origin O. A light ray travels along the line PQ parallel to the x-axis, and reflects off 1C . It then travels along the line QR and reflects off 2C . Finally, it travels along the line RS, going through a small hole near the vertex of 1C to reach a detector located at S on the x-axis. (a) Explain why, if the ellipse and parabola have a common focus, incoming light rays which are parallel to the x-axis will all pass through a common point, S after reflecting off both mirrors. [2] The equation of 1C is 2 37.6yx= . The vertex of 2C is at ( )9.6,0 , and S is at ( )0.2,0− . (b) State the coordinates of F, the common focus of 1C and 2C . [1] (c) Find the equation of 2C . [4] (d) Given that the range of y-coordinates of 1C is 4.2 4.2y−   , find the smallest possible range of y-coordinates of 2C so that all the light reflected off 1C will reach 2C . [5] x y O P Q R S

2024 H2 FM 9649 Prelim Paper 2 (ACJC_EJC_NJC_RVHS) P2 Q5 For 2n , the complex numbers 12, , ..., nz z z are the n roots of the equation 10nz −= with ( ) ( )0 arg arg 2 πijzz   for 1 i j n   so that 1nz = . (a) (i) Express kz in the form ier  where 0r  and 02 π , leaving your answer in terms of k ( )1, 2, ..., kn= . [1] (ii) In the case where n is odd, evaluate the series 1 2 3 1 2 2 2 2 2 ...1 1 1 1 1 nnz z z z z − + + + + ++ + + + + , leaving your answer in terms of n. Full working must be shown. [4] (b) Let 7n= . (i) Sketch the roots of 7 10z −= on an Argand diagram, showing clearly the relationships between their moduli and arguments. [2] (ii) For 1,2,3,4,5,6m= , the locus mL is the set of points z satisfying ( ) ( )arg 1 arg 1 mzz− = − and 1z  . Sketch, on the same Argand diagram in part (b)(i), the six loci 1 2 6, , , L L L . [2] (iii) Show that ( )( )( )