VJC 9758 2025 Prelim P2 Solutions

2025 JC2 H2 Math Prelim P2 Solutions 2025/VJC/Math Dept 1 It is given that ( ) ( )f lnx a x=+ , x , xa− , where a is a constant. (a) Using the standard series from the List of Formulae (MF27), find the series expansion for ( )f x , up to and including the term in 3x . [2] It is given that 1a= . (b) Hence, or otherwise, show that the series expansion of ( )sin f x , up to and including the term in 3x is given by 2311 26 x x x−+ . [2] (c) Deduce the Maclaurin series for ( )cos f x up to and including the term in 2x . [2] (d) Find 3 1 23 d11 26 xx x x −+  . Without the use of a calculator or any further calculation, explain, whether this value is a good approximation to the value of ( ) 3 1 sin f d xx . [2] 1a ( ) 23 23 23 ln ln 1 ln ln 1 11ln ... 23 ln ... 23 xa x a a xa a x x xa a a a x x xa a aa  + = +   = + +      = + − + +         = + − + + 1b ( ) ( ) 23 32 3 2 3 23 3 23 sin ln 1 sin ... 23 1... ... ...2 3 3! 2 3 1 ...2 3 6 ...26 xxxx x x x xxx xxxx xxx + = − + +      = − + + − − + + +          = − + − + = − + + 1c ( ) 2311sin ln 1 ..... 26x x x x+ = − + + Differentiate w.r.t. x: ( ) 211 cos ln 1 1 ...12 x x xx + = − + ++ ( ) ( ) 2211cos ln 1 1 1 ... 1 ... 22x x x x x + = + − + + = − +  

2025/VJC/Math Dept 1d 3 23 1 d326 xxxx− + = For ( ) ( 1,3 1,1x  − , the expansion ( ) 23 sin ln 1 sin ... 23 xxxx + = − + +   is not valid and hence, the approximation is not a good approximation. 2 The following diagram shows the dimensions of a trapezoidal prism with fixed volume 4 3k units3, with variables x and y. The top surface of the prism, ABCD, is an isosceles trapezoid with AB of length 5x units, DC of length 3x units, AD BC= and 60ABC BAD = = o . The rectangular sides ABFE and BCGF are perpendicular to both the top surface ABCD and the bottom surface EFGH, with AE BF DH CG y= = = = units. (a) Show that the total external surface area A of the trapezoidal prism is given by 2 3 128 kAx x+= . [4] (b) Using differentiation, find the value of x in terms of k at which A is a minimum. [4] (c) It is given instead that the volume of the prism is 1000 units3 and its external surface area is 800 units2. Find the two possible values of x. [2] No. Solution Marks 2a Let the height of the isosceles trapezoid be h. tan60 3h x x== and 2cos60 xBC x== o A B C D E F G H 5x 3x y y y y 3x 3x D x x h

2025/VJC/Math Dept ( ) 2 2 1 5 3 32 4 3 4 3 V x x x y k x y ky x = +   = = ( )( ) 2 1 5 3 3 2 3 5 2(2 )2 8 3 12 A x x x xy xy xy x xy = +  + + + =+ 2 2 2 8 3 12 128 3 (shown) kxx x kx x =+  =+ 2b For minimum A, d 0d A x = . 11 333 2 12 12 3 316 3 0 416 3 4 3 k k k kx x xx − =  =  = =    2 2 3 3 d 24 2416 3 0 , 0 0d A k k kxx x x = +      Q Hence, A is minimum when 1 33 4 kx = . Alternative (for 2nd derivative test) At 1 33 4xk = , 2 2 d 24 16 3 48 3 0d 3 4 Ak x k = + =  . Alternative (for 1st derivative test) x 1 30.75k 1 33 4 k  1 30.76k d d A x 1 30.549k− 0 1 30.286k sign (since 0k  ) – 0 + slope

2025/VJC/Math Dept 2c When 1000,V = 1000 250 4 3 3 k == . 2 12 250800 8 3 3 x x =+  From GC, 8.5102x=− (N.A. since 0x ) or 6.10x= or 2.41x= 3 With reference to the point O as the origin and the x-y plane as a horizontal plane, t he pyramid OPQRV has a parallelogram base OPQR and height OV. The position vectors of the points P and R are 3 4 3− + + kij and 52−− kij respectively. (a) Find the coordinates of the point S that lies on the line PR such that the distance from O to S is a minimum. [3] (b) Find the cartesian equations of the planes such that the perpendicular distance from each plane to the base OPQR is 2 86 units. [3] (c) Find the acute angle between OV and the vertical. [2] (d) Given that QV is parallel to the vector 58−+ kj , find the position vector of the point V. Hence find the exact volume of the pyramid OPQRV. [5] [Volume of a pyramid = 1 base area height3 ] No. Solution Marks 3a 5 3 8 4 2 4 6 2 3 1 3 4 2 PR −− = − − = − =− −−                                       Line PR: 54 2 3 ,   12 r  − = − +  −                   OS ⊥ PR for OS to be minimum: 0OS PR = Since S lies on line PR, 5 4 4 2 3 3 0 1 2 2  −− − + = −                             . 28( 20 6 2) (16 9 4) 0 29 − − − + + + =  = 5 4 33 28 12 3 26 29 291 2 27 OS − = − + = −                             Coordinates of S are 33 26 27,, 29 29 29  .

2025/VJC/Math Dept Alternative PO PR PRPS PR PR =   3 4 4 4 3 3 4 3 2 2 30 3 2929 29 2 PS −− − − − −==                                3 4 33 30 14 3 26 29 293 2 27 OS OP PS − −= + = − + = −                             Coordinates of S are 33 26 27,, 29 29 29  . 3b 3 5 2 1 4 2 12 2 6 3 1 14 7 OP OR −  =  − = = − − −                                       Let the equation of a plane be 1 6 7 rt  =− , and note that ( ),0,0t is a point on the plane. ( ) 01 0 0 6 0 0 7 2 86 1 36 49 2 86 172 t t t            −           −      = ++ = = OR ( ) ˆ 2 86 1 6 7 2 86 1 36 49 1 6 2 86 172 7 rn r r =  − = ++   = =− The planes have equations 6 7 172x y z+ − = and 6 7 172x y z+ − =− . Alternative 3 5 2 1 4 2 12 2 6 3 1 14 7 OP OR −  =  − = = − − −                                       A normal to the plane OPQR = 1 6 7−    Let T and U be different points on the two planes such that 2 86OT OU== .

2025/VJC/Math Dept 1 6 2 72 86 12 1 36 49 14 OT −=  = ++ −       1 6 2 72 86 12 1 36 49 14 OU − − −=  = − ++       The equations of the planes are 1 2 1 6 12 6 7 14 7 r = − − −                             and 1 2 1 6 12 6 7 14 7 r − =− −−                             , i.e. 6 7 172x y z+ − = and 6 7 172x y z+ − =− . 3c Let the acute angle between OV and the vertical be . 10 60 71 7cos 86 86  −==                   41.0 = 3d 3 5 2 4 2 2 3 1 2 OQ OP OR − = + = + − = −                             2 1 0 2 6 5 2 7 8 OQ OV VQ a b            = = + = + −           −      , for some ,ab  By comparing LHS and RHS, 2a= (and 2b= ). 12 2 6 12 7 14 OV         = =       −−   

2025/VJC/Math Dept ( ) 3 1Volume of pyramid 3 1 3 1 3 1 3 units 3 5 2 4 2 12 3 1 14 22 12 12 14 14 4 144 196 344 3 OP OR OV=    =      = −           −          −−             =       −−    = + + Alternative (to find position vector of V) 3 5 2 4 2 2 3 1 2 OQ OP OR − = + = + − = −                             Line QV: 0   2 2 5 , 28 r = +  −                   Line OV: ,  1 6 7 r = −    V is the intersection of lines QV and OV: 1 2 0 6 2 5 , for  some , 7 2 8    = +  −−                             Solving gives 2, 2== . 12 2 6 12 7 14 OV == −−                  

2025/VJC/Math Dept 4 Do not use a calculator in answering this question. The complex number z has modulus 1 and argument  , where π π2  ,