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 ,