VJC_9758_2025_Prelim_P1_Solutions

2025 JC2 H2 Math Prelim P1 Solutions 2025/VJC/Math Dept 1 Using an algebraic method, solve the inequality 2 43 14 3 1 x xx − +− . [3] Hence, find the set of values of x that satisfy 2 4ln 3 14(ln ) 3ln 1 x xx − +− . [2] No. Solution 1 2 2 2 2 2 43 14 3 1 43 104 3 1 4 3 4 3 1 0( 1)(4 1) 42 0( 1)(4 1) 42 0( 1)(4 1) x xx x xx x x x xx xx xx xx xx − +− − −+− − − − + +− − + − +− −+ +− 22 2 114 2 4 2 88x x x    − + = − − +       2 1 314 8 16x= − + Since 2 1 08x − for all real values of x, 2 1 3140 8 16x − +  OR Since coefficient of 2x is 40 and the discriminant of 242xx−+ is 2( 1) 4(4)(2) 31 0,− − =−  24 2 0xx− +  for all x . We have ( 1)(4 1) 0xx+ −  . 1x− or 1 4x Replace x with ln x : ln 1x− or 1ln 4x 10e x − or 1 4ex –1 1 4

2025/VJC/Math Dept 2 The function f is defined by 1f : , xx x − x , 0,x 1x . (a) Show that ( ) ( ) 21ff xx −= . [3] (b) Find ( ) 3f x in simplified form. [1] (c) Find ( ) 2030f5 . [2] Functions g and h are defined by 1g : , xx x − x , 1x , h : sin ,x ax − x , where a is a positive constant. (d) Find the value of a given that the range of hg is ( 1,0− . [2] 2a 11Let 1 xy xx −= = − 1 1 1 1 yx x y =− = − Hence ( ) 1 1f 1x x − = − ( ) 2 1ff 1 1 1 1 1 1 1 xx x x x x x xx x x −=  − − = − −−= − = − ( ) ( ) 21ff xx −= 2b ( ) ( ) 21ff xx −= ( ) ( ) ( ) 32 1 f f f ff xx x x − = = = 2c ( ) ( ) ( ) 2030 3(676) 2 2 11f 5 f 5 f 5 1 5 4 += = = =− −

2025/VJC/Math Dept 2d Range of g =  )0,1 = new domain of h Given: range of hg is ( 1,0− ππ 122 aa =  = 3 Referred to the origin O, points A and B have position vectors a and b respectively. Point C lies on OA, such that : 2:1OC CA = . Point D lies on OB, such that ::OD DB = . It is given that the area of triangle ABD is half the area of triangle ABC. (a) Show the area of triangle ABD is given by ( )2   + ab . Hence find the ratio : . [4] (b) The point E has position vector 15 48 +ab . Show that A, E and D are collinear. [3] It is further given that the angle AOB is π 4 and O lies on the perpendicular bisector of the line segment AB. (c) Find the length of projection of a on b , giving your answer in terms of b . Hence find the position vector of the point F, the foot of perpendicular from A to OB. [3] No. Solution 3a ( ) ( ) ( ) ( ) 1Area of triangle 2 1 2 2 02 2 ABD DB AB b b a b b b a ab ab         = =  −+ =  − + = + + = + ( ) 1Area of triangle 2 11 23 1 6 1 6 ABC CA AB a b a a b a a ab = =  − =  −  = π 2a

2025/VJC/Math Dept Area of triangle ABD = 1/2 (area of triangle ABC) ( ) ( ) 11 2 2 6 1 2 12 6 5 : 5:1 a b a b          =  +  =+ =+ = = 3b 15 48 35 48 AE OE OA=− =+− =−