EJC 9758 2025 Prelim P1 Solutions

EJC 9758 2025 Prelim P1 Solutions Solution 1(a) 11 1x x+ − 110 1x x+ − − ( )( )1 1 1 01 xx x + − − − ( )( ) 0 22 1 xx x +− − 21 x−  or 2x (b) 0x or 12lln n22x =

Solution 2 Perimeter of outline = 180 ( )13 2 2 180 2 5 2 1802 590 24 a a b a ab ba    + + + =  + + =  = − + Area enclosed by wire, 2 2 22 2 13 22 53 90 2 4 8 5 15270 82 5 15270 82 aA ab a a a a a a aa     =+   = − + +   = − − = − +  d5 270 15d4 dWhen =0, d 270 or 14.2655 154 A aa A a a   = − +  = + 2 2 d5 15 0d4 A a =− +  Hence the area is maximum when 270 5 154 a = + . Maximum area 2 2 270 5 15 270270 55 8215 1544 1925.82 cm           = − +       ++    =

Solution 3(a) 2213 5 12AC= − = 12cos 13 ACBAC AX = = (b) ( ) 12 cos 12 cos cos sin sin 12 12 5cos sin13 13 156 (shown)12cos 5sin AX BAC BAC BAC     = − =  +  = + = + (c) Using small angle approximation,   2 12 12 1 222 22 2 156 12 1 5 2 5156 12 1 12 2 5156 12 1 12 2 5513 1 ... 12 2 12 2 5513 1 ...12 2 12 65 126113 12 144 AX        − − −  −+  = + −    = + −        = − − + − +         = − + + +   − + 65 1261, 12 144pq=− = (shown)

Solution 4 (a) Translation of 2 units in the negative x-direction. (b) (c) x y y = k (0, 0) x y y = x = 2 (0, 0) x y (4, 0) x = –2 (–4 , 0)

Solution 5(a) 2 1 2 1 1 7 2 3 2 2 12S      = + =           Area under curve between 1x= and 2x= is   2 2 1 1 1 d ln ln 2xxx == Since the two rectangles lie under the curve, the area of the two rectangles must be less than the area under curve, so 7 ln 212 (b) We can use any integer larger than 2 for n. Some possible answers are: n nS 3 37 60 4 533 840 5 1627 2520 6 15797 27720 (c) ( ) ( ) ( )1 2 2 1 1 1 1 1 1 n n n nn n n S n n n ++           = + + +                           1 1 1 1 1 2 2 1 2n n n n= + + + ++ + −  1 1n r nr= = + (d) y x 2 … 1

Consider n rectangles of equal width in the same interval, drawn above the curve. Then ln 2 Area of rectangles 1 2 1 1 1 1 1 1 1 ...n n n n n nn n n +−           = + + +                     1 1 1 ...1 2 1n n n= + + ++− 2 1 1 nS nn = + −  11 2 n nnS +−= 1 2 nS n= +

Solution 6(a) AP=−pa  BP=+pa  ( ) ( ) 22 22 (since and are radius, ) 0 AP BP OA OP = − + =− = −  = = . p a . p a pa a a p a  (b) By Ratio Theorem, ( ) ( ) ( ) ( ) ( ) 1 1 1 12      −+ −− − − − = = − = = ab ba aa c c c a (c) ( )12PC −−= ap  ( )( ) ( ) ( ) ( ) ( ) 2 22 1 2 0 1 2 0 1 2 cos120 0 1 1 2 02 1 1 2 12 3 2 0OP PC       = − − = − − = − − = − − − = − − = = p a p pa. pp p a p a . . . a   A B O P a p A B O P a p C 1

Solution 7(a) 13 2 9 22 uu bu + == + (b) 2 93bbv = = (c) 22 8u v−= 22 8vu= − ( ) 22 22 8vu −= 2 14 499 4 bbb −+= 2 50 49 0bb − + = ( )( )1 49 0bb− − = 1b= or 49b= If 49b= , 49 7 93r = = but we need 1r  since the sequence is convergent. So 1b= . 2 91 52u +== 2 5 8 3v = − =−

Solution 8(a) 2 22 22 ( i) 2i a 2 i 2i a 0 (1) 2 2 (2) ab b ab b ab += − + = − = − =− From (2), 1a b= 22 4 1( ) 0 10 bb b −= −= 22 2 (1 )(1 ) 0 1 ( b is real) bb b + − = =  So 1, 1ba== or 1, 1ba=− =− (reject since 0 arg( ) 2w  ) 1iw=+ (b) Since 1i+ is a root, we can write ( ) ( ) 3 2 2 (1 i) 1 iz z z s z z Bz C− + − + = − +  + + for some (possibly complex) constants B and C. Comparing coefficients, ( ) 2 : 1 1 i i :1 i i 0 constant: 1 i 0 z B B z B B C C s C s − =− − +  = − =− − +  = =− +  = ( ) 2 i0 i0 0 or i zz zz zz += += = =− (c) Replace z with iv: ( ) ( ) ( ) 32 i i (1 i) i 0v v v s− + − + = 32i (1 i) 0v v v s− + + + + = Hence, iiz v v z=  =− i(1 i) 1 iv=− + = − , i(0) 0v=− = or i( i) 1v=− − =−

Solution 9(ai) From GC, coordinates are ( 1,10)− and (3, 22)− (aii) Largest possible value is 1a=− Domain of 1f − is equal to range of f, which is ( ,10− (bi) 2 1212 2g ( ) 12 22 2 2(1 2 ) 1 2 2( 2) 5 5 x xx x x xx xx x x −−  += − + + + − −= − + + = = (bii) if is even g ( ) 12 if is odd2 n xn x x nx = −  + (ci) h gg R [0, ) D \{ 2} OR D ( , 2) ( 2, ) = = − = − −  −   Since hgRD , then gh exists. (cii) From graph of g( )yx= with domain restricted to [0, ) ( 1 gh 2R 2,=− x y y = –2 (0, 0.5)