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 ln2 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