MI 9758 2023 Prelim P2 Solution

1 © Millennia Institute 9758/02/PU3/Prelim/23 Solutions PU3 MATHEMATICS Paper 9758/02 Section A: Pure Mathematics Qn Solution 1(i) [2] ( ) ( ) ( ) ( ) 2 2 2 3 2 32 2 23 3 3 2 3 33 (1) dd 2dd dd 2 (2)dd dSubst (1) and (2) into 2 ,d d22 d d22 d d d d 1 (shown)d y vx yv vx xxx yv vx xxx yx x xy yx vx vx x x x vx vxx vvx x x vx vxx vx x vxx v vx = −−−−−− = + = + −−−−−− =++  += ++ + = ++ = + = + 1(ii) [3] Method 1 2 2 2 1 2 d 1d 1 d 1 d1 ln 1 1e 1e 1 e where e 1e Subst 1 and 1, 11e 1 2 e 21e e 1 2e xc xc xc x x x v vx vxv v xc v v vA A y Ax yx A A y x y x + + − = + =+ +=+ += += ± += = ± += = = += = +=  += ⌠⌡ ∫

2 © Millennia Institute 9758/02/PU3/Prelim/23 Solutions Qn Solution Method 2 2 2 2 d 1d 1 d 1 d1 ln 1 ln 1 Subst 1 and 1, 1ln 1 11 ln 2 1 ln 1 ln 2 1 v vx vxv v xc y xcx yx c c y xx = + =+ +=+ += + = = += + = − += +− ⌠⌡ ∫ Qn Solution 2(a) [5] ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) 2 2 2 4 2 22 f sec f 2sec sec tan 2 f 2 sec 2sec s s ta ec tan 2sec 4 n tan s ec se e c c tan x x x xx x x xx x xx x xx x xx = ′ = = ′′ = + = + When 0x= , ( ) ( ) ( ) f0 1 f0 f0 2 x = ′ = ′′ = ( ) 2 2 2f 1 0 ... 2! 1 ... x xx x ∴ = ++ + = ++

3 © Millennia Institute 9758/02/PU3/Prelim/23 Solutions 2(b) [3] ( ) ( )( ) ( )( )( ) 42 22 451 1 4 ...2! 1 4 10 ... ax ax ax ax a x − −−+ =+− + + = −+ + Since the coefficients of the x and 2x terms in the expansion are equal, ( ) 2 2 4 10 10 4 0 25 2 0 aa aa aa −= += += 20 (rejected since 0) or 5a aa= ≠= − Qn Solution 3(a) [3] Method 1 (most used this) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) 234 1 2 2 2 ... 1ln 1 ln 1 ln 1 ln1 ln 3 ln 2 ln 4 ln 3 ln 5 ln 3 ln 1 ln 2 ln ln 1 ln 1 ln1 ln 2 ln ln 1 ln 2 ln 1 N NN N n n N n N n S uuu u u u n n nn NN NN NN NN NN − = = = =++ + + = −=  + = −− + = − +− +− + −− − + −− + −− + =+− − + = −+ ∑ ∑ ∑  Method 2 ( )( ) 234 1 ... 123 2 1ln ln ln ... ln ln345 1 1 2 3 ... ( 2) ( 1)ln 3 4 5 ... ( 1) ( ) ( 1) 2ln ( 1) ln 2 ln 1 N NNS uuu u u NN NN NN N NN NN NN −=++ + + −−      = + + ++ +       +        × × × ×− ×−=  × × × ×− × ×+ =  + = −+

4 © Millennia Institute 9758/02/PU3/Prelim/23 Solutions Qn Solution 3(b) (i) [3] 123 1 3 333 ...1 111 n n x xxx x xxx ∞ =    =+++   ++++   ∑ is a geometric series with 3 1 xr x= + . If the series has a finite sum, S∞ exists and 1r < . 1 3 11 r x x < <+ Method 1 (Graphical): 0.25 0.5x− << Method 2 (Algebraic): ( ) ( ) ( )( ) 22 22 2 3 11 31 31 9 21 8 2 10 4 12 1 0 0.25 0.5 x x xx xx xx x xx xx x <+ <+ <+ <++ − −< + −< − <<

5 © Millennia Institute 9758/02/PU3/Prelim/23 Solutions Qn Solution 3(b) (ii) [2] Method 1 ( ) 11 1 123 3 0.253 1 0.25 1 3 5 333 ...555 3 5 31 5 3 2 nn nn n n x x ∞∞ = = ∞ =  =  ++  =    =+++     = − = ∑∑ ∑ Method 2 ( ) ( ) 1 3 3 1 31 1 1 3 1 12 1 3 12 3 0.25 since 0.251 2 0.25 1.5 n n x x x xx x x x x x x x x ∞ =  += + − + += − + = − = =− = ∑ Qn Solution 4(i) [2] ( ) ( ) ( ) ( ) 22 22 22 22 22 22 2 20 110 20 1 10 20 1 10 2 5 xy xy xy xy += += += +=

6 © Millennia Institute 9758/02/PU3/Prelim/23 Solutions Qn Solution 4(ii) [1] 222 20 d42 0 d d24 d d42 (shown)d2 xy yxy x yyx x yxx x yy += += =− = −= − 4(iii) [5] Method 1 ( ) ( ) Gradient of normal 2 At , , Gradient of normal 2 0 21 2 0 10 0 1or y x bab a bb aa ab b ab ab b ba ba = = −= − −= += += = =− y x O ( )10,0 ( )10,0− ( )0, 2 5 ( )0, 2 5−

7 © Millennia Institute 9758/02/PU3/Prelim/23 Solutions Qn Solution ( ) ( ) ( ) ( ) ( ) ( ) 22 2 2 2 , 2 0 20 10 or 10 , 2 1 20 18 18 or 18 32 o r 32 The four coordina Wh tes are 10,0 , 10,0 , 1, 3 2 a en 0 When nd 1 1 3 2. or , b a a a b b b b += =− −+= = = = =− − =− − −− − Method 2 Gradient of normal 2 y x= ( ) ( ) At , , Gradient of normal 2 Equation of normal at : 2 22 22 bab a bPy b x a a bbyxb a bbyx a = −= − = −+ = + Since normal passes through (1, 0), ( ) ( ) 01 22 0 10 0 o 1r bb a b ab ba ba = + = + += = =−

8 © Millennia Institute 9758/02/PU3/Prelim/23 Solutions Qn Solution ( ) ( ) ( ) ( ) ( ) ( ) 22 2 2 2 , 2 0 20 10 or 10 , 2 1 20 18 18 or 18 32 o r 32 The four coordina Wh tes are 10,0 , 10,0 , 1, 3 2 a en 0 When nd 1 1 3 2. or , b a a a b b b b += =− −+= = = = =− − =− − −− − Qn Solution 5(i) [2] Method 1 Perpendicular distance from the origin to p ( ) 222 4 12 5 4 units 30 = + +− = Method 2 Note that A(4, 0, 0) lies on p. Perpendicular distance from the origin to p ( ) 222 1 24 5 40 units 3012 50    − = •= + +− 5(ii) [3] Method 1 l lies on p means l is parallel to p and a point on l lies on p. l is parallel to p means direction vector of l is perpendicular to normal of p:

9 © Millennia Institute 9758/02/PU3/Prelim/23 Solutions Qn Solution 1 220 52 4 10 0 10 2 (shown)5 a a aa a      •=    −   +−= = = We need 3 2 b  − to lie on p: 31 2 24 5 345 4 55 1 (shown) b b b b      −• =     −   −− = =− =− Method 2 Equation of plane p: 1 254 2 4 5 xyz r  + − =⇒• = − Sub 3 22 2 a a b λ    = −+    r into 1 24 5 r  •=− : ( ) ( ) 31 22 2 4 25 3 4 4 5 10 4 5 1 05 50 5 10 5 5 0 10Comparing coefficient of :5 10 0 2 (shown) 5 Comparing constants: 5 5 0 1 (shown) a a b a ab ab ab aa bb λ λ λ λλ λ λλ λ λ +     −+ • =    +−   + −+ − − = − −− = − +−− = − =⇒= = −− =⇒= − 5(iii) [5] Let 32 24 12 OR λ      = −+    −     be the position vector of a point on l for some λ∈  . 3 2 4 72 2 4 7 94 1 2 3 42 QR λ λλ λ  − +         = − + − =−+        − −+      

10 © Millennia Institute 9758/02/PU3/Prelim/23 Solutions Qn Solution ( ) ( ) ( ) ( )( ) ( ) ( ) 222 2 22 2 2 7 2 9 4 4 2 110 49 28 4 81 72 16 16 16 4 110 24 60 36 0 2 5 30 2 3 10 2 3 0 or 1 0 3 or 12 QR λλλ λλ λ λ λλ λλ λλ λλ λλ λλ = + +−+ +−+ = + + +− + +− + = − += − += − −= −= −= = =   When 3 ,2λ = 3 26 32 4421 22 OC      = −+ =    −     When 1,λ = 3 25 2 14 2 1 21 OD      = −+ =    −     5(iv) [1] Let F be the foot of perpendicular from Q to l. Then QC and QD are the hypothenuses of the right-angled triangles QFC and QFD respectively. Therefore the shortest distance from Q to l, QF, has to be smaller than QC and QD, which is 110 . OR: Assume the shortest distance from Q to l is greater than 110 . But there are points C and D found on l that are a distance of 110 away from Q, which contradicts the original assumption. Therefore the shortest distance from Q to l, has to be smaller than 110 . OR: There should only be one unique point on l , the foot of perpendicular, that is the shortest distance from Q. Currently there are 2 points C and D found on l that are a distance of 110 away from Q. As the distance is shortened, C and D will converge to the foot of perpendicular. Q F O C D