NYJC 2021 FM CT2 P2 - modified (Solutions)

2021 NYJC JC2 CT2 9649/2 Marking Guide Page 1 of 13 Qn Key Points Marks Guidance 1 Source: NJC Promo 2025 Question 4 ( ) ( ) ( ) ( ) ( ) ( ) ( ) 23 3 22 2 2 f , 3, f , 2 3, f , 3 3, f , 2, f , 2 3 3, f , 2 3 3, f , 6, x y xx xy yx yy x y x y xy xy x y xy y y x y xx yx xy y xy x y xy x y x y xy =+− = +− = +− = = +− = +− = ( ) ( ) ( ) ( ) ( ) ( ) ( ) f 1, 2 4. f 1, 2 6. f 1, 2 10. f 1, 2 4. f 1, 2 11. f 1, 2 11. f 1, 2 12. x y xx xy yx yy = = = = = = = ( ) ( )( ) ( )( ) ( )( ) ( )( )( ) ( )( ) ( ) ( ) ( )( ) ( )( ) ( )( ) 22 22 , 4 f 1,2 1 f 1,2 2 11 f 1,2 1 f 1,2 1 2 f 1,2 222 4 6 1 10 2 11 4 1 11 1 2 12 222 6 1 1 4 11 1 14 10 2 2 11 12 22 614 10 2 xy xx xy yy TT T Qxy xy x xy y xy x xy y xx x yy y x y =+ −+ − + −+ − −+ − =+ −+ − + −+ − −+ − −− −      = ++      −− −         = +−   1 4 11 11 2 11 12 22 T xx yy      +− −            where 4k = , 6 10 = w , 1 2 = a and 4 11 11 12 = H .

2021 NYJC JC2 CT2 9649/2 Marking Guide Page 2 of 13 Qn Key Points Marks Guidance ( ) ( ) ( ) ( ) f 1, 26 f 1, 2 f .f 1, 210 x y = = = ∇= ∇   wa ∴w is the gradient of f at a, i.e., ( )f∇ a . Note that 2∈a  , not 3 . 2(i) ( ) ( ) f 0.5 0.035533 0 f1 3 0 ≈> = −< ∴there is a root in ( )0.5,1 . ( )f 13 3.2990 0≈> ∴there is a root in ( )1,13 . 2(ii) ( )f 12 0.2842= − ∴there is a root in ( )12,13 . 1 13 0.2842 12 3.2990 12.07930.2842 3.2990x += ≈+ ( )f 12.0793 0.0157 0≈− < ∴there is a root in ( )12.0793,13 . 2 13 0.0157 12.0793 3.2990 12.08360.0157 3.2990x += ≈+ Note that ( )f 12.0793 0.0157 0≈− < and ( )f 12.085 0.0036 0≈> ∴there is a root in ( )12.0793,12.085 . ∴root is 12.08, correct to 2 decimal places. 2(iii) ( ) 13 22f' 3 7xx x − = −− 31 22 1 13 22 2 72 37 n nn nn nn x xxxx xx − + − −+= − −− Using GC,

2021 NYJC JC2 CT2 9649/2 Marking Guide Page 3 of 13 Qn Key Points Marks Guidance n nx 1 0.1 2 0.2509 3 0.4358 4 0.5010 5 0.5046 ( )f 0.495 0.0742 0≈> ( )f 0.505 0.0028 0= −< ∴there is a root in ( )0.495,0.505 . ∴root is 0.50, correct to 2 decimal places. 2(iv) For the recurrence to work, 1 0nx + > 31 22 13 22 31 22 13 22 2 72 0 37 3 0 37 n nn n nn nn x xxx xx xx xx − − − − −+−> −− − > −− Using GC, 03 x<< or 5.5636x > To solve for a, 31 22 30 3 1.7320 1.73 (3sf) xx x − −= = ≈= To solve for b, ( ) 13 22f' 3 7 0 Using GC 5.5636 5.56 (3sf) xx x x − = −− = ≈=

2021 NYJC JC2 CT2 9649/2 Marking Guide Page 4 of 13 Qn Key Points Marks Guidance Q3 (a) [3] Source: RI Promo 2025 Qn 10 (b) [4] By symmetry, ( ) ( ) 2 3 2 3 2222 0 (1 2 cos ) 2 sin d (1 2 cos ) 2 sin dk π π π θ θθ θ θθ++ = ++∫∫ ( ) ( ) 2 3 2 3 22 0 22 (1 2cos ) 2sin d (1 2cos ) 2sin d k π π π θ θθ θ θθ ++ ∴= ++ ∫ ∫ 3.98= (3 sf) (c) [5] Since the flower bed is symmetrical about 0θ = , the optimal position lies on 0θ = . The mid-point of ( )1,π and ( )3, 0 is ( )1, 0Q , where the sprinkler should be positioned. For any point P on curve C, by cosine rule, 0θ = ( )3, 0 O ( )21,π ( )3 21, π ( )1,π 2 3θπ= 4 3θπ= A B

2021 NYJC JC2 CT2 9649/2 Marking Guide Page 5 of 13 Qn Key Points Marks Guidance ( ) ( ) ( ) ( ) ( ) ( ) 2 22 2 2 2 2 24 33 237 2 8 83 1 2 cos 1 2cos 1 2cos 1 2cos 1 2cos 1 2cos 1 2cos , if lies on small loop 1 2cos 1 2cos 1 2cos , if lies on big loop 2 8cos 6cos , 2 2cos , otherwise 8 cos , QP r r P P θ θ θθ θ θθ θ θθ θ θ πθ π θ θ =+− = + +− +  + ++ += + +− +  + + ≤≤= + ++= 4 3 2 2cos , otherwise πθ π θ  ≤≤ + If P lies on the small loop, QP is largest when θπ= , and ( ) 22 37 8881 4AP ≤ −+ + = . If P lies on the big loop, QP is largest when 0θ = , and 2 224AP ≤+= In both cases, 2QP≤ . Thus minimum 2a= . 4 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 22 2 i2 1 π 21 πi 2 21 π 2 1πii 2 21 πi 2 21 πi 2 21 π 2 1π 2 1πii i 44 4 21 π 2 1π 2 1πii i 44 4 1 10 1 1e1 1 e where 0,1, , 2 11 1e e e1 e1 ee e ee e nn n k k n kk nn k n k n kk k nn n kk k nn n zz z z z knz zz z + + ++ + + ++ + − ++ + − + +− = + = −= − + = = −− += − −= +  −=  + 

2021 NYJC JC2 CT2 9649/2 Marking Guide Page 6 of 13 Qn Key Points Marks Guidance ( ) ( ) ( ) 21 π2i sin 4 21 π2cos 4 21 πi tan , 0,1, 2, , 2 14 k n k n k knn + = + += = − ( ) ( ) ( )22 1 1 π 21 π 2 1πtan tan π tan4 44 nk kk n nn −− + ++ = −= −  . Thus ( ) ( )22 1 1 π21 πtan t an44 nkk nn −− ++ =− . By above result and the factor theorem, ( ) ( ) ( ) ( ) ( ) ( ) ( ) 22 21 0 1 21 0 1 21 0 11 21 π2 i tan 4 21 π 2 1π2 i tan i tan44 22 1 1 π21 π2 i tan i tan44 2 i ta nn n k nn k kn nn k kn zz kz n kkzz nn nkkzz nn z − = −− = = −− = = + +− += −     ++ = −−            −− ++ = −+       = − ∏ ∏∏ ∏∏ ( ) ( ) ( ) ( ) ( ) 11 00 1 0 1 22 0 21 π 2 1πn i tan44 21 π 2 1π2 i tan i tan44 21 π2 tan 4 nn kk n k n k kk znn kkzz nn kz n −− = = − = − =   ++   +                ++   = −+         += +    ∏∏ ∏ ∏

2021 NYJC JC2 CT2 9649/2 Marking Guide Page 7 of 13 Qn Key Points Marks Guidance ( ) ( ) 22 234 2 1 2 2 3 4 21 2 24 11 22 2 2 21 1 2 3 4 21 22 2 2 2 1 1 2 3 4 21 22 222 2 22 4 22 nn nn nn zz nn n n nz z z z zz n nn n n nz z z z zz n nn nzz n − − + +−     = + ++++ + +     −         + − +−+− − +     −       = +++ +   −      22 2 2nnzz − + On the other hand, ( ) 1 22 0 21 π2 tan 4 n k kz n − =  + +  ∏ ( )22 22 22 22 21 ππ 3π 5π2 tan tan tan tan444 4 nzzz z nnn n −  = +++ +        . Equating coefficients of 22nz − ,                222 2 2 1 2 1 2 1 22 1 2 21 ππ 3π 5π2 2 tan tan tan tan22 444 4 2 21 πtan2 4 221 21 πsec 12! 4 21 πsec 2 14 21 πsec 24 n k n k n k n k n n n nnn n n k n nn k n k n nnn k nn                                     

2021 NYJC JC2 CT2 9649/2 Marking Guide Page 8 of 13 Qn Key Points Marks Guidance 5 (ai) Let µmm be the population mean distance. A 99% confidence interval for µ is 0.432.5758x n  ±  We require 0.432 2.5758 0.60 n ×≤  13.631n≥ Least value of n = 14 (aii) Since population standard deviation is known and sample size is small, we need to assume that the distance between pairs of holes drilled by machine is normally distributed. (b) Let p be the population proportion of parked cars fitted with an anti-theft device. Let 72ˆ 0.29032248p= ≈ be the unbiased estimate for p A 95% confidence interval for p is ( )( ) ( ) ( ) 0.29032 1 0.290320.29032 1.9600 248 0.23383, 0.34682 0.234, 0.347 (3 d.p)  − ± ⇒ ⇒ Only cars in the car park were sampled which may not constitute a random sample. 6 ln 2 ln 2 ln 2fl 2( ) e e n 2e2 xx xx x a aaxa −− −−= = == = . Thus ln2a= . 01 2 2 X x Y x y≥⇒= ≥ =

2021 NYJC JC2 CT2 9649/2 Marking Guide Page 9 of 13 Qn Key Points Marks Guidance ln 2( ) ln ln ln 2 ) (2 ) ( ln 2 ln ) () 1 ( ln l 1 n2 11 y X y PY y yP y PX y P e ye X − −= ≤= −= ≤ = ≤ ≤ = − = − For 1y≥ 2 1f() ( 11 )Y y dy dy y= − = 2 1 1 , f( ) 0 o.w , yyy ≥=  7(a) (i) 1 0 0 d aiia i xaxx ii − = =∫ 7(a) (ii) ( ) 11 1 1 1 1 1 0 1 0 0 0 d E 11 1 P( ) d 1 d sum to infinity of GP 1 ln 1 ln(1 ) ln( ) 1 q ii i i i q i i q i q i X i qpX pq q p yyq p yyq p yqy p yq p q pp qp ii i ∞∞ − ∞ ∞ − ∞ − = = = = =  = = − = =   −   =−− −−= = −  = = = ⌠ ⌡ ⌠⌡ ∑∑ ∑ ∑ ∫ ∑ ( ) 1 01 0 1 0 0 00 1 0 1 1 d d d. .. d .. d . q i q qq i i q i y yy yy y y yy yy ∞ = − = ∞ − + = += =++ ⌠⌡ ∑ ∫ ∫∫ ∫∑ Note that since q is a probability, 01 iq i<< for all i . Thus 1 0 0 d1 q iyy−<<∫ . Hence 1y < as otherwise the