NYJC 2025 FM CT (Solutions)

1 A rectangular box is placed in the region ( ){ },, :,, 0R xyz xyz= ≥ with one corner at the origin, three faces in the coordinate planes and the corner P diagonally opposite the origin on the plane 3xyz++= . Use partial differentiation to find the maximum possible surface area for such a box, showing your working clearly and justifying your conclusion. [5] Qn Solution 1 Surface area of box, ( )2S xy xz yz= ++ where 3xyz++= . Substitute 3z xy=−− into the expression for S gives ( ) ( ) ( ) ( ) ( ) ( ) 22 set set 2 23 3 23 3 23 2 0 2 3 23 2 0 2 3 S xy xz yz xy x x y y x y x x y y xy S xy xyx S yx x yy = + + = + −− + −− = −+−− ∂ = − − =⇒ +=∂ ∂ = − − =⇒+ =∂ Solving gives 11xy z==⇒= . ( )( ) ( ) 2 22 22 22 2 2 22 4 0, 4, 2, 4 4 2 12 0 S SS yxxy SS SD yxxy ∂ ∂∂= −< = − = − ∂∂∂∂ ∂∂ ∂= ⋅ − =− − −− = >∂∂∂∂  By the second derivative test, the box which is a unit cube has maximum surface area 6 units2.

2 The curve C has polar equation ( )2sin 1 cosr θθ= − for 0 πθ≤≤ . (a) Find d d r θ and hence find the polar coordinates of the point P of C that is furthest from the pole. [3] (b) Sketch C, indicating the polar coordinates of the point P. [2] (c) Find the exact area of the region bounded by C and the two lines 0θ = and π 4θ = . [4] Qn Solutions 2(a) ( ) ( ) ( )d 2 1 cos cos sin sin 2 cos cos 2d r θθ θθ θ θθ = −+ = − . ( ) [ ]d2 π0 2 cos cos 2 0 cos cos 2 0, πd3 r θ θ θ θθθ =⇒ − =⇒ = ⇒= ∈ . Coordinates of point furthest from O is 2π 2π 2π 3 3 2π2sin 1 cos , ,3 3 3 23  −=    . 2(b) 332 π,23  

2(c) Area of the region bounded by C and the two lines 0θ = and π 4θ = ( ) ( ) π 24 2 0 π 4 22 0 1 4sin 1 cos d2 2 sin 1 2cos cos d θ θθ θ θ θθ = − = −+ ∫ ∫ ππ π 44 4 2 22 00 0 12sin d 4 cos sin d sin 2 d 2θθ θ θθ θθ= −+∫∫ ∫ ( ) ( ) ππ π 44 4 2 00 0 ππ π44 3 4 000 11 cos 2 d 4 cos sin d 1 cos 4 d 4 1 4 11sin 2 sin si n 42 3 44 π 1 422 1 π 4 2 3 8 44 5π1 2 16 2 3 θ θ θ θθ θ θ θθ θ θθ = − − +−   = − − +−     =−− +    = −− ∫∫∫

3 The sequence { }nu is given by the recurrence relation ( )21 1 3 42 2n nnu uu++ = + − for 0,n≥ together with initial terms 1u p= and 2 ,u q= where p and q are positive integers. The sequence { }nv is given by 6nnv un= + for 0.n≥ (a) Find an expression nv in terms of n, p and q. [5] (b) Given that the limit for which nv converges to is 13, find the possible values of p and q. [2] (c) Hence, find nu in terms of n, p and q. State what happens to nu for large values of n. [2] Qn Solution (a) ( ) ( )( ) 21 21 21 21 1 3 1 3 21 33 2 6( 2) 2 6( 1) 6 6 12 2 0 24 24 44 8 32 n nn n nn n nn n nn u uu v n v n vn v nn v vn v vv ++ ++ ++ ++ − − = −− = + −+ = −+ + − −− +− −− − = Characteristic equation: 2 1 3 20 (3 1)( 1) 0 or 1 3λ λλ λλ λ − −= + −= =− General solution: 11 33() ( 1 ) ()nn n n AB ABv − ==−+ + Initial conditions: 11 22 211 33 6(1) 6 6(2) 12 ) 6 (1) ) 12 (2)(( A pq Bp B v q u vu A = = −− += + + = + + = + −−− + = + −−−

( ) ( ) ( ) 9 4 9 4 3 1 44 1 3 6 6 Subt into (1) : 6 6 6 3 42 (2) (1) : Aqp A qp B Ap qp p qp =−+ = −+ = ++= −+ − ++= ++ ( ) ( )1 3 9 1 44 6 ( ) 3 42n nv qp qp∴= −+ − + ++ (b) ( )1 4 3 42 13 3 10 for positive integers and , possible va lues : 1, 7 2, 4 3, 1 qp qp pq qp qp qp ++ = += = = = = = = (c) ( ) ( )1 3 9 1 44 6 6 ( ) 3 42 6 nn n u n vn qp qp = − = −+ − + ++ − As n→∞ , nu →−∞

4 The curve with equation lnyx= and the line with equation 2yx= − intersects at x α= and x β= , where αβ< . (a) It is given that β lies in the interval ( ),1kk + , where k is an integer, state the value of k . Justify that there is indeed a root in the interval ( ),1kk + . [2] (b) Show that one application of linear interpolation using (b) will obtain a first approximation for β as 3.138439, correct to 6 decimal places. [2] (c) Hence, use the Newton-Rhapson method to approximate β , correct up to 4 decimal places. Justify that the approximation is correct up to 4 decimal places. [3] (d) It is known that the Newton- Rhapson method, when applied to ( )f x can approximate the root α using, as long as the first approximation lies in the interval ( )0,c . Find the largest value of c, leaving your answer in exact form. [3] Qn Solution 4(a) k = 3 ( )f 3 0.098612 0= > ( )f 4 0.61370 0= −< By the Intermediate Value Theorem, there exist a root in ( )3, 4 4(c) ( ) ( ) ( ) ( ) 1 3f 4 4f 3 f3 f4 3.13843858 3.138439 (6dp) x += + ≈ =

4(d) ( ) 1f' 1x x= − ( ) ( ) 1 21 1 f f' ln 3.138439 3.138439 23.138439 1 13.138439 3.146197 3.1462 (4dp) xxx x= − −+= − − ≈ = To justify 3.1462β = (correct to 4dp), ( )f 3.14615 0.000029483 0= > , ( )f 3.14625 0.000038732 0=−< By the Intermediate Value Theorem, ( )3.14615,3.14625β∈ . Any number in this interval is rounded to 3.1462. 4(e) Consider the tangent to ( )fyx= at point ( )( ),fcc : ( ) ( )( ) ( ) ( ) f f' 1ln 2 1 y c cxc y cc xcc −= − − −+ = − −  Sub in ( )0, 0 , 1 0 ln 2 1 ln 1 e cc c c c − − +−=− =− =

5 The country of Ganyan is looking to export timbre to boost its economy. In 2025, the government of Ganyan acquired a timbre plantation and decided to let the plantation grow. After t number of years, the amount of timbre, T thousand tonnes, can be modelled as d 1d TT kTta = −  , where a and k are positive constants. (a) State the significance of a, in the context of the question. [1] For the rest of the question, let 1000a= . It is also given that when 0t = , 200T = , d 16d T t = . (b) Find the time taken for the timbre to double in amount. [4] Upon reaching 800T = , the government of Ganyan starts harvesting a fixed amount of timbre, h thousand tonnes per year. (c) Given that 0.1k = , find the range of values of h for the harvest to be sustainable in the long run. [2] (d) For 15h= , find the equilibrium amount of timbre in the long run. [3]

Qn Solution 5(a) a is the carrying capacity, which is the maximum possible amount of timbre the plantation can have. 5(b) 0t = , 200T = , d 16d T t = 20016 200 1 1000 0.1 k k = −  = 0.1 d 0.1 1d 1000 1 d 0.1d 1 1000 11 d 0.1d1000 ln 0.11000 e , e1000 tC TT Tt TtTT TtTT T tCT T AAT = −  =−  += − = +− = =±− ∫∫ ∫∫ 200 800 1 4 A A = = 400, T = 0.1400 1 e600 4 810ln 3 9.8082 9.81 years (3sf) t t = = ≈ =

5(c) Let q years be the number of years after the harvesting starts. d 0.1 1d 1000 TT Thq =−−  ( ) ( ) 2 d1 1000d 10000 1 500 2510000 T T Thq Th = −− = − − −+ For d 0d T q ≥ , 25 0 25 h h −+ ≥ ≤ 0 25h∴≤≤ 5(d) ( ) 2 d 0.1 1 15d 1000 1 500 1010000 TT Tq T =−−  = − −+ Let d 0d T q = 500 100000T = − or 500 100000T = + Since 800 500 100000T =>− , 500 100000T →+