RI 2022 H3 Test 2 (Questions and Solutions)

RI 2022 RAFFLES INSTITUTION 2022 Year 6 H3 9820 Test 2 Time allocated: 1 hour 50 minutes Total Marks: 60 Instructions: Write your name and CT group on all the work you hand in. Answer all questions. 1 (a) Expand and simplify ( )( ) 1 2 2 1 ...n n n n na b a a b a b ab b− − −− + + + + + . [2] (b) The prime number 3 has the property that it is one less than a perfect square. Determine all prime numbers with this property, justifying your answer. [2] (c) Find all prime numbers that are one more than a perfect cube, justifying your answer. [3] (d) Is 2021 202132 − a prime number? Explain your reasoning carefully. [2] (e) Is there a positive integer k for which 32 2 2 1k k k+ + + is a perfect cube? Explain your reasoning carefully. [3] 2 (a) Suppose that a, b and c are positive real numbers such that the polynomial 32f ( ) 3 3x x ax bx c= − + − has three positive real roots , and    . (i) Express a, b and c in terms of , and    . [3] (ii) Show that 3bc . [2] (iii) Use the graph of f ( )yx= to e xplain why the polynomial f '( )x has 2 positive roots. [2] (iv) Hence by considering f '( )x , show that ab . [2] (b) Let A, B and C be the angles of a triangle. (i) Show that tan tan tan tan tan tan 12 2 2 2 2 2 A B B C C A+ + = . [3] (ii) Hence, using the results established in (a) and (b)(i), show that tan tan tan 32 2 2 A B C+ +  and 3tan tan tan2 2 2 9 A B C  . [3]

RI 2022 3 A sequence 12, ,...xx of real numbers is defined by 2 1 2nnxx+ =− for 1n and 1xa= . (a) Show that if 2a then ( ) 12 4 2 .n nxa − + − [5] (b) Show also that nx → as n→ if and only if 2a  . [5] 4 Throughout this question, no marks will be awarded for any use of the exponential or the logarithmic function. For positive real numbers x, define 1 1F( ) d x xt t= . (a) Show that F is a strictly increasing function. [1] (b) Show in any order, that for all positive real numbers ,ab , (i) F( ) F( ) F( )ab a b=+ , (ii) F F( ) F( )a abb  =− . [4] (c) If there exists a real number L such that lim F( ) x xL → = , state lim F(2 ) x x → . Hence deduce that lim F( ) x x → =+ . [3] (d) Show that 1F F( ) xx  =− for all positive real numbers x and hence find 0 lim F( ) x x +→ , explaining your reasoning clearly. [2] (e) Show also that F(2) 1 F(3) . [3] 5 Let 1, 2,3,...,2 1 .Sn=− Remove at least 1n− numbers from S using the following rules: • If the number sS is removed and 2sS , then 2s must be removed, • If the numbers ,s t S are removed and s t S+ , then st+ must be removed. After all the possible numbers are removed from S, let T denote the sum of the remaining numbers. (a) Find the smallest possible value of T. [3] (b) Find the largest possible value of T. [7] [END OF TEST]

RI 2022 y x 1 (a) [2] ( )( ) 1 2 2 1 1 1 2 1 2 1 11 ... ... n n n n n n n n n n n n n nn a b a a b a b ab b a a b a b a b a b ab ab b ab − − − + − − + ++ − + + + + + = − + − + − − + − =− (b) [2] Since 2( 1)( 1) 1n n n p+ − = − = , we must have 11n−= since 11nn−  + . This means that n = 2 and the only prime with this property is 22 1 3−= . (c) [3] We want 32 1 ( 1)( 1)n n n n p+ = + − + = and therefore 11n+= or 2 11nn− + = . In the first case we must have n = 0 which doesn’t lead to a prime, and in the second we have either n = 0 or n = 1. For n = 1 we get 31 1 2+= , a prime as desired. (d) We have ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) 47 472021 2021 43 43 46 45 45 4643 43 43 43 43 43 43 43 3 2 3 2 3 2 3 3 2 ... 3 2 2 − = − = − + + + + and since each term is greater than 1, 2021 202132 − is not a prime number. (e) Note that for positive k, 3 3 2 3 2 2 2 1 3 3 1k k k k k k k + + +  + + + and since the expression lies between 2 consecutive cubes, it cannot be a perfect cube. 2 (a) (i) [3] 32f ( ) ( )( )( ) 3 3x x x x x ax bx c  = − − − = − + − Compare the coefficients of 2x : 3 3aa      +++ + =  = x: 3 3bb      +++ + =  = constant term: c = (a) (ii) [2] Since , and    are positive, , and    are positive. By AM-GM inequality, 3223 ()3bc    ++=  = Since b and c are positive real numbers, 3bc . (a) (iii) [2] The possible cases for the graph of polynomial f ( )x are (i) (ii) (iii) (iv) y x y x y x

RI 2022 Since the polynomial f ( )x has 3 roots, the polynomial f '( )x which is quadratic, has 2 roots. In addition, the roots of the polynomial f '( )x are the x-coordinates of the stationary points of the polynomial f ( )x . As the stationary points occur between the roots ( , and    ) of the polynomial f ( )x (as seen in (i)) or are one of the roots themselves (as seen in (ii), (iii) and (iv)) and since these roots , and    are positive, the roots of the polynomial f '( )x which are the x-coordinates of the stationary points of the polynomial f ( )x are positive. Hence the polynomial f '( )x has 2 positive roots. (a) (iv) [2] 2f '( ) 3 6 3x x ax b= − + Let the roots of polynomial f '( )x be p and q. Then 2p q a+= and pq b= . Since p and q are positive, by AM-GM inequality, 2 pqa pq b+=  = . Alternative 22f '( ) 3 6 3 3( 2 )x x ax b x ax b= − + = − + Since the roots of polynomial f '( )x are real, Discriminant = 24 4 0a b a b−    . (b) (i) (b) (ii) Since A, B and C be the angles of a triangle, 22 A B C A B C   + + = ++ = tan tan2 2 2 2 tan tan22 cot 21 tan tan22 tan tan tan tan 1 tan tan2 2 2 2 2 2 tan tan tan tan tan tan 12 2 2 2 2 2 A B C AB C AB A C B C A B A B B C C A    + = −       + = − + = − + + = Let tan , tan and tan2 2 2 A B C  = = = . As A, B and C be the angles of a triangle, , and 2 2 2 A B C are acute angles and hence tan , tan and tan2 2 2 A B C  = = = are positive.

RI 2022 From (a)(iv), 33 tan tan tan tan tan tan tan tan tan 12 2 2 2 2 2 2 2 2 33 3 tan tan tan 32 2 2 ab A B C A B B C C A A B C        + + + + + + + + = + +  From (a)(ii), 3bc 3 3 3 3 tan tan tan tan tan tan2 2 2 2 2 2 tan tan tan3 2 2 2 1 tan tan tan2 2 23 A B B C C A A B C A B C    ++  ++   3 13tan tan tan2 2 2 9 3 A B C = 3 (a) Let Pn be the statement ( ) 12 4 2n nxa − + − for 2a . Base case 1P : When 1,n= ( ) 1 12 4 2 2 2n a a a x−+ − = + − = = . So 1P is true. WTS Pk is true  1Pk+ is true. Suppose ( ) 12 4 2k kxa − + − for some k + (and 0kx  as 2a ) Then ( ) ( ) ( ) ( ) ( ) ( ) 221 1 21 2 2 222 2 2 4 2 2 4 2 2 4 2 4 2 2 2 4 2 4 2 2 4 2 k kk kk kk k x x a aa aa a − + −− − = −  + − − = +   − + − − = + − + −  + − Hence 1Pk+ is true. Thus by induction ( ) 12 4 2n nxa − + − for positive integer n.

RI 2022 (b) If 2kx  , then 2 04 kx , so 2 2 2 2 kx−  −  , that is 122 kx +−   . If 2a  , 1 2x  and thus by induction 22 nx−   , that is nx → For the case where 2a  : Regardless of whether a is positive or negative, 22 22aa− = − , hence it suffices to consider 2a for the behavior of all terms after 1x . Therefore, from part (i), we know ( ) 12 4 2n nxa − + − for 2n , and thus nx → as n→ ; Hence we have shown nx → as n→ if and only if 2a  . 4 (a) [1] By the Fundamental Theorem of Calculus, 1F ( ) 0x x  = Alternatively, if ab , since 1 0t  for positive t, then 1F( ) F