2021 A Levels H3 Math Paper 1 Soln

Solutions to 2021 GCE ‘A’ Level H3 Mathematics (9820/01) 1 (a) (i) As shown in the diagram, let R be the region between the x-axis and the curve 2 ln 1 xy x = + for 01 x . Let S be the region between the x- axis and the curve 2 ln 1 xy x = + for 1x . y O S R x Use the substitution 1x t= to prove that R and S have equal areas. [3] (ii) For a > 0, use the substitution x at= to find 220 ln dx x ax  + . [5] (b) For a > 0, use integration by parts to find 22 20 ln dax xx  + [5] Solution: 1 (a) (i) y O S R 1 x Area of R = 1 20 ln d 1 x x x − + (since region lies below the x-axis) 1 1 2 2 1ln 1 d11x t t tt t =  = − − + 21 ln d = Area of 1 t tS t  = + .

2 (ii) ( )2 2 2 2 200 ln ln d dx atx at x a t a x a a t  = = ++ 1 20 20 2200 1 2 2 20 0 1 ln d 1 11 1 220 00 1 ln ln d 1 1 ln ln d d 11 ln 1 1 ln ln d d d 1 1 1 ln 1 ln lntan d d by (i) 11 ln= t t t at ta t at tta tt a t t t t taa t t t a t t t t taa tt a a    − + − += + =+  ++   = + + + + +    = + −  ++        ( ) π 2 ln π 2 a a   = (b) ( ) 2 2 2 2 2 2 2 3 22 22ln 1 d d d dd a x a au u x x x a x x a x x v x v x    = +  = − =−    + +    =  = 2 2 2 2200 ln d ln 1 da x a xxxx    + =+       22 2 2 2 00 0 2 220 1 0 2ln 1 d 12d 2 tan π aaxx x a x ax ax xa a a     −  = + +  +  = +  =   =    Note: We have assumed that 2 2 0 ln 1 0 ax x   +=   . The justification is as follows:

3 ( ) 22 2222 L'Hopital 20, 0, 0, 2 2 220, 2 ln 1 lim ln 1 lim lim 11 2 lim x x x x aa a x xxax x x x ax ax →  →  →  →  −+ + + = =  −  =  + 0= 2 Let a, b, c and r be positive real numbers. (a) Prove that ( )( ) ( )( ) ( )( ) 0r r ra a b a c b b c b a c c a c b− − + − − + − −  . [4] (b) Hence, or otherwise, prove that (i) ( ) ( ) ( ) 3 3 3 2 2 2 3a b c abc a b c b c a c a b+ + +  + + + + + , (ii) 2 2 2 2 2 2 5 5 5 2 2 2 3 2 2 2 3 2 2 2 3 1 1 1 a b c b c c a a b a b c a b c a b c a b c a b c + + + + ++ + +  + + . [3] [3] Solution: 2 (a) Since the expression is symmetric in a, b, c, without loss of generality, we may assume 0abc   . Then , , 0a b b c a c− − −  . Sum of first two terms is ( )( ) ( )( ) ( ) ( ) ( ) r r r ra a b a c b b c b a a b a a c b b c − − + − − = − − − −  Since 0abc   and r > 0, 0rrab (since where , 0rx x r  , is an increasing function) and 0a b a c b c  −  −  . Hence, on multiplying corresponding terms, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 0 since 0 r r r r ab rr a a c b b c a a c b b c a b a a c b b c a b  −  −  −  − − −   − − − − −  Last term is ( )( ) ( )( ) 0rrc c a c b c a c b c− − = − −  since 0, , 0c a c b c − −  . Hence ( )( ) ( )( ) ( )( ) 0r r ra a b a c b b c b a c c a c b− − + − − + − −  . ---- (*) (b)(i) Put 1r= in (*) gives

4 (ii) ( )( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 3 3 3 2 2 2 3 3 3 2 2 2 0 0 30 3 a a b a c b b c b a c c a c b a a ab ac bc b b ab bc ac c c bc ac ab a b c abc a b c b c a c a b a b c abc a b c b c a c a b − − + − − + − −  − − + + − − + + − − +  + + + − + − + − +  + + +  + + + + + Put 1 02r = and effect the replacement 2 1 0, , , 0i i a b c i →  =  in (*) gives 1 2 2 2 2 2 2 1 2 2 2 2 2 2 1 2 2 2 2 2 2 4 2 2 2 2 2 2 4 2 2 2 2 2 2 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 11 a a b a c b b c b a c c a c b a a a c a b b c b b a b b c a c c c     −−             + − −             + − −           − − + + − − + + 2 2 2 2 2 2 5 5 5 2 2 2 2 2 2 3 2 3 2 2 3 3 2 2 3 2 3 2 2 2 2 2 2 5 5 5 2 2 2 3 2 2 2 3 2 2 2 3 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 b c a c a b a b c ab c a bc a b c a c a b a b b c b c a c a b c b c c a a b a b c a b c a b c a b c a b c  − − +   + + + + +  + + + + + + + + + + + + +  + +

5 3 Let u and v be quadratic functions of x and let uy v= . (a) Use mathematical induction to prove that ( ) 2 1 2 2 1 2 2d d d d d 20 2dd d d dn n nn n n ny v y v yvn xx x x x++++ ++ + + =  , for 1n . [8] (b) Now assume that ( ) 2 vx =− for some real number  and, for all positive integers n, define ( ) 2 d ! d n n n n x yz n x + −= . Use the result of part (a) to prove that 1 2 3, , ,z z z  is an arithmetic progression. By writing y as partial fractions, or otherwise, show that the common difference is ( )u  . [7] Solution: 3 Let u and v be quadratic functions of x and let uy v= . (a) To prove: ( ) 2 1 2 2 1 2 2d d d d d 20 2dd d d dn n nn n n ny v y v yvn xx x x x++++ ++ + + =  for 1n . For n = 1, LHS = 3 2 2 3 2 2 d d d d d33ddd d dy v y v yv xxx x x++ . uy vy uv=  = . Differentiate w.r.t. x: d d d d d d y v uvy x x x+= . Differentiate w.r.t. x: 2 2 2 2 2 2 d d d d d2 ddd d dy v y v uvy xxx x x+ + = . Differentiate w.r.t. x: 3 2 2 3 3 3 2 2 3 3 d d d d d d d33ddd d d d dy v y v y v uvy xxx x x x x+ + + = . ---- (1) Since u and v are quadratic functions of x, i.e. of the form 2ax bx c++ , 33 33 dd 0 dd uv xx == . Therefore the above equation (1) gives

6 3 2 2 3 2 2 LHS d d d d d3 3 0ddd d dy v y v yv xxx x x++=  = RHS. Hence the statement holds for n = 1. Assume the statement holds for some positive integer k. That is ( ) 2 1 2 2 1 2 2d d d d d 20 2dd d d dk k kk k k ky v y v yvk xx x x x++++ ++ + + =  . Differentiate above w.r.t. x: ( ) ( ) 3 2 2 2 1 3 2 2 2 1 2 1 3 2 1 3 0 3 2 2 1 3 2 2 1 d d d d d d d 2ddd d d d d 2 d d d d 0 2 d d d d 2d d d d d 3 2 0 2dd d d d k k k k k k k k kk kk k k k k k k y v y v y v yvk xxx x x x x k v y v y x x x x ky v y v yv k k xx x x x + + + + + + + + + + + + + + + + + + + +    + + + =    +  + + + + + =    ---- (*) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( ) 2 2 ! 1 2 !2 2 2 2 2! ! 2! 1 ! 32! 2 ! 1 2 ! 3 2 ! 3 ! 22! 1 ! 2! 1 ! 2! 1 ! k k k kk k k kk kk k k k k k k k k + + + ++ + = + + = + + + ++ + + + + + + = = = = + + +  Hence (*) gives ( ) 3 2 2 1 3 2 2 1 3d d d d d 30 2dd d d dk k kk k k ky v y v yvk xx x x x+ + ++ + + ++ + + =  proving that the statement holds for n = k + 1 and hence for all 1n . (b) ( ) ( ) 22 2 dd 22d dvvv x x x x= −  =− −  = . ( ) 2 d ! d n n n n x yz n x + −= . From (a), we have for 1n , ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) 2 1 2 2 1 2 212 21 2 1 12 2 1 1 2d d d d d 20 2dd d d d 2d d d 2 2 2 0 2d d d 2d d d d 2 2 2 2d d d d n n n n n n n n n n n n n n n n n n n n ny v y v yvn xx x x x ny y yx n x x x x ny y y yx n x n x x x x x     ++ ++ ++ ++ + + + + + + ++ + + =  +− + + − − + =    +− − + − = + − − 

7 Multiply throughout by ( ) ( ) 2 2! n x n  + − + gives ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 43 21 21 32 1 1 dd 2 ! 1 !dd 2!dd 1 ! ! 2 !dd nn nn nn nn nn nn xx yy nn xx x n x yy n n n xx   ++ ++ ++ ++ + + −− −++ − + −=− ++ ( ) ( ) ( ) ( ) ( ) ( ) ( ) 4 3 3 2 2 1 1 2 1 1 2 1 1 d d d d 2 ! 1 ! 1 ! !d d d d n n n n n n n n n n n n n n n n x x x x y y y y n n n n x x x x z z z z     + + + ++ + + + + + + + + − − − − − = −+ + +  − = −  1 2 3, , ,z z z  is an arithmeti