2023 H2 Paper 2 Solutions Students

A level H2 Mathematics Paper 2 2023_Suggested Solutions Section A: Pure Mathematics (40 marks) 1 (a) Find the set of values of x for which 3 2 5 2xx−  − . [2] Solutions 1 2.2x   (b) Express 2 25 345 x xx + +−− as a single fraction. Hence, without using a calculator, solve exactly the inequality 2 25 345 x xx + −−− . [4] Solutions 2 22 2 2 25 25 3 12 1534 5 4 5 3 11 10 45 x x x x x x x x xx xx + + + − −+=− − − − −+= −− ( )( ) ( )( ) ( ) ( ) ( )( )( )( ) 22 2 2 2 2 25 345 25 3045 3 11 10 045 3 5 2 0, 51 Multiplying by 5 1 : 3 5 2 5, 1 5 1 0 x xx x xx xx xx xx xxx xx x x x x + −−− + +−− −+ −− −− −+ −+ − − − + −   –1 5/3 2 5

A level H2 Mathematics Paper 2 2023_Suggested Solutions 51 or 2 or 53x x x −    2 (a) Given that ln(sec )yx= , show that 32 32 d d d 2.d d d y y y x x x  =   [3] Solutions d1 (sec tan )d sec y xxxx= tanx= ( ) 2 2 2 2 2 d secd 1 tan d 1 1 d y xx x y x = =+ = + −−− Differentiating with respect to x: 32 32 d d d 2 (2)d d d y y y x x x   = −−−  (shown) 2 (b) Hence, or otherwise, obtain the Maclaurin expansion of y in terms of x up to and including the term in 4x . [3] Solutions Differentiating (2) with respect to x: 4 3 2 2 4 3 2 2 d d d d d22d d d d d y y y y y x x x x x     =+          When 0,x= 0,y= d 0,d y x = 2 2 d 1,d y x = 3 3 d 0,d y x = 4 4 d 2.d y x = 24 2 ...2 4! xxy  = + + 

A level H2 Mathematics Paper 2 2023_Suggested Solutions 24 2 12 xx+

A level H2 Mathematics Paper 2 2023_Suggested Solutions 2 (c) By putting 1 ,4x = find an approximation for ln 2 in terms of  . [2] Solutions 24 ln(sec ) 2 12 xxx + Substituting :4x = 24 44ln sec 4 2 12             + ( ) 24 ln 2 32 3072 + ( ) 241 ln 22 32 3072 + ( ) 24 ln 2 16 1536   + 2 (d) Using your answer to part (b), find an approximation to 1 10 0 ln(sec ) dxx   . Give your answer correct to 4 significant figures. [1] Solutions 11 24 10 10 00 ln(sec ) d + d 2 12 xxx x x    0.005219= (to 4 sig fig)

A level H2 Mathematics Paper 2 2023_Suggested Solutions 3 The points A and B have position vectors 1 2 5 −   and 1 2 8  − respectively. The point C is such that 2BC AB= . (a) Find the position vector of C. [2] Solutions 11 2 , 2 58 OA OB −       = = −          1 1 2 2 2 4 8 5 3 AB OB OA −           = − = − − = −                Given 2 , 2 24 3 4 1 5 8 2 10 6 8 14 BC AB OC OB OC =  − = −             = − + − = −                The points D has position vector 1 2 d    and is such that AD BD= . (b) Find the value of d. [2] Solutions 1 1 2 2 2 0 55 1 1 0 2 2 4 88 AD OD OA dd BD OD OB dd −           = − = − =            −                 = − = − − =           −     

A level H2 Mathematics Paper 2 2023_Suggested Solutions ( ) ( ) ( ) ( ) 2222 22 22 22 2 5 4 8 4 5 16 8 4 10 25 16 16 64 10 29 16 80 6 51 17 8.52 AD BD dd dd d d d d d d d d d d = + − = + − + − = + − + − + = + − + − + = − + = == (c) Use a scalar product to find angle ADB. [3] Solutions 1 1 0 2 2 4 8 8.5 0.5 DB            = − − = −           −      1 1 2 2 2 0 5 8.5 3.5 DA −−           = − =           −      02 40 0.5 3.5cos 02 40 0.5 3.5 1.75 16.25 1.5642 1.56 rad or 83.8 −       −      −−   = −       −      −−    = = =  

A level H2 Mathematics Paper 2 2023_Suggested Solutions (d) Find exactly the position vector of the point P, where P is the centre of the circle that passes through A, B and D. [5] Solutions Method 1 1 1 2 2 2 0 5 8.5 3.5 1 1 0 2 2 4 8 8.5 0.5 DA OA OD DB OB OD −−           = − = − =            −                 = − = − − = −           −      Let M be the midpoint of A and B, 2 20 04 3.5 0.5 2 1 2 2 DA DBDM += −       +−      −−   = − =−− Equation of line DM is 11 2 2 , 8.5 2 r         = +           Since P is on line DM, 11 2 2 , for some 8.5 2 OP        = +            A B P M D

A level H2 Mathematics Paper 2 2023_Suggested Solutions 1 1 2 2 2 2 2 8.5 2 5 3.5 2 AP    + − +           = + − =          ++      11 2 2 2 2 8.5 2 8.5 2 DP    +           = + − =          +      Since DP and AP are radius of the circle, AP DP= ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3.5 2 2 2 2 4 4 12.25 14 4 8 18 16.25 0 65 72 + + + + = + + + + + + + + = + += =−                11 6522 728.5 2 7 1 1472 482 OP        =−           =  Method 2 Let M be the midpoint of A and B, A B P M D E

A level H2 Mathematics Paper 2 2023_Suggested Solutions 2 20 04 3.5 0.5 2 1 2 2 DA DBDM += −       +−      −−   = − =−− Equation of line DM is 11 2 2 , 8.5 2 r         = +           Since P is on line DM, 11 2 2 , for some 8.5 2 OP        = +            ( ) 2 2 DE DP OE OD OP OD = − = − ( )2 2 1 1 1 2 2 2 2 8.5 2 8.5 11 2 2 2 8.5 2 OE OD OP OD OP OD = + − =−            = + −                       =+            Note that 90DBE=

A level H2 Mathematics Paper 2 2023_Suggested Solutions 0 0 1 1 1 4 2 2 2 2 0 0.5 8.5 2 8 0 0 1 4 4 2 2 0 0.5 0.5 2 116 16 2 04 65 72 DB BE =                − + − − =              −                   − + =          −      − − − − = =−     11 6522 728.5 2 7 1 1472 482 OP        =−           =  Method 3 P is of equal distance from A, B and D and P lie in plane ABD. 2 4 3 AB  =− , 2 0 3.5 AD  = Normal vector of plane ABD, ( ) ( ) 22 40 3 3.5 14 0 76 08 14 1 8 n        = −           −− = − −  −− − =−