2023 H2 Paper 2_Solutions_Students
Uploaded by passionfruit · 6 November 2024
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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 v
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