29)Sn=4n-n2 S1 = 3 ½ mark S2= 4 ½ mark a2 = S2 - S1 =1 1 mark a10 = S10 S9 = -15 2 mark – x = /3 and x = -√5 /3 are zeroes of p(x) 30)Answer :As √5 Therefore (x - ) ( x + ) is factor of p(x) ½ mark ( x2 5/3) is the required factor – Or 3 x2 5 is the required factor – Now On Dividing 3x4 + 6x3 2x2- 10x 5 by 3 x2 5 – – – 3 x2 5 3x4 + 6x3 2x2- 10x 5 x2 + 2x +1 – – – 1 mark 3x4 5x2 – - + 6x3+ 3x2- 10x 5 – 6x3 - 10x - + + 3x2 5 – + 3x2 5 – - + 0 1 mark we get the quotient x2 + 2x +1 Other zeros are given by x2 + 2x +1= 0 x2 + x + x + 1 =0 1 mark x( x +1) +1 ( x +1) =0 x +1 =0 or x +1 = 0
29)Sn=4n-n2 S1 = 3 ½ mark S2= 4 ½ mark a2 = S2 - S1 =1 1 mark a10 = S10 – S9 = -15 2 mark 30)Answer :As x = √5 /3 and x = -√5 /3 are zeroes of p(x) Therefore (x - ) ( x + ) is factor of p(x) ½ mark ( x2 – 5/3) is the required factor Or 3 x2 – 5 is the required factor Now On Dividing 3x4 + 6x3 – 2x2- 10x – 5 by 3 x2 – 5 3 x2 – 5 3x4 + 6x3 – 2x2- 10x – 5 x2 + 2x +1 1 mark 3x4 – 5x2 - + 6x3+ 3x2- 10x – 5 6x3 - 10x - + + 3x2 – 5 + 3x2 – 5 - + 0 1 mark we get the quotient x2 + 2x +1 Other zeros are given by x2 + 2x +1= 0 x2 + x + x + 1 =0 1 mark x( x +1) +1 ( x +1) =0 x +1 =0 or x +1 = 0
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