Edexcel Mathematics Higher Tier, November 2009 (1380/4H) (Paper 4, calculator)


Question 28 a = 6.43 correct to 2 decimal places. a could really be as small as 6.425 or as large as 6.435 and still get rounded to 6.43


b = 5.514 correct to 3 decimal places. b could really be as small as 5.5135 or as large as 5.5145 The largest value of v would happen when a is as large as possible and b is as small as possible. v = √(6.435 ÷ 5.5135) = 1.08034… The smallest value of v would happen when a is as small as possible and b is as large as possible. v = √( 6.425 ÷ 5.5145) = 1.07940…


if we round both of these figures to 2 decimal places we get the same answer of 1.08 so we can say that v = 1.08 correct to 2 decimal places.


Question 29 first multiply both sides by (x + 3)


4 +    = (x + 3)


multiply both sides by (2x – 1) 4(2x – 1) + 3(x + 3) = (x + 3)(2x – 1) expand 8x – 4 + 3x + 9 = 2x2 + 6x – x – 3 group terms 11x + 5 = 2x2 + 5x – 3 subtract 11x from both sides 5 = 2x2 – 6x – 3 subtract 5 from both sides 0 = 2x2 – 6x – 8


rewrite 2x2 – 6x – 8 = 0 divide by 2


x2 – 3x – 4 = 0 factorise (x – 4)(x + 1) = 0 so if two brackets multiply to give 0 then one or both of the brackets must equal 0 x = 4 or x = -1


to check: put back x = 4 into original equation to get  and put x = -1 into original equation to get 


 +   +   =   = 1  = 2 + -1 = 2 – 1 = 1


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