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Core 1 maths question
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ruby_zara
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#1
Hi, please could someone help me with the last part of this question. I don't quite understand from the mark scheme how you get to the answer. I'll just put the other parts of the question in before for context.
In the previous parts of the question you have to complete the square on 2x^2+6x+5 which I unmderstand and then write down the minimum value for the second part, which would be 1/2.
And then it says if point A has co-ordinates (-3,5) and point B has the co-ordinates (x,3x+9) show that AB^2 = 5(2x^2 +6x +5), which I understand how to do but then the last part says
Using your answer from part a).ii (which was 1/2) find the minimum value of the length AB as x varies giving your answer in the form 1/2(square root of n) where n is an integer.
Sorry if that is really confusing . If it helps the links to the past paper and mark scheme are here.
http://filestore.aqa.org.uk/subjects...1-QP-JUN13.PDF - question paper
http://filestore.aqa.org.uk/subjects...W-MS-JUN13.PDF - mark scheme
Thanks
In the previous parts of the question you have to complete the square on 2x^2+6x+5 which I unmderstand and then write down the minimum value for the second part, which would be 1/2.
And then it says if point A has co-ordinates (-3,5) and point B has the co-ordinates (x,3x+9) show that AB^2 = 5(2x^2 +6x +5), which I understand how to do but then the last part says
Using your answer from part a).ii (which was 1/2) find the minimum value of the length AB as x varies giving your answer in the form 1/2(square root of n) where n is an integer.
Sorry if that is really confusing . If it helps the links to the past paper and mark scheme are here.
http://filestore.aqa.org.uk/subjects...1-QP-JUN13.PDF - question paper
http://filestore.aqa.org.uk/subjects...W-MS-JUN13.PDF - mark scheme
Thanks
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tajtsracc
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#2
(Original post by ruby_zara)
Hi, please could someone help me with the last part of this question. I don't quite understand from the mark scheme how you get to the answer. I'll just put the other parts of the question in before for context.
In the previous parts of the question you have to complete the square on 2x^2+6x+5 which I unmderstand and then write down the minimum value for the second part, which would be 1/2.
And then it says if point A has co-ordinates (-3,5) and point B has the co-ordinates (x,3x+9) show that AB^2 = 5(2x^2 +6x +5), which I understand how to do but then the last part says
Using your answer from part a).ii (which was 1/2) find the minimum value of the length AB as x varies giving your answer in the form 1/2(square root of n) where n is an integer.
Sorry if that is really confusing . If it helps the links to the past paper and mark scheme are here.
http://filestore.aqa.org.uk/subjects...1-QP-JUN13.PDF - question paper
http://filestore.aqa.org.uk/subjects...W-MS-JUN13.PDF - mark scheme
Thanks
Hi, please could someone help me with the last part of this question. I don't quite understand from the mark scheme how you get to the answer. I'll just put the other parts of the question in before for context.
In the previous parts of the question you have to complete the square on 2x^2+6x+5 which I unmderstand and then write down the minimum value for the second part, which would be 1/2.
And then it says if point A has co-ordinates (-3,5) and point B has the co-ordinates (x,3x+9) show that AB^2 = 5(2x^2 +6x +5), which I understand how to do but then the last part says
Using your answer from part a).ii (which was 1/2) find the minimum value of the length AB as x varies giving your answer in the form 1/2(square root of n) where n is an integer.
Sorry if that is really confusing . If it helps the links to the past paper and mark scheme are here.
http://filestore.aqa.org.uk/subjects...1-QP-JUN13.PDF - question paper
http://filestore.aqa.org.uk/subjects...W-MS-JUN13.PDF - mark scheme
Thanks
But they wanted the minimum value of AB, not (AB)2, so square root this value. √5/2
√5/2 = √5/√2 Now rationalise the denominator. √5 x √2 / √2 x √2 = √10/2 = 1/2√10.
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ruby_zara
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#3
(Original post by tajtsracc)
The minimum value for 2x2 + 6x + 5 is 1/2, hence the minimum point for AB2 is equal to 5 times the minimum value of 2x2 + 6x + 5 (which is a 1/2), hence 5 x 1/2 = 5/2.
But they wanted the minimum value of AB, not (AB)2, so square root this value. √5/2
√5/2 = √5/√2 Now rationalise the denominator. √5 x √2 / √2 x √2 = √10/2 = 1/2√10.
The minimum value for 2x2 + 6x + 5 is 1/2, hence the minimum point for AB2 is equal to 5 times the minimum value of 2x2 + 6x + 5 (which is a 1/2), hence 5 x 1/2 = 5/2.
But they wanted the minimum value of AB, not (AB)2, so square root this value. √5/2
√5/2 = √5/√2 Now rationalise the denominator. √5 x √2 / √2 x √2 = √10/2 = 1/2√10.
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