Mathematics: Post your doubts here!

Muhammad Amer

Muhammad Amer

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Greetings,
Yes, I'm stuck with question 8 in Exercise 16B in Pure Mathematics 2 & 3.
I'd be very thankful if somebody helped me out with it.
 

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Muhammad Amer

Muhammad Amer

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DDanielAWE@ said:
Cant solve these , can anyone help in this paper by showing full working
http://freeexampapers.com/A-Level/Maths/CIE/2011-Nov/9709_w11_qp_11.pdf
Question 4 (ii)
question 5
Question 8
9(i) Thanks !!
Click to expand...

While I, myself, am waiting for help, I might as well help you DDanielAWE@

As for Q 4 (ii):

to get the expression for f(x), you will first need to integrate on f'(x), so f(x)=x^2-6x+c
Now, you know this is a parabola and its stationary point is a minimum because the coefficient of x^2 is positive, therefore the x-coordinate of the minimum point is 3
You also know that the range of the function in f(x)>=-4 , that is; the lowest value the function could have is -4, so y minimum is -4
You have a point (3,-4) substitute it in the expression f(x)=x^2-6x+c : (3)^2-6(3)+c=-4 to get c=5
Therefore the expression is f(x)=x^2-6x+5
 
Muhammad Amer

Muhammad Amer

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DDanielAWE@ said:
Cant solve these , can anyone help in this paper by showing full working
http://freeexampapers.com/A-Level/Maths/CIE/2011-Nov/9709_w11_qp_11.pdf
Question 4 (ii)
question 5
Question 8
9(i) Thanks !!
Click to expand...

Q 5:
To get the perimeter, you must find all the external lengths of the plate
For the arc, l=rθ the length of an arc is equal to the radius of the circle multiplied by the angle opposite to the arc in radians.
Let CB=X & OC=Y
Using basic trigonometric knowledge, sinθ=Y/r so Y=rsinθ
also, cosθ=X/r so X=rcosθ
Thus, the perimeter is equal to r+rθ+rsinθ+rcosθ by taking r as a common factor: perimeter=r(1+θ+sinθ+cosθ)
 

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DDanielAWE@

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Muhammad Amer said:
While I, myself, am waiting for help, I might as well help you DDanielAWE@

As for Q 4 (ii):

to get the expression for f(x), you will first need to integrate on f'(x), so f(x)=x^2-6x+c
Now, you know this is a parabola and its stationary point is a minimum because the coefficient of x^2 is positive, therefore the x-coordinate of the minimum point is 3
You also know that the range of the function in f(x)>=-4 , that is; the lowest value the function could have is -4, so y minimum is -4
You have a point (3,-4) substitute it in the expression f(x)=x^2-6x+c : (3)^2-6(3)+c=-4 to get c=5
Therefore the expression is f(x)=x^2-6x+5
Click to expand...


Thanks a lot !!!! Muhammad Amer

https://www.xtremepapers.com/community/members/muhammad-amer.125724/
 
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ashcull14

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nov 2003 maths p4.PNG
Can any one explain D part fully PLEASE Ive gotten the values in marking scheme BUT imunable to figure out how to bring the answer from certain values
 
The Godfather

The Godfather

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DDanielAWE@ said:
Cant solve these , can anyone help in this paper by showing full working
http://freeexampapers.com/A-Level/Maths/CIE/2011-Nov/9709_w11_qp_11.pdf
Question 4 (ii)
question 5
Question 8
9(i) Thanks !!
Click to expand...
9i)
x^2 + 3x + 2k = kx + 6
x^2 + 3x - kx + 2k - 6 = 0
x^2 + x - 2 =0
x = 1 and -2
Substitute this values in any ewn to get value of y
A(1,8) and B(-2,2)
distance = sqroot[(-2-1)^2 + (2-8)^2] = sqroot of 45
Midpoint :
x : (-2 + 1 )/2 = -0.5
y : (2 + 8)/2 = 5
(-0.5,5)
 
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