Does anyone know how to answer Q15 from 8H.
if a and b are the zeros of the function y = 3x^2 - 5x - 4, find the value of a^3 + b^3.
Thanks.
Thanks for that help.
Also having troubles with 3c) from 8D.
Solve the following simultaneous equations:
x^2 + y^2 -2x + 6y - 35 = 0 and 2x + 3y = 5
How do you get rid of the horrible fractions??
Also 11c) from 8H
If a and b are the roots of the equation 3x^2 + 2x + 7 = 0 , find a + b and ab. Hence form the equation with integer coefficients having roots:
a +2b and b + 2a.
Thank for all your help.
Thanks for that help.
Do you know how to solve Q24 and Q25 from 8H
24)the line x + y -1 = 0 intersects the circle x^2 + y^2 = 13 at A (a1, a2) and B (b1, b2). Without finding the corordinates of A and B, find the length of the chord AB. ( HINT: form a quadratic equation in x and evaluate |a1 - b1| and similarly find |a2 - b2| )
25) for what values of m are the roots of x^2 + 2x + 3 = m(2m +1) real and positive?
Thanks.
Thanks for the help.
Do you know how to to 8E Q 26.
OAB is a triangle in which OA is perpendicular to OB. OA and OB have lengths of 60cm and 80cm respectively. A rectangle inscribed inside the triangle so that one of its sides lies along the base OA of the triangle. By using similar triangles find the size of the rectangle of maximum area that may be inscribed in the triangle.
Last edited by appleibeats; 9 Jun 2015 at 8:26 PM.
Desperately need help with Question 26 from 8H.
if the equations mx^2 + 2x + 1 = 0 and x^2 + 2x + m = 0 have a common root, find the possible values of m and the value of the common root in each case.
The answer in the book is:
When m = 1, x = -1, and when m = -3 , x = 1
Thank you for helping me answer this question!
Could somebody please post a picture of the cover of the 2015 year 11 extension 1 cambridge textbook ??
6N 18. [A rather difficult proof] CAMBRIDGE
Need help with question 12 from 10B
the function y = ax^3 + bx^2 + cx + d has a relative maximum at the point (-2,27) and a relative minimum at the point (1,0). Find the values of a,b,c,d.
Thank you for your help.
We have several simultaneous equations here:
From the fact that the points (-2,27) and (1,0) pass through the graph of y=ax^3+bx^2+cx+d
27 = a(-2)^3 + b(-2)^2 + c(-2) + d ---------eqn 1
0 = a(1)^3 + b(1)^2 + c(1) + d ----------eqn 2
dy/dx = 3ax^2 + 2bx + c
Since (-2,27) and (1,0) are relative minima and maxima respectively, we know that when x=-2 and x=1 we have stationary points.
At a stationary point, dy/dx = 0
So we also have:
0 = 3a(-2)^2 + 2b(-2) + c ----------eqn 3
0 = 3a(1)^2 + 2b(1) + c -----------eqn 4
My advice is to do eqn 1 - eqn 2 first. Then you have another equation in terms of a, b, and c. Call that eqn 5.
Then solve eqn 3, 4 and 5 to get a, b and c. Which you can sub in to 2 to get d.
Thanks for that help.
Do you also know how to solve Q 13 part b from 10B.
b)
Show that y = x^a(1-x)^b has a turning point whose x coordinate divides the interval between the points (0,0) and (1,0) in the ration a:b
It is obviously related to part a) which I have done but am unsure how to go about this question.
Thank you for your help.
Need help with 10C.
Question 8:
y = |x| +3
a) find dy/dx when x < 0 and when x > 0
I know what the curve look like, and that when < 0 it has a negative gradient and when x > 0 it is positive gradient.
But how do you show that working out with the absolute value. Isn't any number ( positive or negative) inside the absolute value become positive??
Also Q10)
Part a) can do : differentiate f(x) = ( x - 2) ^ 1/5
Part b) show that there are no stationary points, but that a critical value occurs at x = 2 .
I thought stationary points is when the derivate is 0. But that is wrong as that would mean 2 is a stationary point but it isn't. It is instead a critical point. Please help clarify the meanings.
Thanks for helping and explaining.
Cheers.
Graphically, the function's slope tends to plus or minus infinity as you approach x = 2: http://www.wolframalpha.com/input/?i...%5E%281%2F5%29
Last edited by InteGrand; 17 Nov 2015 at 10:52 AM. Reason: LaTeX
Thanks for your help. Really appreciate it. It now makes more sense.
Also do you know how to do question 10b from 10D.
If y = (2x - 1)^ 4, prove that d/dx(y dy/dx) = y d^2y/dx^2 + (dy/dx)^2.
What is most troubling me is how to find d/dx of y dy/dx which equals i believe 8y(2x -1)^6 ... what do you do with the y at the front??
Thanks for your help in advance.
Cheers
could you explain how you used the product rule? I don't quite seem to get that. Thanks.
Disregard my last question, I figured it out. But could you explain what you mean by 'both functions of x' in your working out. Thank you.
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