So do you think that in this particular question they just want you to do the algebra rather than worry about ranges of validity?
Tey would want you to realise that it only works for x > y or x < y, et ; and in the off chance that condition is not given you should say that condition
But my t2 does yield t3 as cot(28pi/9) just the same, which does give the right answer. So I'm quite confused about Tom's post now. Perhaps you could clarify this:
tommm
STEP III 2003, Q4 We can also apply the formula similarly to the cotangent function, so
Thanks for the attempt but i am afraid u have misread the question quite a bit. I can manage eveything u have done but the question is asking for the necessary and sufficient condition for the cubic equation to have 3 distinct positive roots. and thats where i got stuck. (such as how u might compare the smallest root with a? ) I
Thanks for the attempt but i am afraid u have misread the question quite a bit. I can manage eveything u have done but the question is asking for the necessary and sufficient condition for the cubic equation to have 3 distinct positive roots. and thats where i got stuck. (such as how u might compare the smallest root with a? ) I
Your right. But I think only a slight alteration is required.
The curve y=(x−a)3−3b2(x−a)+c will have max point (a−b,c+2b3) and min (a+b,c−2b3) and y-intercept c.
To ensure the roots are all positive we want the y-intercept below the x-axis c<0 and the x-coordinate of the max point greater than zero a>b. This with my previous condition for 3 distinct roots should do.
More formally, to prove sufficiency consider sign changes and apply Intermediate value theorem (i.e sign change rule); to prove necessity use sign change / IVT to show that if one of the conditions is not meet then there'll be a negative root
1. y-intercept negative 2. turning points different sign of y 3. turning points both have positive x values
A sketch might help to explain? Worth stating that the graph goes from negative infinity to positive infinity and make it clear how you get the conditions I think.
I think you've used the wrong trig identity in the second part, it should be sin(a + 1/2)x - sin(a - 1/2)x + .... sin(b - 1 + 1/2)x, which gives the answer as x(a + b - 1) = pi(2n + 1) or x(b - a) = 2npi
I think you've used the wrong trig identity in the second part, it should be sin(a + 1/2)x - sin(a - 1/2)x + .... sin(b - 1 + 1/2)x, which gives the answer as x(a + b - 1) = pi(2n + 1) or x(b - a) = 2npi
I did using 'otherwise' and I got that answer as well.
I might be misinterpreting the question but the question asks us to 'show that T = (1 - t^2)/2t and 3t^2 =/ 1'. So this means we should give some justification as to why 3t^2 =/1 right? If so then what would the justification be?
If 3t^2 = 1 then it means that T = (1 + t^2)/4t which I can't see a problem with...
I might be misinterpreting the question but the question asks us to 'show that T = (1 - t^2)/2t and 3t^2 =/ 1'. So this means we should give some justification as to why 3t^2 =/1 right? If so then what would the justification be?
If 3t^2 = 1 then it means that T = (1 + t^2)/4t which I can't see a problem with...
I think it's because you need T=t to be a repeated root, since its a tangent. Not sure though.