The line 2x-y = k is a tangent to the curve x^2 - xy + y^2 = 1. Find the values of k. Hence find the range of values of k such that 2 the line meets the curve.
i think the phrasing of the 2nd is kinda odd really. can someone explain to me what the question wants in order to get the answer below?
2</= k </= 2 (answer)
I think the question is asking
Hence find the range of values of k such that the line meets the curve.
Sub y = 2x - k (line) into curve
so you get x^2 - x (2x - k) + (2x - k)^2 = 1
x^2 -2x^2 +kx + 4x^2 - 4kx + k^2 -1 = 0
3x^2 -3kx + (k^2 - 1) = 0
To meet the curve, you must have solutions right?
Hence, discriminant >= 0
(3k)^2 - 4(3)(k^2 - 1) >= 0
9k^2 - 12k^2 + 12 >= 0
k^2 - 4 <= 0 ==> divide by -3 throughout, remember to change sign
(k+2)(k-2) <= 0
Sketch the curve
2<= k <= 2
Thanks eagle.
I think there's another erroneous question, can you please check if there's missing info?
The polynomial x^3 + ax^2 + bx - 3 leaves a reaminder of 27 when divided by x-2 and a remainder when divided by x +1. Calculate the remainder when the polynomial is divided by x-1. Answer being 3.
For the underlined part they didnt tell me what the other remainder is. So it is impossible to form an equation to find a and b and subsequently obtain the final answer. Am I correct? If not, then maybe someone can help me out.
you are correct
So let's change the question instead since you have the answer
The polynomial x^3 + ax^2 + bx - 3 leaves a reaminder of 27 when divided by x-2 and a remainder of 3 when divided by x -1. Calculate the remainder when the polynomial is divided by x+1.
There u go :)