Solutions:Maths HL 2009 Paper 1

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Question 1

(a.)

(b.)

(i.)
(ii.)

(c.) Given that x-c+1 is a factor of x^{2}-5x+5cx-6b^{2} express c in terms of b

If x-c+1 is a factor then x\ =\ c-1 is a solution to x^{2}-5x+5cx-6b^{2}\ =\ 0

(c-1)^{2}-5(c-1)+5c(c-1)-6b^{2}\ =\ 0

c^2-2c+1-5c+5+5c^2-5c-6b^{2}\ =\ 0

6c^2-12c+6-6b^{2}\ =\ 0

c^2-2c+1-b^{2}\ =\ 0

(c-1)^2\ =\ b^2

c\ =\ (\pm b)+1

Question 2

(a.)

(b.)

(i.)
(ii.)

(c.)

(i.) One of the roots of px^2+qx+r=0 is n times the other root. Express r in terms of p, q and n
Let \alpha and n\alpha be the roots.
\rightarrow \alpha + n\alpha = -\frac{q}{p}
(n+1)\alpha = -\frac{q}{p}
\alpha = -\frac{q}{p(n+1)}


\rightarrow (\alpha).(n\alpha) = \frac{r}{p}
n\alpha^2 = \frac{r}{p}
n(-\frac{q}{p(n+1)})^2 = \frac{r}{p}
r = \frac{q^2n}{p(n+1)^2}
(ii.) One of the roots of x^2+qx+r=0 is 5 times the other. If q and r are positive integers, determine the set of possible values of q
r = \frac{q^2(5)}{(1)(5+1)^2} = \frac{5q^2}{36}
Therefore q must be of the form 6n with n \in N for r to be an integer.
Therefore the possible values of q are 6, 12, 18, 24...

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Lauraholden, Zorba

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