Let $ heta$ be the complex number satisfying $ar{ heta}^3 = 1$ and $ ext{Im}( heta) > 0$ - HSC - SSCE Mathematics Extension 2 - Question 6 - 2008 - Paper 1
Question 6
Let $ heta$ be the complex number satisfying $ar{ heta}^3 = 1$ and $ ext{Im}( heta) > 0$. The cubic polynomial, $p(z) = z^3 + az^2 + bz + c$, has zeros $1, - heta$ ... show full transcript
Worked Solution & Example Answer:Let $ heta$ be the complex number satisfying $ar{ heta}^3 = 1$ and $ ext{Im}( heta) > 0$ - HSC - SSCE Mathematics Extension 2 - Question 6 - 2008 - Paper 1
Step 1
Find $p(z)$
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Answer
To find the cubic polynomial p(z) with given roots, we can use Vieta's formulas. The polynomial can be expressed as:
So, combining with $(z - 1)$ gives us:
$$p(z) = (z - 1)(z^2 - 2 ext{Re}( heta)z + 1)$$
When simplified, we will match the coefficients to find $a$, $b$, and $c$.
Step 2
Show that the line $ ext{l}$ has equation $bx ext{ sec } heta - ay an heta - ab = 0$
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To find the equation of the tangent line at point P:
Differentiate the hyperbola equation to find the slope at P.
Use the point-slope form of the line equation.
By substituting the coordinates of point P, we derive the equation to be:
bxextsecheta−ayanheta−ab=0.
Step 3
Show that $SR = rac{ab( ext{sec } heta - 1)}{rac{1}{ an^2 heta} + b^2 ext{sec}^2 heta}$
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To show the length of segment SR:
Apply the Pythagorean theorem.
Use trigonometric identities to express the distances in terms of a, b, and heta.
Simplify to arrive at:
SR = rac{ab( ext{sec } heta - 1)}{rac{1}{ an^2 heta} + b^2 ext{sec}^2 heta}.
Step 4
Show that $SR imes S'R' = b^2$
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To demonstrate this relation, we apply the properties of the hyperbola and the lengths calculated before:
Express both SR and S′R′ in terms of b and simplify.
Verify the product results in b2.
Step 5
Show that $rac{1}{r} - rac{r-1}{r} = rac{1}{n}$
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To prove this identity:
Start with the left-hand side: rac{1}{r} - rac{r-1}{r} = rac{-1}{r}
Rearranging gives rac{1}{n}.
Thus proving the equality.
Step 6
Show that the sum $rac{1}{r} + rac{1}{r} + rac{1}{m} - rac{1}{r} = rac{1}{n}$
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To demonstrate this:
Simplify the left-hand side by adding values.
After rearranging, you end up with the relationship:
rac{1}{n}.
Step 7
What is the limiting value of the sum $rac{1}{inom{m}{n}}$ as $m$ increases without bound?
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As m increases:
The term rac{1}{inom{m}{n}} approaches 0 when considering the growth of binomial coefficients for large m.
Therefore, the limiting value is 0.