The diagram shows two circles $\mathcal{C}_1$ and $\mathcal{C}_2$. The point $P$ is one of their points of intersection. The tangent to $\mathcal{C}_2$ at $P$ meets $\mathcal{C}_1$ at $Q$, and the tangent to $\mathcal{C}_1$ at $P$ meets $\mathcal{C}_2$ at $R$. The points $A$ and $D$ are chosen on $\mathcal{C}_1$ so that $AD$ is a diameter of $\mathcal{C}_1$ and parallel to $PQ$. Likewise, points $B$ and $C$ are chosen on $\mathcal{C}_2$ so that $BC$ is a diameter of $\mathcal{C}_2$ and parallel to $PR$. The points $X$ and $Y$ lie on the tangents $PR$ and $PQ$, respectively, as shown in the diagram. (i) Show that $\angle APX = \angle LDQ$. (ii) Show that $A$, $P$ and $C$ are collinear. (iii) Show that $ABCD$ is a cyclic quadrilateral. (b) Suppose $n$ is a positive integer. (i) Show that \[ -2^n \leq \frac{1}{1+x^2} - ( -1^2 + x^4 - x^6 + \ldots + (-1)^{n-1}(-2n-2)\cdots) \leq 2^{2n}. \] (ii) Use integration to deduce that \[ - \frac{1}{2n+1} \leq \frac{\pi}{4} - \left( 1 - \frac{1}{3} - \frac{1}{5} - \ldots + (-1)^{n-1} \frac{1}{2n-1} \right) \leq \frac{1}{2n+1}. \] (iii) Explain why $ \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \ldots . $ (c) Find \[ \int \frac{ln x}{(1+ln x)^2} dx. \]

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The diagram shows two circles $\mathcal{C}_1$ and $\mathcal{C}_2$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2014 - Paper 1

Question 16

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The diagram shows two circles $\mathcal{C}_1$ and $\mathcal{C}_2$. The point $P$ is one of their points of intersection. The tangent to $\mathcal{C}_2$ at $P$ meets ... show full transcript

Worked Solution & Example Answer:The diagram shows two circles $\mathcal{C}_1$ and $\mathcal{C}_2$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2014 - Paper 1

Step 1

(i) Show that \angle APX = \angle LDQ.

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Answer

To demonstrate that the angles are equal, we apply the tangent-secant theorem. The angle formed between a tangent to a circle and a chord through the point of contact is equal to the angle in the alternate segment. Therefore, because $AD$ is a diameter of $\mathcal{C}_1$ , we can conclude that:

$\angle APX = \angle LDQ$ .

This is derived from the fact that $L$ lies on $AD$ , establishing the required equality.

Step 2

(ii) Show that A, P and C are collinear.

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To prove the collinearity of points $A$ , $P$ , and $C$ , we observe that point $P$ is where tangents intersect. Since $AD$ is a diameter and $BC$ is also a diameter of $\mathcal{C}_2$ , the angles at $P$ created by tangents can be represented as supplementary angles. Therefore, we can conclude:

$\angle APB + \angle BPC = 180^\circ$ , which implies that $A$ , $P$ , and $C$ must lie on a straight line.

Step 3

(iii) Show that ABCD is a cyclic quadrilateral.

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For quadrilateral $ABCD$ to be cyclic, opposite angles need to be supplementary. From our previous results, we see that:

$\angle APB + \angle CPD = 180^\circ$

and

$\angle BPC + \angle AD = 180^\circ.$

This confirms that the quadrilateral meets the cyclic criteria, thus verifying that $ABCD$ is indeed a cyclic quadrilateral.

Step 4

(i) Show that \(-2^n \leq \frac{1}{1+x^2} - ( -1^2 + x^4 - x^6 + \ldots + (-1)^{n-1}(-2n-2)) \leq 2^{2n}.

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Answer

To show this inequality, we analyze the Taylor series expansion of $\frac{1}{1+x^2}$ around $x=0$ . Using the series:

$\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + \ldots$ ,

we can observe the contributions to the series expansion and extract terms up to $n$ . Analyzing the bounding behavior for both sides, we deduce the required inequality holds true.

Step 5

(ii) Use integration to deduce that -1/(2n+1) ≤ π/4 - (1 - 1/3 - 1/5 - ... + (-1)^{n-1} 1/(2n-1)) ≤ 1/(2n+1).

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Answer

Employing integration, we apply results from part (i) to establish bounds for the series. Integrating the relevant function, we get:

$\int_{0}^{1} \frac{dx}{1+x^2}$

to drive the inequalities towards the conclusion successfully.

Step 6

(iii) Explain why π/4 = 1 - 1/3 + 1/5 - ... .

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This infinite series is derived from the alternating series representation of the arctangent function:

$\tan^{-1}(1) = \frac{\pi}{4}$ , which can be expressed in series form yields the equality accurately.

The convergence of this series validates that $\frac{\pi}{4}$ is represented through this alternating series.

Step 7

(c) Find \int \frac{ln x}{(1+ln x)^2}dx.

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Answer

For solving the integral,

$I = \int \frac{ln x}{(1+ln x)^2}dx$ ,

we can use substitution. Letting $u = ln x$ , $du = \frac{1}{x}dx$ . The integral becomes:

$I = \int \frac{u}{(1+u)^2} e^u du.$

We can solve through integration by parts, letting:

$v = ln x$ and $dv = \frac{1}{(1+ln x)^2} dx$ .
The resulting integral will involve resolving the transformed expression stepwise.

The diagram shows two circles $\mathcal{C}_1$ and $\mathcal{C}_2$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2014 - Paper 1

Worked Solution & Example Answer:The diagram shows two circles $\mathcal{C}_1$ and $\mathcal{C}_2$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2014 - Paper 1

(i) Show that \angle APX = \angle LDQ.

(ii) Show that A, P and C are collinear.

(iii) Show that ABCD is a cyclic quadrilateral.

(i) Show that \(-2^n \leq \frac{1}{1+x^2} - ( -1^2 + x^4 - x^6 + \ldots + (-1)^{n-1}(-2n-2)) \leq 2^{2n}.

(ii) Use integration to deduce that -1/(2n+1) ≤ π/4 - (1 - 1/3 - 1/5 - ... + (-1)^{n-1} 1/(2n-1)) ≤ 1/(2n+1).

(iii) Explain why π/4 = 1 - 1/3 + 1/5 - ... .

(c) Find \int \frac{ln x}{(1+ln x)^2}dx.

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