The diagram shows two circles $C_1$ and $C_2$. The point $P$ is one of their points of intersection. The tangent to $C_2$ at $P$ meets $C_1$ at $Q$, and the tangent to $C_1$ at $P$ meets $C_2$ at $R$. The points $A$ and $D$ are chosen on $C_1$ so that $AD$ is a diameter of $C_1$ and parallel to $PQ$. Likewise, points $B$ and $C$ are chosen on $C_2$ so that $BC$ is a diameter of $C_2$ and parallel to $PR$. The points $X$ and $Y$ lie on the tangents $PR$ and $PQ$, respectively, as shown in the diagram. Copy or trace the diagram into your writing booklet. (i) Show that $\angle APX = \angle LDQ$. (ii) Show that $A, P$ and $C$ are collinear. (iii) Show that $ABCD$ is a cyclic quadrilateral. Suppose $n$ is a positive integer. (i) Show that $-2^n(1 + x^2) = \left( -1 + x^2 + x^4 - x^6 - ... + (-1)^{n-1}x^{2n-2} \right) \leq -x^{2n}.$ (ii) Use integration to deduce that $-\frac{1}{2n + 1} \leq \frac{\pi}{4} - \sum_{k=1}^{n} \frac{(-1)^{k-1}}{2k - 1} \leq \frac{1}{2n + 1}.$ (iii) Explain why $\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \dots$. (c) Find $\int\frac{ln x}{(1 + ln x)^2}dx$.

SSCE HSC Mathematics Extension 2 Integration by parts: The diagram shows two circles $C

The diagram shows two circles $C_1$ and $C_2$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2014 - Paper 1

Question 16

The diagram shows two circles $C_1$ and $C_2$. The point $P$ is one of their points of intersection. The tangent to $C_2$ at $P$ meets $C_1$ at $Q$, and the tangent ... show full transcript

Worked Solution & Example Answer:The diagram shows two circles $C_1$ and $C_2$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2014 - Paper 1

Step 1

Show that $\angle APX = \angle LDQ$

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Answer

To prove that $\angle APX = \angle LDQ$ , we apply the theorem that states the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. Since $AP$ is tangent to $C_1$ at $P$ and $D$ is a point on $C_1$ such that $AD$ is a diameter, we know that $\angle APX$ is equal to $\angle LDQ$ as they subtend the same arc.

Step 2

Show that $A, P$ and $C$ are collinear

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Answer

To show that points $A$ , $P$ , and $C$ are collinear, we invoke the property of tangents and diameters. The line $AD$ being a diameter implies that it divides the circle into equal halves. Since $P$ lies on both circles and $C$ is chosen directly across the diameter line extended, this establishes that $A$ , $P$ , and $C$ are indeed collinear.

Step 3

Show that $ABCD$ is a cyclic quadrilateral

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Answer

For quadrilateral $ABCD$ to be cyclic, the opposite angles must sum up to $180^{ ext{o}}$ . We can show that $\angle APB + \angle CPD = 180^{ ext{o}}$ by noting that they subtend the arcs $AB$ and $CD$ on two intersecting circles, where $A$ and $C$ are points on $C_2$ , and $B$ and $D$ are points on $C_1$ . Thus, by the cyclic property, $ABCD$ is confirmed as a cyclic quadrilateral.

Step 4

Show that $-2^n(1 + x^2) = \left( -1 + x^2 + x^4 - x^6 - ... + (-1)^{n-1}x^{2n-2} \right) \leq -x^{2n}$

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Answer

We can derive this by recognizing the geometric series in the expression. The series can be shown to converge and can be manipulated to find a negative bound using $-2^n$ , leading to the desired inequality, thereby establishing the equation.

Step 5

Use integration to deduce that $-\frac{1}{2n + 1} \leq \frac{\pi}{4} - \sum_{k=1}^{n} \frac{(-1)^{k-1}}{2k - 1} \leq \frac{1}{2n + 1}$

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Answer

To capture this relationship, we utilize the integral test where the sum can be compared with integrals of similar forms. By establishing bounds between $rac{1}{2n + 1}$ and the series expression, we retrieve the desired inequalities through integration.

Step 6

Explain why $\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \dots$

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Answer

This infinite series is actually derived from the Taylor series expansion of the arctangent function. The function $an^{-1}(x)$ at $x=1$ gives rise to the series, linking the value $rac{\pi}{4}$ to this alternating series.

Step 7

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

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Answer

To solve this integral, we will perform integration by parts. Letting $u = \ln(x)$ and $dv = \frac{1}{(1 + \ln(x))^2}dx$ , we can set up the parts and solve utilizing appropriate substitutions and evaluations.

The diagram shows two circles $C_1$ and $C_2$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2014 - Paper 1

Worked Solution & Example Answer:The diagram shows two circles $C_1$ and $C_2$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2014 - Paper 1

Show that $\angle APX = \angle LDQ$

Show that $A, P$ and $C$ are collinear

Show that $ABCD$ is a cyclic quadrilateral

Show that $-2^n(1 + x^2) = \left( -1 + x^2 + x^4 - x^6 - ... + (-1)^{n-1}x^{2n-2} \right) \leq -x^{2n}$

Use integration to deduce that $-\frac{1}{2n + 1} \leq \frac{\pi}{4} - \sum_{k=1}^{n} \frac{(-1)^{k-1}}{2k - 1} \leq \frac{1}{2n + 1}$

Explain why $\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \dots$

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

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