13 (15 marks) Use a SEPARATE writing booklet. (a) Using the substitution $t = \tan \frac{x}{2}$, or otherwise, evaluate $$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{1}{3 \sin x - 4 \cos x + 5} \, dx.$$ (b) The base of a solid is the region bounded by $y = x^2$, $y = -x^2$ and $x = 2$. Each cross-section perpendicular to the $x$-axis is a trapezium, as shown in the diagram. The trapezium has three equal sides and its base is twice the length of any one of the equal sides. Find the volume of the solid. Question 13 (continued) (c) The point $S(ae, 0)$ is the focus of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ on the positive $x$-axis. The points $P(at, br)$ and $Q(\frac{a}{t}, -\frac{b}{t})$ lie on the asymptotes of the hyperbola, where $t > 0$. The point $M(\frac{a(t^2 + 1)}{2t}, \frac{b(t^2 - 1)}{2t})$ is the midpoint of $PQ$. (i) Show that $M$ lies on the hyperbola. (ii) Prove that the line through $P$ and $Q$ is a tangent to the hyperbola at $M$. (iii) Show that $OP \times OQ = \sqrt{a^2t^2 + b^2t^2} \times \sqrt{\frac{a^2}{t^2} + \frac{b^2}{t^2}}$. (iv) If $P$ and $S$ have the same $x$-coordinate, show that $MS$ is parallel to one of the asymptotes of the hyperbola.

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13 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 13 - 2014 - Paper 1

Question 13

13 (15 marks) Use a SEPARATE writing booklet. (a) Using the substitution $t = \tan \frac{x}{2}$, or otherwise, evaluate $$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \fra... show full transcript

Worked Solution & Example Answer:13 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 13 - 2014 - Paper 1

Step 1

(a) Using the substitution $t = \tan \frac{x}{2}$, or otherwise, evaluate

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Answer

To solve the integral $I = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{1}{3 \sin x - 4 \cos x + 5} \, dx,$ we can use the substitution method. With the substitution $t = \tan \frac{x}{2}$ , we have:

Using the identities for sine and cosine:
- $\sin x = \frac{2t}{1+t^2}$
- $\cos x = \frac{1-t^2}{1+t^2}$

Substituting into the integral transforms it into a rational function of $t$ that can be integrated using partial fractions. After evaluating the integral, we will substitute back to return to the variable $x$ .

Step 2

(b) Find the volume of the solid.

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Answer

To find the volume of the solid with a trapezium cross-section, we follow these steps:

Calculate the height of the trapezium based on the functions $y = x^2$ and $y = -x^2$ : The height can be expressed as $h = 2x^2$ .
The area $A$ of each trapezium with the given base lengths can be modeled by: $A = \frac{1}{2} (2h)(\text{Base length}) = \frac{1}{2} (2 \cdot 2x^2)(2h)$
We can find the volume by integrating this area from $x=0$ to $x=2$ : $V = \int_0^2 A(x) \, dx.$
Finally, evaluating this integral provides the total volume.

Step 3

(c)(i) Show that $M$ lies on the hyperbola.

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To show that the point $M(\frac{a(t^2 + 1)}{2t}, \frac{b(t^2 - 1)}{2t})$ lies on the hyperbola defined by $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ , substitute the coordinates of $M$ into the hyperbola equation:

$\frac{\left(\frac{a(t^2 + 1)}{2t}\right)^2}{a^2} - \frac{\left(\frac{b(t^2 - 1)}{2t}\right)^2}{b^2} = 1$ .

Upon simplifying, we should find that the equation holds true, thus proving that $M$ does belong to the hyperbola.

Step 4

(c)(ii) Prove that the line through $P$ and $Q$ is a tangent to the hyperbola at $M$.

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To prove that the line through $P(at, br)$ and $Q(\frac{a}{t}, -\frac{b}{t})$ is tangent to the hyperbola at point $M$ , we first compute the slope of the line segment $PQ$ using the formula:

$\text{slope} = \frac{y_Q - y_P}{x_Q - x_P}.$

Next, we utilize the derivative of the hyperbola at point $M$ to verify that this slope matches the slope of the tangent to the hyperbola at point $M$ . If both slopes are equal, then the line is indeed a tangent to the hyperbola at that point.

Step 5

(c)(iii) Show that $OP \times OQ = \sqrt{a^2t^2 + b^2t^2} \times \sqrt{\frac{a^2}{t^2} + \frac{b^2}{t^2}}$

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Answer

To establish the relation, we use the coordinates of points $O$ , $P$ , and $Q$ .

Calculate the lengths:

Length $OP$ : $OP = \sqrt{(x_O - x_P)^2 + (y_O - y_P)^2} = \sqrt{a^2t^2 + b^2t^2}.$
Length $OQ$ : $OQ = \sqrt{\left(x_O - \frac{a}{t}\right)^2 + \left(y_O + \frac{b}{t}\right)^2}.$

The relationship then follows from substituting these values into the product $OP \times OQ$ .

Step 6

(c)(iv) If $P$ and $S$ have the same $x$-coordinate, show that $MS$ is parallel to one of the asymptotes of the hyperbola.

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Answer

To show that $MS$ is parallel to one of the asymptotes, we first calculate the slope of the segment $MS$ using its coordinates. The coordinates of $M$ and $S$ are:

$M\left(\frac{a(t^2 + 1)}{2t}, \frac{b(t^2 - 1)}{2t}\right)$
$S(ae, 0)$

Given the common $x$ -coordinate (which equates), we find the slope of segment $MS$ . If this slope corresponds to the slope of the asymptotes of the hyperbola, we confirm that they are parallel.

13 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 13 - 2014 - Paper 1

Worked Solution & Example Answer:13 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 13 - 2014 - Paper 1

(a) Using the substitution $t = \tan \frac{x}{2}$, or otherwise, evaluate

(b) Find the volume of the solid.

(c)(i) Show that $M$ lies on the hyperbola.

(c)(ii) Prove that the line through $P$ and $Q$ is a tangent to the hyperbola at $M$.

(c)(iii) Show that $OP \times OQ = \sqrt{a^2t^2 + b^2t^2} \times \sqrt{\frac{a^2}{t^2} + \frac{b^2}{t^2}}$

(c)(iv) If $P$ and $S$ have the same $x$-coordinate, show that $MS$ is parallel to one of the asymptotes of the hyperbola.

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