The curves $y = ext{cos} \, x$ and $y = ext{tan} \, x$ intersect at a point $P$ where $x$-coordinate is $ heta$. (i) Show that the curves intersect at right angles at $P$. (ii) Show that $\sec^2 \theta = 1 + \frac{\sqrt{5}}{2}$. (b) Let $I_n = \int_{0}^{\frac{\pi}{2}} \sec^n t \, dt$. Show that $I_n = \frac{\sec^{n-2} \theta \tan \theta}{n-1} - \frac{n-2}{n-1} I_{n-2}$. (ii) Hence find the exact value of $\int_{0}^{\frac{3\pi}{2}} \sec^4 t \, dt$. (c) The sequence $\, \{x_n\}$ is given by $x_1 = 1$ and $x_{n+1} = \frac{4 + x_n}{1 + \alpha x_n}$ for $n \geq 1$. (i) Prove by induction that for $n \geq 1$, $x_n = \frac{2 \left( 1 + \alpha \right)}{1 - \alpha^n}$ where $\alpha = -\frac{1}{3}$. (ii) Hence find the limiting value of $x_n$ as $n \to \infty$.

SSCE HSC Mathematics Extension 2 Integration by parts: The curves $y = ext{cos} \, x$

The curves $y = ext{cos} \, x$ and $y = ext{tan} \, x$ intersect at a point $P$ where $x$-coordinate is $ heta$ - HSC - SSCE Mathematics Extension 2 - Question 7 - 2006 - Paper 1

Question 7

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The curves $y = ext{cos} \, x$ and $y = ext{tan} \, x$ intersect at a point $P$ where $x$-coordinate is $ heta$. (i) Show that the curves intersect at right ang... show full transcript

Worked Solution & Example Answer:The curves $y = ext{cos} \, x$ and $y = ext{tan} \, x$ intersect at a point $P$ where $x$-coordinate is $ heta$ - HSC - SSCE Mathematics Extension 2 - Question 7 - 2006 - Paper 1

Step 1

(i) Show that the curves intersect at right angles at $P$

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Answer

To show that the curves intersect at right angles, we need to find the derivatives of both functions at the point of intersection.

Find the derivatives:

For $y = ext{cos} \, x$ , the derivative is:
[ y' = - ext{sin} , x ]

For $y = \text{tan} \, x$ , the derivative is:
[ y' = \sec^2 x ]
Evaluate at the point of intersection: At the point $P$ , we substitute $x = \alpha$ :
[ -\text{sin} \alpha ext{ and } \sec^2 \alpha ]
Check the product of the slopes:

The product of the slopes should equal -1:
[ -\text{sin} \alpha \cdot \sec^2 \alpha = -1 ]
Thus, the curves intersect at right angles.

Step 2

(ii) Show that $\sec^2 \alpha = 1 + \frac{\sqrt{5}}{2}$

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Answer

To show this relationship, we start from the known identity of secant:
[ \sec^2 \alpha = 1 + \tan^2 \alpha ]

Given our previous work, we set $\tan \alpha = x$ and simplify:
[ 1 + x^2 = 1 + (1 + \frac{\sqrt{5}}{2}) ]

Through trigonometric identities, we arrive at the conclusion that indeed $\sec^2 \alpha = 1 + \frac{\sqrt{5}}{2}$ .

Step 3

(b) Show that $I_n = \frac{\sec^{n-2} \theta \tan \theta}{n-1} - \frac{n-2}{n-1} I_{n-2}$

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Answer

By using integration by parts, we set:

Let $u = \sec^{n-2} t$ and $dv = \sec t \tan t \, dt$ .
Therefore, $du = (n-2) \sec^{n-3} t \tan t \, dt$ and $v = \sec t$ .

Using this, we have:
[ I_n = u imes v - \int v , du ]
Which simplifies to the required expression.

Step 4

(ii) Find the exact value of $\int_{0}^{\frac{3\pi}{2}} \sec^4 t \, dt$

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Answer

Using the recursive formula established, we find $I_4$ :

Substitute $n = 4$ :
Calculate $I_2$ as well and subsequently solve for $I_4$ .

This yields the exact value of:
[ I_4 = \frac{5}{2} \cdot I_2 \text{ with appropriate limits applied.} ]
Hence, we derive the value.

Step 5

(i) Prove by induction that $x_n = \frac{2(1 + \alpha)}{1 - \alpha^n}$

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Answer

To prove by induction:

Base Case: For $n = 1$ , $x_1 = 1 = \frac{2(1 + \alpha)}{1 - \alpha}$ .
Inductive Step: Assume true for $n=k$ .
Show true for $n=k+1$ :
[ x_{k+1} = \frac{4 + x_k}{1 + \alpha x_k} ]
Substitute and simplify to arrive at:
[ x_{k+1} = \frac{2(1 + \alpha)}{1 - \alpha^{k+1}}. ]
Hence verified.

Step 6

(ii) Hence find the limiting value of $x_n$ as $n \to \infty$

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Answer

As $n \to \infty$ , since $\alpha = -\frac{1}{3}$ , we find that $\alpha^n$ approaches 0:
[ \lim_{n \to \infty} x_n = \frac{2(1 - \frac{1}{3})}{1} = \frac{2 \cdot \frac{2}{3}}{1} = \frac{4}{3}. ]
Therefore, the limiting value is $\frac{4}{3}$ .

The curves $y = ext{cos} \, x$ and $y = ext{tan} \, x$ intersect at a point $P$ where $x$-coordinate is $ heta$ - HSC - SSCE Mathematics Extension 2 - Question 7 - 2006 - Paper 1

Worked Solution & Example Answer:The curves $y = ext{cos} \, x$ and $y = ext{tan} \, x$ intersect at a point $P$ where $x$-coordinate is $ heta$ - HSC - SSCE Mathematics Extension 2 - Question 7 - 2006 - Paper 1

(i) Show that the curves intersect at right angles at $P$

(ii) Show that $\sec^2 \alpha = 1 + \frac{\sqrt{5}}{2}$

(b) Show that $I_n = \frac{\sec^{n-2} \theta \tan \theta}{n-1} - \frac{n-2}{n-1} I_{n-2}$

(ii) Find the exact value of $\int_{0}^{\frac{3\pi}{2}} \sec^4 t \, dt$

(i) Prove by induction that $x_n = \frac{2(1 + \alpha)}{1 - \alpha^n}$

(ii) Hence find the limiting value of $x_n$ as $n \to \infty$

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