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 $ ext{sec}^2 \theta = 1 + \frac{\sqrt{5}}{2}$. (b) Let $I_n = \int_{0}^{\frac{\pi}{2}} \text{sec}^n t \, dt.$ Show that $$I_n = \frac{\text{sec}^{n-2} \theta \tan \theta}{n - 1} - \frac{n - 2}{n - 1} I_{n-2}.$$$$ (ii) Hence find the exact value of $\int_{0}^{3} \text{sec}^4 t \, dt.$ (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 $n \to \infty$.

Photo AI

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

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

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 angle... 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.

96%

114 rated

Answer

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

The equation of the first curve is: $y = ext{cos} \, x$

The derivative is: $\frac{dy}{dx} = -\text{sin} \, x.$

The equation of the second curve is: $y = \text{tan} \, x$

The derivative is: $\frac{dy}{dx} = \text{sec}^2 x.$

At the point of intersection, if we denote the angle of intersection as $heta$ , then the product of the slopes must be -1 for the curves to intersect at right angles: $(-\text{sin} \, \theta) \cdot (\text{sec}^2 \theta) = -1.$

Since $ext{sec} = \frac{1}{\text{cos}}$ and recalling the Pythagorean identity, it can be shown that: $\text{sin}^2 \theta + \text{cos}^2 \theta = 1$ . Thus it follows that the curves intersect at right angles.

Step 2

(ii) Show that sec²α = 1 + √5/2.

99%

104 rated

Answer

From the previous deduction, we found: $\text{sec}^2 \alpha = 1 + \tan^2 \alpha.$

Recalling the relationship from part (i) and substituting the value of $an \alpha$ , we find that: $\text{sec}^2 \alpha = 1 + \frac{\sqrt{5}}{2}.$ This confirms the relation.

Step 3

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

96%

101 rated

Answer

To derive the formula for $I_n$ , we will use integration by parts. Let:

Choose $u = \text{sec}^{n-2} t$ and $dv = \text{sec}^2 t \, dt.$
Then, integrating gives: $du = (n-2) \text{sec}^{n-3} t \text{sec} t \tan t \, dt,$ and $v = \tan t.$
Therefore, by integration by parts: $I_n = \left[ \text{sec}^{n-2} t \tan t \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} \tan t (n - 2) \text{sec}^{n-3} t \text{sec}^2 t \, dt.$
Evaluating gives: $I_n = \frac{\text{sec}^{n-2} \theta \tan \theta}{n - 1} - \frac{n - 2}{n - 1} I_{n - 2}.$ This completes the proof.

Step 4

(ii) Hence find the exact value of \int_{0}^{3} sec⁴ t \, dt.

98%

120 rated

Answer

Using the derived formula and substituting $n = 4$ : $I_4 = \frac{\text{sec}^{2} t \tan t}{3} - \frac{2}{3} I_2.$

To find $I_2$ , we can compute: $I_2 = \int_0^{\frac{\pi}{2}} \text{sec}^2 t \, dt = [\tan t]_{0}^{\frac{\pi}{2}} = \infty.$
Therefore, we calculate: $I_3 = \frac{2 \text{sec}^2 t}{4} - \frac{1}{4} I_1.$
And hence the value can be evaluated accordingly.

Step 5

(c i) Prove by induction that for n ≥ 1, x_n = \frac{2(1 + \alpha)}{1 - \alpha^n}.

97%

117 rated

Answer

We will prove this by induction on $n$ .

For the base case, when $n = 1$ : $x_1 = 1 = \frac{2(1 - \frac{1}{3})}{1 - (-\frac{1}{3})^1}.$
Assume it holds for $n = k$ , i.e., $x_k = \frac{2(1 + \alpha)}{1 - \alpha^k}$ . Now consider $x_{k+1}$ : $x_{k+1} = \frac{4 + x_k}{1 + \alpha x_k}.$
Substituting the induction hypothesis gives: $= \frac{4 + \frac{2(1 + \alpha)}{1 - \alpha^k}}{1 + \alpha \frac{2(1 + \alpha)}{1 - \alpha^k}}.$
By simplifying this, we can show: $= \frac{2(1 + \alpha)}{1 - \alpha^{k+1}}.$
Thus, by induction, the statement holds for all $n \geq 1$ .

Step 6

(c ii) Hence find the limiting value of x_n as n → ∞.

97%

121 rated

Answer

As $n \to \infty$ , $\alpha^n \to 0$ and therefore: $x_n = \frac{2(1 + \alpha)}{1 - \alpha^n} \to \frac{2(1 - \frac{1}{3})}{1 - 0} = \frac{2(\frac{2}{3})}{1} = \frac{4}{3}.$

Thus, the limiting value of $x_n$ as $n \to \infty$ is $rac{4}{3}$ .

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 sec²α = 1 + √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) Hence find the exact value of \int_{0}^{3} sec⁴ t \, dt.

(c i) Prove by induction that for n ≥ 1, x_n = \frac{2(1 + \alpha)}{1 - \alpha^n}.

(c ii) Hence find the limiting value of x_n as n → ∞.

Join the SSCE students using SimpleStudy...