Evaluate $$ \lim_{x \to 0} \frac{\sin(\frac{x}{5})}{2x} $$ Find $$ \frac{d}{dx} \cos^{-1}(3x^2) $$. The line $AT$ is the tangent to the circle at $A$, and $BT$ is a secant meeting the circle at $B$ and $C$. Given that $AT=12, BC=7$ and $CT=x$, find the value of $x$. Write $8\cos\alpha + 6\sin\alpha$ in the form $A\cos(\theta - \alpha)$, where $A > 0$ and $0 \leq \alpha \leq \frac{\pi}{2}$. Hence, or otherwise, solve the equation $8\cos\alpha + 6\sin\alpha = 5$ for $0 \leq \alpha \leq 2\pi$. A four-person team is to be chosen at random from nine women and seven men. (i) In how many ways can this team be chosen? (ii) What is the probability that the team will consist of four women?

SSCE HSC Mathematics Extension 1 Rational function inequalities: Evaluate $$ \lim

Evaluate $$ \lim_{x \to 0} \frac{\sin(\frac{x}{5})}{2x} $$ Find $$ \frac{d}{dx} \cos^{-1}(3x^2) $$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2004 - Paper 1

Question 2

$Evaluate--$$-\lim_{x-\to-0}-\frac{\sin(\frac{x}{5})}{2x}-$$--Find--$$-\frac{d}{dx}-\cos^{-1}(3x^2)-$$-HSC-SSCE Mathematics Extension 1-Question 2-2004-Paper 1.png$

Evaluate $$ \lim_{x \to 0} \frac{\sin(\frac{x}{5})}{2x} $$ Find $$ \frac{d}{dx} \cos^{-1}(3x^2) $$. The line $AT$ is the tangent to the circle at $A$, and $BT$ i... show full transcript

Worked Solution & Example Answer:Evaluate $$ \lim_{x \to 0} \frac{\sin(\frac{x}{5})}{2x} $$ Find $$ \frac{d}{dx} \cos^{-1}(3x^2) $$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2004 - Paper 1

Step 1

Evaluate $$ \lim_{x \to 0} \frac{\sin(\frac{x}{5})}{2x} $$

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Answer

To evaluate this limit, we can use L'Hôpital's Rule since it results in a $rac{0}{0}$ form. Differentiating the numerator and denominator gives:

$\lim_{x \to 0} \frac{\frac{1}{5} \cos(\frac{x}{5})}{2} = \frac{1}{2 \cdot 5} \cdot 1 = \frac{1}{10}$

Step 2

Find $$ \frac{d}{dx} \cos^{-1}(3x^2) $$

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Answer

Using the chain rule, we have:

$\frac{d}{dx} \cos^{-1}(u) = -\frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}$

Letting $u = 3x^2$ , we find:

$\frac{du}{dx} = 6x$

Then, applying the chain rule:

$\frac{d}{dx} \cos^{-1}(3x^2) = -\frac{6x}{\sqrt{1 - (3x^2)^2}} = -\frac{6x}{\sqrt{1 - 9x^4}}$

Step 3

Given that $AT=12, BC=7$ and $CT=x$, find the value of $x$.

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Answer

Using the relationship in a circle, we apply Pythagoras' theorem in the triangle:

$AT^2 + BT^2 = AB^2$

Here, let $BT = 7$ , thus:

144 + 49 = x^2 \\ 193 = x^2 \\ \therefore x = \sqrt{193} \approx 13.89 $$

Step 4

Write $8\cos\alpha + 6\sin\alpha$ in the form $A\cos(\theta - \alpha)$, where $A > 0$ and $0 \leq \alpha \leq \frac{\pi}{2}$.

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To rewrite in the required form, first calculate:

$A = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$

Next, find $\alpha$ such that:

\sin(\alpha) = \frac{6}{10} = 0.6 $$ Thus, we can write: $$ 8\cos\alpha + 6\sin\alpha = 10\cos(\theta - \alpha) $$

Step 5

Hence, or otherwise, solve the equation $8\cos\alpha + 6\sin\alpha = 5$ for $0 \leq \alpha \leq 2\pi$.

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We rewrite the equation using our previous result:

\cos(\theta - \alpha) = 0.5 $$ This gives solutions: $$ \theta - \alpha = \frac{\pi}{3} \quad \text{or} \quad \theta - \alpha = \frac{5\pi}{3} $$ Thus: $$ \alpha = \theta - \frac{\pi}{3} \quad \text{or} \quad \frac{5\pi}{3} $$

Step 6

In how many ways can this team be chosen?

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To choose a four-person team from nine women and seven men, we can use the combination formula $C(n, k)$ :

$\text{Total ways} = C(16, 4) = \frac{16!}{4!(16-4)!} = 1820$

Step 7

What is the probability that the team will consist of four women?

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Answer

The number of ways to choose 4 women from 9 is:

$C(9, 4) = \frac{9!}{4!(9-4)!} = 126$

The probability is:

$P = \frac{C(9, 4)}{C(16, 4)} = \frac{126}{1820} \approx 0.06923$

Evaluate $$ \lim_{x \to 0} \frac{\sin(\frac{x}{5})}{2x} $$ Find $$ \frac{d}{dx} \cos^{-1}(3x^2) $$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2004 - Paper 1

Worked Solution & Example Answer:Evaluate $$ \lim_{x \to 0} \frac{\sin(\frac{x}{5})}{2x} $$ Find $$ \frac{d}{dx} \cos^{-1}(3x^2) $$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2004 - Paper 1

Evaluate $$ \lim_{x \to 0} \frac{\sin(\frac{x}{5})}{2x} $$

Find $$ \frac{d}{dx} \cos^{-1}(3x^2) $$

Given that $AT=12, BC=7$ and $CT=x$, find the value of $x$.

Write $8\cos\alpha + 6\sin\alpha$ in the form $A\cos(\theta - \alpha)$, where $A > 0$ and $0 \leq \alpha \leq \frac{\pi}{2}$.

Hence, or otherwise, solve the equation $8\cos\alpha + 6\sin\alpha = 5$ for $0 \leq \alpha \leq 2\pi$.

In how many ways can this team be chosen?

What is the probability that the team will consist of four women?

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