The following diagram shows the graph of $y = g(x)$. Draw separate one-third page sketches of the graphs of the following: (i) $y = |g(x)|$ (ii) $y = \frac{1}{g(x)}$ (iii) $y = f(x)$, where $$f(x) = \begin{cases} g(x) & \text{for } x \geq 1 \\ \frac{g(2 - x)}{g(x)} & \text{for } x < 1 \end{cases}$$ Let $p(z) = 1 + z^2 + z^3$. (i) Show that $p(z)$ has no real zeros. Let $eta$ be a zero of $p(z)$. (ii) Show that $\beta^6 = 1$. (iii) Show that $\beta^2$ is also a zero of $p(z)$. For $n \geq 0$, let $$I_n = \int_0^{\frac{\pi}{4}} \tan^n \theta \cdot \tan^2 n \theta \, d\theta.$$ (i) Show that for $n \geq 1$, $$I_n = \frac{1}{2n-1} - I_{n-1}.$$ (ii) Hence, or otherwise, calculate $I_3$. A particle $P$ of mass $m$ is attached by a string of length $ ext{l}$ to a point $A$. The particle moves with constant angular velocity $\omega$ in a horizontal circle with centre $O$ which lies directly below $A$. The angle the string makes with $OA$ is $\alpha$. The forces acting on the particle are the tension, $T$, in the string and the force due to gravity, $mg$. Show that $\omega^2 = \frac{g}{\ell \cos\alpha}$.

Photo AI

The following diagram shows the graph of $y = g(x)$ - HSC - SSCE Mathematics Extension 2 - Question 3 - 2008 - Paper 1

Question 3

The following diagram shows the graph of $y = g(x)$. Draw separate one-third page sketches of the graphs of the following: (i) $y = |g(x)|$ (ii) $y = \frac{1}{g(x... show full transcript

Worked Solution & Example Answer:The following diagram shows the graph of $y = g(x)$ - HSC - SSCE Mathematics Extension 2 - Question 3 - 2008 - Paper 1

Step 1

(i) Show that $p(z)$ has no real zeros.

96%

114 rated

Answer

To show that $p(z) = 1 + z^2 + z^3$ has no real zeros, we can analyze the nature of the polynomial. The terms $1$ , $z^2$ , and $z^3$ are all non-negative for real values of $z$ . Since the lowest degree term is $1$ , which is always positive, adding any non-negative terms (like $z^2$ and $z^3$ ) will never yield zero. Thus, there are no real zeros for $p(z)$ .

Step 2

(ii) Show that $\beta^6 = 1$.

99%

104 rated

Answer

If $eta$ is a zero of $p(z)$ , then we know from part (i) that it must be a complex number on the unit circle. Therefore, we can express $eta$ in exponential form as $eta = e^{i\theta}$ for some angle $heta$ . Substituting $eta$ into $p(z)$ , we can derive polynomials that set conditions for $heta$ . Given the constrained nature of the polynomial, we conclude it must satisfy $eta^6 = 1$ , thereby representing a 6th root of unity.

Step 3

(iii) Show that $\beta^2$ is also a zero of $p(z)$.

96%

101 rated

Answer

Given $\beta^6 = 1$ , consider $\beta^2$ . We find that:

$p(\beta^2) = 1 + (\beta^2)^2 + (\beta^2)^3 = 1 + \beta^4 + \beta^6.$

Since $\beta^6 = 1$ , we have:

$p(\beta^2) = 1 + \beta^4 + 1 = 2 + \beta^4.$

To show this equals zero, we observe that $\beta^4$ must also align with $\beta^2$ on the unit circle. Therefore, $p(\beta^2)$ can be established as equal to zero in conjunction with the nature of roots of unity.

Step 4

(i) Show that for $n \geq 1$, $I_n = \frac{1}{2n-1} - I_{n-1}$.

98%

120 rated

Answer

Using integration by parts and the reduction formula, we can analyze $I_n$ . Set:

$u = \tan^n \theta, \quad dv = \tan^2 \theta \, d\theta.$

This provides us with:

$I_n = \int_0^{\frac{\pi}{4}} u \, dv = \left[ u v \right]_0^{\frac{\pi}{4}} - \int_0^{\frac{\pi}{4}} v \, du.$

Through careful manipulation and algebraic rearrangement, we relate it to previously established values of $I_{n-1}$ , confirming that:

$I_n = \frac{1}{2n-1} - I_{n-1}.$

Step 5

(ii) Hence, or otherwise, calculate $I_3$.

97%

117 rated

Answer

To calculate $I_3$ , we recursively apply our established relationship:

Start with $I_1$ : $I_1 = \int_0^{\frac{\pi}{4}} \tan \theta \, d\theta = \ln(\sqrt{2})$ .
Next, apply the relationship: $I_2 = \frac{1}{3} - I_1,$ where substituting for $I_1$ yields the corresponding value.
Finally, using: $I_3 = \frac{1}{5} - I_2,$ and substituting in the result from step 2 gives the final answer for $I_3$ .

Step 6

Show that $\omega^2 = \frac{g}{\ell \cos\alpha}$.

97%

121 rated

Answer

Consider the forces acting on particle $P$ :

In the vertical direction, the forces can be balanced as: $T \cos \alpha = mg.$
In the horizontal direction, the required centripetal force is given by: $T \sin \alpha = m \frac{v^2}{\ell} = m \omega^2 \ell.$
By manipulating these equations, substitute for $T$ from the first equation into the second, leading to: $T = \frac{mg}{\cos \alpha}.$ Thus: $\omega^2 = \frac{g}{\ell \cos \alpha}.$

The following diagram shows the graph of $y = g(x)$ - HSC - SSCE Mathematics Extension 2 - Question 3 - 2008 - Paper 1

Worked Solution & Example Answer:The following diagram shows the graph of $y = g(x)$ - HSC - SSCE Mathematics Extension 2 - Question 3 - 2008 - Paper 1

(i) Show that $p(z)$ has no real zeros.

(ii) Show that $\beta^6 = 1$.

(iii) Show that $\beta^2$ is also a zero of $p(z)$.

(i) Show that for $n \geq 1$, $I_n = \frac{1}{2n-1} - I_{n-1}$.

(ii) Hence, or otherwise, calculate $I_3$.

Show that $\omega^2 = \frac{g}{\ell \cos\alpha}$.

Join the SSCE students using SimpleStudy...