2. (a) Sketch the graph of $y=3 ext{cos}^{-1}(2x)$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2003 - Paper 1

To differentiate the function $y = x \tan^{-1} x$, we will use the product rule:

$\frac{d}{dx}(uv) = u'v + uv'$

Letting $u = x$ and $v = \tan^{-1} x$:

• $u' = 1$.
• For $v'$, using the derivative of $an^{-1} x$, we have $v' = \frac{1}{1+x^2}$.

Applying the product rule: $\frac{d}{dx}(x \tan^{-1} x) = 1 \cdot \tan^{-1} x + x \cdot \frac{1}{1+x^2}$

Thus, the derivative is: $\tan^{-1} x + \frac{x}{1+x^2}$

To evaluate the integral $I = \int_0^1 \frac{1}{\sqrt{2 - x^2}} \, dx$, we can use a substitution method. Let $x = \sqrt{2} \sin \theta$, then $dx = \sqrt{2} \cos \theta \, d\theta$.

Changing the limits:

• When $x = 0$, $\theta = 0$.
• When $x = 1$, $\theta = \frac{\pi}{4}$.

Thus, the integral becomes: $I = \int_0^{\frac{\pi}{4}} \frac{\sqrt{2} \cos \theta}{\sqrt{2 - 2\sin^2 \theta}} \, d\theta = \int_0^{\frac{\pi}{4}} \frac{\sqrt{2} \cos \theta}{\sqrt{2(1 - \sin^2 \theta)}} \, d\theta$ This simplifies to: $I = \int_0^{\frac{\pi}{4}} \frac{\sqrt{2} \cos \theta}{\sqrt{2 \cos^2 \theta}} \, d\theta = \int_0^{\frac{\pi}{4}} 1 \, d\theta = \left[ \theta \right]_0^{\frac{\pi}{4}} = \frac{\pi}{4} - 0 = \frac{\pi}{4}$

To find the coefficient of $x^4$ in the expansion of $(2 + x^2)^5$, we can use the binomial expansion formula:

$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

In this case, $a = 2$, $b = x^2$, and $n = 5$. We are interested in the term where the power of $x$ is $4$, which occurs when $k = 2$:

$\binom{5}{2} (2)^{5-2} (x^2)^2 = \binom{5}{2} (2)^{3} x^4$

Calculating:

• $\binom{5}{2} = 10$ and $(2)^3 = 8$.

Thus, the coefficient of $x^4$ is: $10 \times 8 = 80$

To express $\cos x - \sin x$ in the form $R \cos(x + \alpha)$, we start by finding $R$ and $\alpha$ using the identities:

1. $R = \sqrt{a^2 + b^2}$ where $a = 1$ and $b = -1$. Thus, $R = \sqrt{1^2 + (-1)^2} = \sqrt{2}$

2. To find $\alpha$, we use $\tan \alpha = \frac{-b}{a} = \frac{-(-1)}{1} = 1$, thus: $\alpha = \frac{3\pi}{4}$

So, we can write: $\cos x - \sin x = \sqrt{2} \cos\left(x + \frac{3\pi}{4}\right)$

The function $y = \cos x - \sin x$ represents a periodic wave. Given that we can express this in terms of $R$ and $\alpha$, we know:

• The amplitude is $\sqrt{2}$, and the phase shift is rac{3\pi}{4}.

To sketch the graph:

1. Start from the intersection with the x-axis, find critical points where the derivative (which can be derived) is equal to zero.
2. The maximum value occurs at $\frac{3\pi}{4}$, which will be the peak of the wave.
3. Mark the behavior in the range $0 \leq x \leq 2\pi$ with characteristics of oscillation:

• The function will oscillate between $-\sqrt{2}$ and $\\sqrt{2}$.

4. Be sure to label important points including intercepts at $0$, rac{\pi}{2}, and $\pi$. The plot will reflect these oscillations with appropriate peaks and troughs.