2. Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 2 - 2003 - Paper 1

Using the product rule, differentiate: $\frac{d}{dx}(x \tan^{-1}(x)) = \tan^{-1}(x) + x \cdot \frac{1}{1+x^2}.$
Thus the answer is: $\tan^{-1}(x) + \frac{x}{1+x^2}.$

To solve this integral, we can use the 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 $x = 1$, $\theta = \frac{\pi}{4}$.

This transforms the integral to: $\int_0^{\frac{\pi}{4}} \frac{\sqrt{2} \cos \theta}{\sqrt{2 - 2 \sin^2 \theta}} \, d\theta = \int_0^{\frac{\pi}{4}} \cos \theta \, d\theta = [\sin \theta]_0^{\frac{\pi}{4}} = \frac{\sqrt{2}}{2}.$
Hence, the value is $\frac{\sqrt{2}}{2}$.

Using the binomial theorem, the expansion is: $(a + b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^k$ Let $a = 2$, $b = x^2$, and $n = 5$, To find the coefficient of $x^4$, set $k = 2$: $\text{Coefficient} = {5 \choose 2} (2)^{5-2} = 10 \cdot 8 = 80.$

To express in this form:

1. Recognize that $R = \sqrt{a^2 + b^2}$ where $a = 1$ (coefficient of $\cos x$) and $b = -1$ (coefficient of $\sin x$): $R = \sqrt{1^2 + (-1)^2} = \sqrt{2}.$
2. Find $\tan \alpha = \frac{-b}{a} = \frac{-(-1)}{1} = 1 \Rightarrow \alpha = -\frac{\pi}{4}.$
   Thus, the expression becomes: $\sqrt{2}\cos\left(x - \frac{\pi}{4}\right).$

1. Identify amplitude and period:

   • Amplitude = $\sqrt{2}$.
   • Period = $2\pi$.

2. Determine critical points:

   • The function crosses zero when: $\cos x - \sin x = 0 \Rightarrow \tan x = 1 \Rightarrow x = \frac{\pi}{4}, \frac{5\pi}{4}.$

3. Sketch the graph from $0$ to $2\pi$, ensuring it oscillates between $-\sqrt{2}$ and $\sqrt{2}$, marking key points: