Evaluate $$\lim_{x \to 0} \frac{\sin 3x}{x}$$ - HSC - SSCE Mathematics Extension 1 - Question 1 - 2002 - Paper 1

To differentiate the function $f(x) = 3x^2 \ln x$, we use the product rule, which states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.

Let $u = 3x^2$ and $v = \ln x$:

1. Differentiate $u$: $u' = 6x$
2. Differentiate $v$: $v' = \frac{1}{x}$

Applying the product rule:

$f'(x) = 6x \ln x + 3x^2 \cdot \frac{1}{x} = 6x \ln x + 3x$.

To evaluate the integral $\int \sec 2x \tan 2x \, dx$, we recognize that the integral of $\sec(x)$ is $\ln |\sec(x) + \tan(x)| + C$. Thus:

Using the substitution $u = 2x$, then $du = 2dx$ or $dx = \frac{du}{2}$. Change the limits accordingly:

• When $x = 0$, $u = 0$.
• When $x = \frac{\pi}{3}$, $u = \frac{2\pi}{3}$.

Now, $\int_{0}^{\frac{\pi}{3}} \sec 2x \tan 2x \, dx = \frac{1}{2}\int_{0}^{\frac{2\pi}{3}} \sec u \tan u \, du$, which equals $\frac{1}{2}\left[\ln |\sec u + \tan u| \right]_{0}^{\frac{2\pi}{3}}$.

Calculating this gives: $\frac{1}{2}\left(\ln |\sec(\frac{2\pi}{3}) + \tan(\frac{2\pi}{3})| - \ln |\sec(0) + \tan(0)|\right).$

The function $f(x) = 3\sin^{-1}\left(\frac{x}{2}\right)$ is defined for the input values of $x$ that satisfy the condition of the arcsine function, where $-1 \leq \frac{x}{2} \leq 1$. Therefore:

• The domain of $f(x)$ is: $-2 \leq x \leq 2$.

The range of $f(x)$ can be determined by considering the output of $\sin^{-1}(y)$, which produces values in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Thus, multiplying by 3 gives:

• The range of $f(x)$ is: $[-\frac{3\pi}{2}, \frac{3\pi}{2}]$.

We can express the points as $x = 3t$ and $y = 2t$. To eliminate the parameter $t$:

1. Solve for $t$ in terms of $x$: $t = \frac{x}{3}$.
2. Substitute $t$ back into the equation for $y$: $y = 2\left(\frac{x}{3}\right) = \frac{2x}{3}$.

Thus, the Cartesian equation relating $x$ and $y$ is: $y = \frac{2}{3}x$.

Using the substitution $u = 1 - x^2$ gives us:

• Then, differentiate: $du = -2x \, dx$ or $dx = -\frac{du}{2x}$.
• Change the limits:
  • When $x = \frac{3}{2}$, $u = 1 - \left(\frac{3}{2}\right)^2 = 1 - \frac{9}{4} = -\frac{5}{4}$.
  • When $x = 2$, $u = 1 - 2^2 = 1 - 4 = -3$.

Thus, the integral becomes: $-\int_{-\frac{5}{4}}^{-3} \frac{2x}{u^2} \left(-\frac{du}{2x}\right) = \int_{-3}^{-\frac{5}{4}} \frac{1}{u^2} \, du$.

Which evaluates to: $\left[-\frac{1}{u}\right]_{-3}^{-\frac{5}{4}} = \left(-\frac{1}{-\frac{5}{4}} + \frac{1}{-3}\right) = \frac{4}{5} - \frac{1}{3} = \frac{12}{15} - \frac{5}{15} = \frac{7}{15}$.