Evaluate $$\lim_{x \to 0} \frac{\sin 3x}{x}$$ (b) Find $$\frac{d}{dx}(3x^{2} \ln x)$$ for $x > 0$ - HSC - SSCE Mathematics Extension 1 - Question 1 - 2002 - Paper 1

Using the product rule for differentiation:
$\frac{d}{dx}(uv) = u'v + uv'$, where $u = 3x^{2}$ and $v = \ln x$.
Calculating:

• $u' = 6x$,
• $v' = \frac{1}{x}$.
  Thus,
  $$\frac{d}{dx}(3x^{2} \ln x) = 6x \ln x + 3x^{2} \cdot \frac{1}{x} = 6x \ln x + 3x.$

The integral can be evaluated using the standard result:
$\int \sec x \tan x \, dx = \sec x + C$.
Applying this,
$\int \sec 2x \tan 2x \, dx = \frac{1}{2} \sec 2x + C.$
Evaluating from $0$ to $\frac{\pi}{3}$:
$$\left[ \frac{1}{2} \sec 2x \right]_{0}^{\frac{\pi}{3}} = \frac{1}{2} \sec\left(\frac{2\pi}{3}\right) - \frac{1}{2} \sec(0) = \frac{1}{2} \left(-2\right) - \frac{1}{2} \cdot 1 = -1 - \frac{1}{2} = -\frac{3}{2}.$

For the function $f(x) = 3 \sin^{-1}(\frac{x}{2})$, the argument $\frac{x}{2}$ must lie within $[-1, 1]$.

• Setting up the inequalities:
  $-1 \leq \frac{x}{2} \leq 1$
  leads to
  $-2 \leq x \leq 2.$
  Thus, the domain is $[-2, 2]$.
• The range of the inverse sine function is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, so multiplying by 3 gives a range of $[-\frac{3\pi}{2}, \frac{3\pi}{2}]$.

Given the parametric equations: $x = 3t$ and $y = 2t^{2}$, we can express $t$ in terms of $x$:
$t = \frac{x}{3}.$
Substituting into the equation for $y$:
$y = 2\left(\frac{x}{3}\right)^{2} = 2\cdot\frac{x^{2}}{9} = \frac{2}{9} x^{2}.$
Thus, the Cartesian equation of the parabola is
$y = \frac{2}{9} x^{2}.$

Using the substitution $u = 1 - x^{2}$:

• Therefore, $\frac{du}{dx} = -2x \Rightarrow dx = \frac{du}{-2x}$.
• Changing the limits:
  • When $x = \frac{2}{3}$, $u = 1 - \left(\frac{2}{3}\right)^{2} = \frac{5}{9}$;
  • When $x = 2$, $u = 1 - 2^{2} = -3$.
    Substituting into the integral:
    $\int_{\frac{5}{9}}^{-3} \frac{2x}{u^{2}} \cdot \frac{du}{-2x} = -\int_{\frac{5}{9}}^{-3} \frac{1}{u^{2}} du.$
    This leads to:
    $$-\left[ -\frac{1}{u} \right]{\frac{5}{9}}^{-3} = \left[\frac{1}{u}\right]{\frac{5}{9}}^{-3} = \left(\frac{1}{-3} - \frac{9}{5}\right) = -\frac{1}{3} - \frac{9}{5} = -\frac{1}{3} - \frac{27}{15} = -\frac{1}{3} - \frac{3}{5} = -\frac{5 + 9}{15} = -\frac{14}{15}.$