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

To differentiate (3x^2 \ln x), we will apply the product rule. Let (u = 3x^2) and (v = \ln x), hence:

$\frac{du}{dx} = 6x, \quad \frac{dv}{dx} = \frac{1}{x}.$

Using the product rule:

$\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} = 3x^2 \cdot \frac{1}{x} + \ln x \cdot 6x = 3x + 6x \ln x.$

Therefore, ( \frac{d}{dx}(3x^2 \ln x) = 3x + 6x \ln x ).

The integral can be solved using the fact that the derivative of ( \sec 2x ) is ( \sec 2x \tan 2x \cdot 2 ), which leads us to:

[\int \sec 2x \tan 2x , dx = \frac{1}{2} \sec 2x + C.]

Now we evaluate:

[ \int_{0}^{\frac{\pi}{3}} \sec 2x \tan 2x , dx = \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).]

Calculating the values gives us:

[ \frac{1}{2} \left(-2\right) - \frac{1}{2} (1) = -1 - \frac{1}{2} = -\frac{3}{2}. ]

The function ( \sin^{-1}(x) ) has a domain of ([-1, 1]) and a range of ([-\frac{\pi}{2}, \frac{\pi}{2}]). Therefore, for ( \frac{x}{2} ) to lie within this interval, we have:

• Domain: ( -2 \leq x \leq 2 ).

After multiplying by 3 in the function, the range becomes:

• Range: ( -\frac{3\pi}{2} \leq f(x) \leq \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 this into the equation for ( y ):

[ y = 2\left(\frac{x}{3}\right)^2 = \frac{2x^2}{9}.]

Thus, the Cartesian equation of the parabola is:

[ y = \frac{2}{9} x^2.]

Using the substitution ( u = 1 - x^2 ), then ( du = -2x , dx ) or ( dx = -\frac{du}{2x} ). This changes the limits of integration:

• When ( x = 2 ), ( u = 1 - 4 = -3 ).
• When ( x = 3 ), ( u = 1 - 9 = -8 ).

Transforming the integral:

[ \int \frac{2x}{(1 - x^2)^2} , dx = -\int_{-3}^{-8} \frac{1}{u^2} , du = -\left[-\frac{1}{u}\right]_{-3}^{-8} = -\left(-\frac{1}{-8} + \frac{1}{-3}\right). ]

Evaluating gives:

[ = \frac{1}{8} - \frac{1}{3} = \frac{3 - 8}{24} = -\frac{5}{24}.]