The point P divides the interval joining A(-1,-2) to B(9,3) internally in the ratio 4 : 1 - HSC - SSCE Mathematics Extension 1 - Question 1 - 2011 - Paper 1

To differentiate ( \frac{\sin^2 x}{x} ), we apply the quotient rule:

$$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$$

Let (u = \sin^2 x) and (v = x).

First, we find (u') and (v'):

• (u' = 2\sin x \cos x = \sin(2x))
• (v' = 1)

Now applying the quotient rule:

$$\frac{d}{dx} \left( \frac{\sin^2 x}{x} \right) = \frac{\sin(2x) \cdot x - \sin^2 x}{x^2}$$

To solve the inequality ( \frac{4 - x}{x} < 1 ), we can start by manipulating the inequality:

1. Multiply both sides by (x) (assuming (x \neq 0) and considering the sign of x later): (4 - x < x) (4 < 2x) (x > 2)

Now, we also have to consider the constraint that (x > 0), thus the solution to the inequality is:

$$\text{The solution is } x > 2.$$

Using the substitution ( u = \sqrt{x} ), we have:

• (x = u^2 \Rightarrow dx = 2u , du).
• Changing the limits: when (x = 1), (u = 1); when (x = 4), (u = 2).

Now substituting into the integral:

$$\int_{1}^{4} \frac{\sqrt{x}}{\sqrt{x}} \, dx = \int_{1}^{2} 1 \cdot (2u) \, du = 2 \int_{1}^{2} u \, du = 2 \left[ \frac{u^2}{2} \right]_{1}^{2} = 2 \left( 2 - \frac{1}{2} \right) = 2 \cdot \frac{3}{2} = 3.$$

Thus, the value of the integral is 3.

The value of ( \cos^{-1}( -\frac{1}{2} ) ) corresponds to the angle in the second quadrant where the cosine of the angle equals (-\frac{1}{2} ).

Thus, the exact value is:

$$\cos^{-1}( -\frac{1}{2} ) = \frac{2\pi}{3} \, ext{or } 120\degree.$$

To find the range of the function ( f(x) = \ln(x^2 + e) ), we analyze the expression inside the natural logarithm. Since (x^2 \geq 0), the minimum value of (x^2 + e) is (e) when (x = 0).

Thus, the function can take values from (\ln(e) = 1) to infinity as (x^2) increases:

Therefore, the range is:

$$[1, +\infty).$$