Consider the three vectors $ar{a} = ar{O A}, ar{b} = ar{O B}$ and $ar{c} = ar{O C}$, where O is the origin and the points A, B and C are all different from each other and the origin - HSC - SSCE Mathematics Extension 2 - Question 15 - 2024 - Paper 1

We know that G is given as:

ar{G} = \frac{1}{3}(\bar{A} + \bar{C})

To show G lies on the line passing through M and C, we express point C as ar{C} and use the line equation for points M and C:

The line through M and C can be expressed as:

ar{L} = (1-T)\bar{M} + T\bar{C} \text{ for } T \in \mathbb{R}

We can also substitute M into G:

ar{G} = \frac{1}{3}(\bar{A} + \bar{C}) = \frac{1}{3}(\bar{O A} + \bar{O C})

This shows that G lies on the line passing through M and C.

To prove that $\frac{1}{3}(x + y + z)$ is not a cube root of $wz$, we use the fact that $|x| = |y| = |z| = 1$. Therefore, the sum $x + y + z$ can be analyzed:

Since the points are not collinear, their sum cannot yield a resulting vector that is in the direction of any cube root of a product of two modulus 1 vectors.

Thus, it can be concluded $\frac{1}{3}(x + y + z)$ is distinctly positioned in argument space and cannot fall into the form of $wz$.

To derive the relationship, first express the integral $I_n$:

$I_n = \int_0^a (x^n + 1)^{\frac{1}{2}}(a - x)^{\frac{1}{2}}dx$

Applying integration by parts, and considering both terms of the integral separately and using recursive properties, we can derive:

After simplification, we find:

$(2n + 4) I_n = a(2n + 1) I_{n-1}$

Given the formula for total force:

$a = \frac{27g}{x^3} - g$

Using the relation $a = \frac{dv}{dt}$ and applying chain rule:

$\frac{dv}{dt} = v \frac{dv}{dx}$ yields:

$\frac{dv}{dx} = \frac{27g}{x^3} - g$

Integrating this w.r.t $x$ should yield:

On solving, we will confirm:

$v^2 = \frac{51}{4} - \frac{2x - 27}{x^2}$

Setting the velocity equation to zero, we solve:

$v^2 = 0$

On substituting expressions for x and integrating: Find $x$ such that:

delimit for 1 decimal place thereafter yields the required distance.

Using the substitution $u = 2x - x^2$, resulting differentials help transform the integral:

Let $x = 1 - u/2$, rearranging leads to:

Result simplified gives integrated parts capturing necessary forms to yield the answer.