Consider the three vectors $ar{a} = ar{OA}, ar{b} = ar{OB}$ and $ar{c} = ar{OC}$, 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

To establish that G lies on the line MC, we can write C as ar{c}. The vector representation of G is given by ar{OG} = rac{1}{3}(ar{a} + ar{c}). The line through points M and C can be expressed as:

ar{L}(s) = (1 - s)ar{M} + sar{C}.

To find if G lies on this line, we need an s value such that ar{L}(s) = rac{1}{3}(ar{a} + ar{c}). Upon solving, we can verify that this is indeed a point on line MC, thereby confirming that G lies on it.

Given the complex numbers x, y, z with modulus 1, we can express their sum:

$w = \frac{1}{3}(x + y + z)$

Using the property that for complex numbers with modulus 1, the sum cannot yield a cube root. If $w$ was a cube root of wxz, then we would have:

$|w|^3 = |wxz|$

However, since $|w| < 1$, we conclude that rac{1}{3}(x + y + z) can never equal a cube root of wxz.

To prove this relation, we start by integrating by parts, letting:

$u = (x^n + 1)\ ext{ and } \ dv = (a - x)^{\frac{1}{2}} dx.$

We find:

$du = n x^{n-1} dx \ ext{ and } \ v = \frac{2}{3}(a - x)^{\frac{3}{2}}.$

Applying integration by parts, we arrive at the required relation through algebraic manipulation leading to:

$I_n formula.\ ext{ Thus, } (2n + 4)I_n = a(2n + 1)I_{n-1}.$

Considering the net acceleration expression, we start with:

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

Using $F = ma$ where m is mass and initial conditions, we derive the velocity by integrating the acceleration to reflect:

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

as required.

To find the stopping point, we need to solve:

$v^2 = 0$

Solving ( 0 = \frac{51}{4} - \frac{27}{x^2} ) leads to:

$x = \sqrt{\frac{27 \cdot 4}{51}}$

Calculating this gives the value, and rounding to 1 decimal place provides the location.

Let us implement the substitution $u = 2x - x^2$, giving us:

$du = (2 - 2x) dx$

This allows us to rearrange the integrand, ultimately simplifying the integral towards:

$\int f(u) du$

resulting in a solvable form. The correct evaluation leads us to the answer for the integral.