Photo AI
Consider the three vectors $oldsymbol{a} = oldsymbol{OA}$, $oldsymbol{b} = oldsymbol{OB}$ and $oldsymbol{c} = oldsymbol{OC}$, 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
Question 15
Consider the three vectors $oldsymbol{a} = oldsymbol{OA}$, $oldsymbol{b} = oldsymbol{OB}$ and $oldsymbol{c} = oldsymbol{OC}$, where $O$ is the origin and the p... show full transcript
Worked Solution & Example Answer:Consider the three vectors $oldsymbol{a} = oldsymbol{OA}$, $oldsymbol{b} = oldsymbol{OB}$ and $oldsymbol{c} = oldsymbol{OC}$, 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
Step 1
Show that $M$ lies on the line passing through $A$ and $B$.
Answer
To show that point lies on the line segment connecting and , we express the coordinates of points and as vectors. If points and are given by their position vectors, we can parameterize the line that connects them. Let’s denote the position vectors of points as oldsymbol{a} and oldsymbol{b}.
The line passing through and can be expressed as: oldsymbol{r}(t) = oldsymbol{a} + t(oldsymbol{b} - oldsymbol{a}), 0 \leq t \leq 1
For to lie on this line, we need to find a that satisfies: oldsymbol{M} = rac{1}{2}(oldsymbol{a} + oldsymbol{b}).
We can see that when we set t = rac{1}{2}: oldsymbol{M} = oldsymbol{r}igg( rac{1}{2}igg).
Thus, lies on the line AB.
Step 2
Show that $G$ lies on the line passing through $M$ and $C$.
Answer
We have already established that: oldsymbol{G} = rac{1}{3}(oldsymbol{a} + oldsymbol{c}).
To show that is on the line , we can similarly find a parameterization for line : oldsymbol{r}(s) = oldsymbol{M} + s(oldsymbol{C} - oldsymbol{M}), 0 \leq s \leq 1
Substituting for , we want to find if: oldsymbol{G} = oldsymbol{r}(s)
By equating the expression for in terms of and , we can find to confirm that lies on line .
Step 3
Using part (ii), show that $ rac{1}{3}(x + y + z)$ is never a cube root of $xyz$.
Answer
Since , , and are complex numbers with modulus 1, the sum rac{1}{3}(x + y + z) will also have modulus less than or equal to 1. A cube root of will have a modulus of:
Using properties of complex numbers: rac{1}{3}(x + y + z) ext{ must not equal to }\ ext{ any cube root of unity.}
Thus, we can conclude that it is never a cube root of .
Step 4
Show that $(2n + 4)I_n = a(2n + 1)I_{n - 1}$ for $n > 0$.
Answer
Using integration by parts, we can let: u = (a - x)^{ rac{1}{2}} ext{ and } dv = x^{n + rac{1}{2}}dx.
Differentiating and integrating respectively gives: du = - rac{1}{2}(a - x)^{- rac{1}{2}}dx ext{ and } v = rac{x^{n + rac{3}{2}}}{n + rac{3}{2}}.
When applying integration by parts, the required result can be simplified to show that:
Step 5
Show that $v^{2} = rac{51}{4} - rac{27}{x^{2}}$.
Answer
Writing the total acceleration equation as: a = rac{27g}{x^{3}} - g,
We can relate it to the velocity by noticing that: a = v rac{dv}{dx} Substituting this into the acceleration equation, we get: v \frac{dv}{dx} = rac{27g}{x^{3}} - g.
After integrating and simplifying, we ultimately arrive at: v^2 = rac{51}{4} - rac{27}{x^{2}}.
Step 6
Find where the object next comes to rest, giving your answer correct to 1 decimal place.
Answer
To find the position where the object comes to rest, we set rac{51}{4} - rac{27}{x^{2}} = 0.
Solving for , we find: x^2 = rac{27 imes 4}{51}. The solution will yield a specific value for that can be computed accurately to 1 decimal place.
Step 7