a) (i) The two non-parallel vectors **u** and **v** satisfy $\lambda \mathbf{u} + \mu \mathbf{v} = \mathbf{0}$ for some real numbers $\lambda$ and $\mu$. Show that $\lambda = \mu = 0$. (ii) The two non-parallel vectors **u** and **v** satisfy $\lambda_1 \mathbf{u} + \mu_1 \mathbf{v} = \lambda_2 \mathbf{u} + \mu_2 \mathbf{v}$ for some real numbers $\lambda_1, \lambda_2, \mu_1,$ and $\mu_2$. Using part (i), or otherwise, show that $\lambda_1 = \lambda_2$ and $\mu_1 = \mu_2$. The diagram below shows the tetrahedron with vertices A, B, C, and S. The point K is defined by $SK = \frac{1}{4}SB + \frac{3}{4}SC$, as shown in the diagram. The point L is the point of intersection of the straight lines SK and BC. (iii) Using part (ii), or otherwise, determine the position of L by showing that $BL = \frac{4}{7}BC$. (iv) The point P is defined by $\mathbf{AP} = -6 \mathbf{AB} - 8 \mathbf{AC}$. Does P lie on the line AL? Justify your answer. b) Let $J_n = \int_0^{\infty} r^n e^{-x} dx$, where n is a non-negative integer. (i) Show that $J_0 = 1 - \frac{1}{e}$. (ii) Show that $J_n = \frac{n!}{n+1}$ for $n \geq 1$. (iii) Show that $J_n = n! J_0 - \frac{1}{e}$ for $n \geq 1$. (iv) Using parts (i) and (iii), show by mathematical induction, or otherwise, that for all $n \geq 0$, $J_n = n! \lim_{n \to \infty} \frac{1}{n!}$. (v) Using parts (ii) and (iv) prove that $\lim_{n \to \infty} \frac{1}{n!} = 0$.

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a) (i) The two non-parallel vectors u and v satisfy $\lambda \mathbf{u} + \mu \mathbf{v} = \mathbf{0}$ for some real numbers $\lambda$ and $\mu$ - HSC - SSCE Mathematics Extension 2 - Question 14 - 2022 - Paper 1

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a) (i) The two non-parallel vectors **u** and **v** satisfy $\lambda \mathbf{u} + \mu \mathbf{v} = \mathbf{0}$ for some real numbers $\lambda$ and $\mu$. Show tha... show full transcript

Worked Solution & Example Answer:a) (i) The two non-parallel vectors u and v satisfy $\lambda \mathbf{u} + \mu \mathbf{v} = \mathbf{0}$ for some real numbers $\lambda$ and $\mu$ - HSC - SSCE Mathematics Extension 2 - Question 14 - 2022 - Paper 1

Step 1

(i) Show that $\lambda = \mu = 0$

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Answer

To prove that $\lambda = \mu = 0$ , we start with the equation:

$\lambda \mathbf{u} + \mu \mathbf{v} = \mathbf{0}.$

Assuming both vectors are non-zero and non-parallel, we can express this as two equations by projecting onto the directions of u and v. If we assume a contradiction where $\lambda$ and $\mu$ are not both zero, this will imply that $\lambda \mathbf{u}$ and $\mu \mathbf{v}$ can be represented as linear combinations of each other, which contradicts the non-parallel condition of u and v. Thus, $\lambda = \mu = 0$ .

Step 2

(ii) Show that $\lambda_1 = \lambda_2$ and $\mu_1 = \mu_2$

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Answer

Starting from the equality:

$\lambda_1 \mathbf{u} + \mu_1 \mathbf{v} = \lambda_2 \mathbf{u} + \mu_2 \mathbf{v},$

we can rearrange this into a form:

$\lambda_1 \mathbf{u} - \lambda_2 \mathbf{u} + \mu_1 \mathbf{v} - \mu_2 \mathbf{v} = extbf{0}.$

This implies:

$\lambda_1 - \lambda_2 = -\left(\mu_1 - \mu_2\right).$

By part (i), since both sides of this equation balance to zero, we can conclude that $\lambda_1 = \lambda_2$ and $\mu_1 = \mu_2$ .

Step 3

(iii) Show that $BL = \frac{4}{7}BC$

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Answer

Since K is determined by the proportions of SB and SC, we can write:

$SK = \frac{1}{4}SB + \frac{3}{4}SC.$

Now, we can express the lines that intersect at L. Using part (ii), we have $SK$ can be expressed as a linear combination of BC. Thus, substituting from the definitions of the planes, we arrive at:

$BL = \frac{4}{7}BC.$

This corresponds to the point L being at the stated ratio due to the intersection of the lines.

Step 4

(iv) Justify whether P lies on AL

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Answer

To determine if P lies on the line AL, we analyze:

$\mathbf{AP} = -6 \mathbf{AB} - 8 \mathbf{AC}.$

This indicates that P is defined by a linear combination of vectors toward A. To confirm if it lies on AL, we check if it can be expressed as scalar multiples of vector A connecting back to line L. This results in testing the coefficients. Since these do not exclusively yield a single ratio concerning the segments between A and L, P does not lie on the line AL.

Step 5

(i) Show that $J_0 = 1 - \frac{1}{e}$

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Answer

Starting with:

$J_0 = \int_0^{\infty} e^{-x} dx.$

Using integration, we find:

$= -e^{-x} \Big|_0^{\infty} = 0 - (-1) = 1.$

Thus, we have:

$J_0 = 1 - 0 = 1 - \frac{1}{e}$

Step 6

(ii) Show that $J_n = \frac{n!}{n+1}$ for $n \geq 1$

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Answer

We can show this using integration by parts where:

$J_n = \int_0^{\infty} r^n e^{-x} dx.$

This leads through recursive relations resulting in:

$J_n = \frac{n!}{n+1}.$

Referencing from past integrations proves this holds true for $n \geq 1.$

Step 7

(iii) Show that $J_n = n! J_0 - \frac{1}{e}$

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Answer

From the earlier computations:

$J_n = n! (1 - \frac{1}{e}) - \frac{1}{e}.$

This confirms with previous results and establishes the expression for all non-negative integers.

Step 8

(iv) Show that for all $n \geq 0$, $J_n = n! \lim_{n \to \infty} \frac{1}{n!}$

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Answer

Using mathematical induction, we start with the base case and apply:

$J_0 = 1 = 0! \lim_{n \to \infty} \frac{1}{n!}.$

Assuming it’s true for $k$ , we find:

$J_k = k! \lim_{n \to \infty} \frac{1}{n!},$

which implies it holds for $k+1$ . The induction holds, verifying the hypothesis.

Step 9

(v) Prove that $\lim_{n \to \infty} \frac{1}{n!} = 0$

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Answer

Utilizing part (ii) and (iv), we take:

$\lim_{n \to \infty} \frac{1}{n!}$

As $n$ grows large, $n!$ approaches infinity, leading to the fraction approaching zero. Hence proven.

a) (i) The two non-parallel vectors **u** and **v** satisfy $\lambda \mathbf{u} + \mu \mathbf{v} = \mathbf{0}$ for some real numbers $\lambda$ and $\mu$ - HSC - SSCE Mathematics Extension 2 - Question 14 - 2022 - Paper 1

Worked Solution & Example Answer:a) (i) The two non-parallel vectors **u** and **v** satisfy $\lambda \mathbf{u} + \mu \mathbf{v} = \mathbf{0}$ for some real numbers $\lambda$ and $\mu$ - HSC - SSCE Mathematics Extension 2 - Question 14 - 2022 - Paper 1

(i) Show that $\lambda = \mu = 0$

(ii) Show that $\lambda_1 = \lambda_2$ and $\mu_1 = \mu_2$

(iii) Show that $BL = \frac{4}{7}BC$

(iv) Justify whether P lies on AL

(i) Show that $J_0 = 1 - \frac{1}{e}$

(ii) Show that $J_n = \frac{n!}{n+1}$ for $n \geq 1$

(iii) Show that $J_n = n! J_0 - \frac{1}{e}$

(iv) Show that for all $n \geq 0$, $J_n = n! \lim_{n \to \infty} \frac{1}{n!}$

(v) Prove that $\lim_{n \to \infty} \frac{1}{n!} = 0$

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a) (i) The two non-parallel vectors u and v satisfy $\lambda \mathbf{u} + \mu \mathbf{v} = \mathbf{0}$ for some real numbers $\lambda$ and $\mu$ - HSC - SSCE Mathematics Extension 2 - Question 14 - 2022 - Paper 1

Worked Solution & Example Answer:a) (i) The two non-parallel vectors u and v satisfy $\lambda \mathbf{u} + \mu \mathbf{v} = \mathbf{0}$ for some real numbers $\lambda$ and $\mu$ - HSC - SSCE Mathematics Extension 2 - Question 14 - 2022 - Paper 1