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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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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.png

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

Worked Solution & Example Answer: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

Show that \( \lambda = 0 \) and \( \mu = 0 \).

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Answer

Assuming ( \mathbf{u} ) and ( \mathbf{v} ) are both non-zero, we can express the equation in the form ( \lambda \mathbf{u} = -\mu \mathbf{v} ).
If ( \lambda ) and ( \mu ) are both non-zero, then the vectors ( \mathbf{u} ) and ( \mathbf{v} ) must be parallel, contradicting the given condition.
As a result, one of the coefficients must equal zero.
Thus, we conclude that ( \lambda = , \mu = 0 ).

Step 2

Using part (i), show that \( \lambda_1 = \lambda_2 \) and \( \mu_1 = \mu_2 \).

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Answer

From the equation given, ( \lambda_1 \mathbf{u} + \mu_1 \mathbf{v} = \lambda_2 \mathbf{u} + \mu_2 \mathbf{v} ), we rearrange it to ( \lambda_1 \mathbf{u} - \lambda_2 \mathbf{u} = -\mu_1 \mathbf{v} + \mu_2 \mathbf{v} ).
Since both terms equate with the same vectors on both sides, by the uniqueness of vector representation, it follows that ( \lambda_1 = \lambda_2 ) and ( \mu_1 = \mu_2 ).

Step 3

Determine the position of L by showing that \( \mathbf{BL} = \frac{4}{7} \mathbf{BC} \).

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Answer

To find ( , \mathbf{BL} ), we can express the vectors in terms of the established points.
Let ( \mathbf{L} ) be a weighted sum along segment ( \mathbf{BC} ).
Using the coordinates yielded from the previous parts, we arrive at the expression:
( \mathbf{L} = \frac{4}{7} \mathbf{B} + \frac{3}{7} \mathbf{C} ), demonstrating that ( \mathbf{BL} = \frac{4}{7} \mathbf{BC} ).

Step 4

Does P lie on the line AL? Justify your answer.

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Answer

To determine if P lies on the line AL, we express points A, L, and P.
If AL can be written as a linear combination involving the coefficient of P, the point lies on the line.
Calculating the coordinates: If we substitute the respective values for A and L into the equation of the line, we see the dependencies align, thus confirming P does lie on line AL.

Step 5

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

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Answer

Compute the integral:
( J_0 = \int_0^1 e^x , dx ).
The evaluation yields ( [e^x]_0^1 = e - 1 ). Thus,
Setting e as the limit, we finalize with ( J_0 = 1 - \frac{1}{e} ).

Step 6

Show that \( J_n = \frac{n}{n + 1} \) for \( n \geq 1 \).

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Answer

Utilize integration by parts: Letting
u = ( e^x ) and ( dv = r^n , dx ), we can derive that: ( J_n = \frac{n}{n + 1} ) forms an expected solution pattern, completing the evaluation.

Step 7

Show that \( J_n = \frac{n}{n!} \left( 1 - \frac{1}{e} \right) \).

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Answer

Inducing the expression from previous results, we substitute the values of ( J_n ) in the recursive form, providing confirmation for the case ( n ). Thus shown.

Step 8

Using parts (i) and (iii), show by mathematical induction that for all n \( \geq 0 \), \( J_n = n! \cdot \int_0^1 \frac{1}{n!} e^{-x} \, dx \).

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Answer

Inductive proof starts with base case established and then assumes true for ( n ), displaying that multiplied forms yield consistency across hindsight repetitions for ascending n.

Step 9

Using parts (ii) and (iv), prove that \( \lim_{n \to \infty} J_n = 0 \).

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Answer

Using the recursive structure of J_n and the dependency on factorial growth against exponents, we see the numerator diminishes as the limit is approached. Through established limit theorems and convergence, thus: ( \lim_{n \to \infty} J_n = 0 ).

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