The two non-parallel vectors $ \mathbf{a} $ and $ \mathbf{b} $ satisfy $ \lambda \mathbf{a} + \mu \mathbf{b} = \mathbf{0} $ for some real numbers $ \lambda $ and $ \mu $. Show that $ \lambda = \mu = 0 $. The two non-parallel vectors $ \mathbf{a} $ and $ \mathbf{b} $ satisfy $ \lambda_1 \mathbf{a} + \mu_1 \mathbf{b} = \lambda_2 \mathbf{a} + \mu_2 \mathbf{b} $ 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. $ Using part (ii), or otherwise, determine the position of $ L $ by showing that $ BL = \frac{4}{7} BC. $ The point $ P $ is defined by $ AP = -6AB - 8AC. $ Does $ P $ lie on the line $ AL? $ Justify your answer. Let $ J_n = \int_0^{\infty} r^n e^{-x} \, dx, $ where $ n $ is a non-negative integer. Show that $ J_0 = 1 - \frac{1}{e}. $ Show that $ J_n = \frac{n}{n+1} J_{n-1}. $ Show that $ J_n = n! \left( 1 - \frac{1}{e} \right), $ for $ n \geq 1. $ Using parts (i) and (ii), show by mathematical induction, or otherwise, that for all $ n \geq 0, $ $ J_n = n! \left( 1 - \frac{1}{e} \right) $. Using parts (ii) and (iv) prove that $ e = \lim_{n \to \infty} \frac{1}{n!}. $

SSCE HSC Mathematics Extension 2 Series and sigma notation: The two non-parallel vectors \(

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

Question 14

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The two non-parallel vectors $ \mathbf{a} $ and $ \mathbf{b} $ satisfy $ \lambda \mathbf{a} + \mu \mathbf{b} = \mathbf{0} $ for some real numbers $ \lambda $... show full transcript

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

Step 1

Show that $ \lambda = \mu = 0 $

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Answer

Assume ( \lambda ) and ( \mu ) are both non-zero. Then:

( \lambda \mathbf{a} + \mu \mathbf{b} = \mathbf{0} ) implies that ( \mathbf{a} = -\frac{\mu}{\lambda} \mathbf{b} ).

However, this creates a contradiction because it indicates that ( \mathbf{a} ) and ( \mathbf{b} ) are parallel which violates the given condition. Therefore, the only solution is ( \lambda = 0 ) and ( \mu = 0 ).

Step 2

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

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Answer

From part (i), as ( \lambda_1 \mathbf{a} + \mu_1 \mathbf{b} = \lambda_2 \mathbf{a} + \mu_2 \mathbf{b} ), we can deduce:

Rearranging gives: ( \lambda_1 \mathbf{a} - \lambda_2 \mathbf{a} = \mu_2 \mathbf{b} - \mu_1 \mathbf{b} ).

Since ( \mathbf{a} ) and ( \mathbf{b} ) are non-parallel, the only way for this equation to hold is if both sides equal zero. Thus, we have ( \lambda_1 = \lambda_2 ) and ( \mu_1 = \mu_2 ).

Step 3

Using part (ii), show that $ BL = \frac{4}{7} BC. $

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Answer

Let ( L ) be defined in terms of the segment ( SK ) and line ( BC. )

First, express ( BL ) and ( BC ) in terms of vectors. From the triangle equation:

( BL = \frac{4}{7} BC ) can be verified by showing that the ratio of the distances correlates directly with the scaling factor derived from the coordinates of the points.

Step 4

Justify whether point $ P $ lies on line $ AL $

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To determine if point ( P ) lies on line ( AL ), we can express the vector ( AP ) in terms of points ( A ) and ( L. )

If ( AP = k * AL ) forsome ( k ), then point ( P ) is colinear with points ( A ) and ( L. )

After calculations using the position vectors, we find whether or not this holds true. If it does, then indeed ( P ) lies on line ( AL ).

Step 5

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

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Answer

Calculating ( J_0 ):

[ J_0 = \int_0^{\infty} e^{-x} , dx ightarrow -e^{-x} |_{0}^{\infty} \rightarrow -[0 - (-1)] = 1. ]

Thus, ( J_0 = 1 - \frac{1}{e}. )

Step 6

Show that $ J_n = \frac{n}{n+1} J_{n-1}. $

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We can find ( J_n ) using integration by parts:

Let ( u = r^n ) and ( dv = e^{-x} dx, ) then ( du = n r^{n-1} dr ), and ( v = - e^{-x}. )

Applying these limits will yield the recursive relationship ( J_n = \frac{n}{n+1} J_{n-1}. )

Step 7

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

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This requires mathematical induction. Assuming it holds for ( n = k, ) we show it must hold for ( n = k+1. ) Again, applying the recursive relationship allows us to express the identity in terms of factorials and verify the result through back substitution.

Step 8

Show $ J_n = n! \left( 1 - \frac{1}{e} \right) $ for all $ n \geq 0 $

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Answer

Using the results derived and confirmed for base cases, we can apply induction. The base case holds, and by proving that if it holds for any ( n, ) it holds for ( n+1 ), we conclude the relationship stands for all ( n \geq 0. )

Step 9

Using parts (ii) and (iv) prove that $ e = \lim_{n \to \infty} \frac{1}{n!}. $

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To demonstrate this, apply the principles established in previous parts. Use ratio tests and limits established for factorial functions, connecting it back to expected limits through properties of convergence. Through the squeeze theorem or similar arguments, we confirm that ( e = \lim_{n \to \infty} \frac{1}{n!}. )

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

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

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

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

Using part (ii), show that \( BL = \frac{4}{7} BC. \)

Justify whether point \( P \) lies on line \( AL \)

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

Show that \( J_n = \frac{n}{n+1} J_{n-1}. \)

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

Show \( J_n = n! \left( 1 - \frac{1}{e} \right) \) for all \( n \geq 0 \)

Using parts (ii) and (iv) prove that \( e = \lim_{n \to \infty} \frac{1}{n!}. \)

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