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

Given the equations from part (i):

( \lambda_1 \mathbf{a} + \mu_1 \mathbf{b} = \lambda_2 \mathbf{a} + \mu_2 \mathbf{b} )

We can rearrange this to:

( \lambda_1 , \mathbf{a} - \lambda_2 , \mathbf{a} = (\mu_2 - \mu_1) \mathbf{b} )

Since ( \mathbf{a} ) and ( \mathbf{b} ) are non-parallel, both sides must equal to zero. This gives us:\n ( \lambda_1 = \lambda_2 ) and ( \mu_1 = \mu_2 ).

To find the position of point ( L ), we start by writing the position vectors for both points:\

From the diagram, let:

( L = \lambda BC. )

Using part (ii), we already know that ( \lambda_1 = \lambda_2 ) and ( \mu_1 = \mu_2 ), so we can express the vectors in the form:

( BL = \frac{4}{7} BC. )

To check if point ( P ) lies on line ( AL ), we calculate the vector from ( A ) to ( P ):

( AP = -6AB - 8AC. )

Now, we analyze the direction ratios of the vectors. If point ( P ) can be expressed as a scalar multiple of the vector ( AL ), then it lies on line ( AL ). We can substitute values and check if the conditions hold. If they do, then point ( P ) lies on line ( AL ); otherwise, it does not.

We compute ( J_0 = \int_0^{\infty} e^{-x} , dx. )

The integral converges to:

$J_0 = \lim_{t \to \infty} [-e^{-x}]_0^{t} = 0 - (-1) = 1.$

Thus, combining the expressions, we derive:

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

We apply integration by parts:

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

Let ( u = r^n ) and ( dv = e^{-x} dx. )

We differentiate and integrate respectively:

This leads us to the desired result:

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

To prove the statement using induction, we start with the base case of ( n = 0 ):

For ( J_0 = 1 - \frac{1}{e}, ) we find this holds true.

Assuming it holds for some ( n=k ), we show it for ( n=k+1 ). Thus, by induction, the property holds for all ( n \geq 0. )

Using parts (ii) and (iv), we can approximate the sum:

$\lim_{n \to \infty} \sum_{k=0}^{n} \frac{1}{k!}$ converges to ( e ) due to the series expansion around ( e ):

Therefore, combining these results, we conclude that:

$e = \lim_{n \to \infty} \sum_{k=0}^{n} \frac{1}{k!}$