The parametric equations of a line are given below - HSC - SSCE Mathematics Extension 1 - Question 11 - 2023 - Paper 1

To find the number of arrangements of the letters in "CONDOBOLIN":

1. Count the letters:

   • The word has 10 letters in total.
   • The letters are: C, O, N, D, O, B, O, L, I, N.

2. Identify repetitions:

   • The letter 'O' appears 3 times.

3. Use the formula for permutations of multiset:

   $\text{Number of arrangements} = \frac{n!}{n_1! \cdot n_2! ...} = \frac{10!}{3!}$

4. Calculate:

   $10! = 3628800$ $3! = 6$ So, $\frac{3628800}{6} = 604800$

5. Therefore, the total number of different arrangements is 604800.

To find the coefficients $a$ and $b$:

1. Use the Factor Theorem:

   • We know that $x + 1$ is a factor of $P(x)$.
   • Therefore, we can set $P(-1) = 0$:

   $P(-1) = (-1)^3 + a(-1)^2 + b(-1) - 12 = -1 + a - b - 12 = 0$

   This simplifies to: $a - b - 13 = 0 \Rightarrow a - b = 13 ag{1}$

2. Now for the Remainder Theorem:

   • Given that when $P(x)$ is divided by $x - 2$, the remainder is -18:

   $P(2) = 2^3 + a(2^2) + b(2) - 12$

   which simplifies to: $8 + 4a + 2b - 12 = -18$ Rearranging gives: $4a + 2b - 4 = -18 \Rightarrow 4a + 2b = -14 \Rightarrow 2a + b = -7 ag{2}$

3. Now, solve (1) and (2):

   • Substitute (1) into (2): $2a + (a - 13) = -7$
   • Thus: $3a - 13 = -7 \Rightarrow 3a = 6 \Rightarrow a = 2$

4. Substitute $a = 2$ back into (1): $2 - b = 13 \Rightarrow -b = 11 \Rightarrow b = -11$

5. Therefore, the values are:

   • $a = 2$
   • $b = -11$.