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Let $P(x) = x^3 - ax^2 + x + b$ be a polynomial, where $a$ is a real number - HSC - SSCE Mathematics Extension 1 - Question 2 - 2011 - Paper 1

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Let-$P(x)-=-x^3---ax^2-+-x-+-b$-be-a-polynomial,-where-$a$-is-a-real-number-HSC-SSCE Mathematics Extension 1-Question 2-2011-Paper 1.png

Let $P(x) = x^3 - ax^2 + x + b$ be a polynomial, where $a$ is a real number. When $P(x)$ is divided by $x - 3$ the remainder is 12. Find the remainder when $P(x)$ ... show full transcript

Worked Solution & Example Answer:Let $P(x) = x^3 - ax^2 + x + b$ be a polynomial, where $a$ is a real number - HSC - SSCE Mathematics Extension 1 - Question 2 - 2011 - Paper 1

Step 1

Let $P(x) = x^3 - ax^2 + x + b$ be a polynomial, where $a$ is a real number.

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Answer

To find the remainder when dividing by x+1x + 1, we apply the Polnomial Remainder Theorem. We know that:

P(3)=12P(3) = 12

Substituting x=3x = 3 into P(x)P(x):

33a(32)+3+b=123^3 - a(3^2) + 3 + b = 12 279a+3+b=1227 - 9a + 3 + b = 12 b9a+30=0b - 9a + 30 = 0 b=9a30b = 9a - 30

Now to find the remainder when dividing by x+1x + 1, we compute:

P(1)=(1)3a(1)2+(1)+bP(-1) = (-1)^3 - a(-1)^2 + (-1) + b =1a1+(9a30)= -1 - a - 1 + (9a - 30) =8a32= 8a - 32

Thus, the remainder when P(x)P(x) is divided by x+1x + 1 is 8a328a - 32.

Step 2

The function $f(x) = ext{cos}(2 heta) - x$ has a zero near $x = rac{1}{2}$.

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Answer

To find another approximation using Newton's method, we first calculate:

f(x)=2extsin(2heta)f'(x) = -2 ext{sin}(2 heta)

Let x_0 = rac{1}{2} and compute:

f(x_0) = ext{cos}(2( rac{1}{2})) - rac{1}{2}

Using Newton’s method formula:

xn+1=xnf(xn)f(xn)x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}

g Compute x1x_1. Simplifying down to two decimal places gives us the next approximate zero.

Step 3

Find an expression for the coefficient of $x^2$ in the expansion of $\left(3x - 4 - \frac{8}{x}\right)^8$.

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Answer

Using the binomial expansion, we need to focus on the term that contributes x2x^2:

The general term=(8k)(3x)8k(48x)k\text{The general term} = \binom{8}{k} (3x)^{8-k}(-4 - \frac{8}{x})^k

Setting up for x2x^2, we find terms such that combined they yield x2x^2.

Step 4

Sketch the graph of the function $f(x) = 2 ext{cos}^{-1}(x)$. Clearly indicate the domain and range of the function.

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Answer

The domain of f(x)=2extcos1(x)f(x) = 2 ext{cos}^{-1}(x) is [1,1][-1, 1] since the function is defined only in that interval. The range, as yy, will be [0, 2oldsymbol{ rac{ ext{ ext{Pi}}}}{2}]. A sketch would show the parabola intersecting the yy-axis at 00 and 2extPi2 ext{Pi} and having the flip of symmetry.

Step 5

How many arrangements are there for the 40 songs?

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Answer

The total number of arrangements is given by 40!40! which is computed as:

40!=40×39×38×...×140! = 40 \times 39 \times 38 \times ... \times 1

Step 6

Alex decides that she wants to play her three favourite songs first, in any order. How many arrangements of the 40 songs are now possible?

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Answer

First, the 3 songs can be arranged in 3!3! ways. The rest of the 37 songs can be arranged in 37!37! ways. Thus, the total arrangements are:

3!×37!3! \times 37!

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