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
Question 2
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. When $P(x)$ is divided by $x - 3$ the remainder is 12. Find the remainder when $P(x)$ is divided by $x + 1$.
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Answer
Using the Remainder Theorem, when dividing a polynomial P(x) by x−r, the remainder is simply P(r).
Start by finding P(3):
P(3)=33−a(32)+3+b=27−9a+3+b
Thus, P(3)=30−9a+b=12.
This leads to the equation:
30−9a+b=12
Therefore,
b=9a−18.
Now find P(−1):
P(−1)=(−1)3−a(−1)2+(−1)+b=−1−a−1+(9a−18)
Simplifying, we get:
P(−1)=−2+8a−18=8a−20.
Step 2
The function $f(x) = ext{cos}(2x) - x$ has a zero near $x = rac{1}{2}$. Use one application of Newton's method to obtain another approximation to this zero. Give your answer correct to two decimal places.
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Answer
Newton's method states that:
xn+1=xn−f′(xn)f(xn)
where f′(x)=−2sin(2x)−1.
Calculate f(0.5):
f(0.5)=cos(1)−0.5.
(Using a calculator, extcos(1)≈0.5403, so)
f(0.5)≈0.5403−0.5≈0.0403.
Calculate f′(0.5):
f′(0.5)=−2sin(1)−1.
(Using a calculator, extsin(1)≈0.8415)
f′(0.5)≈−2(0.8415)−1≈−2.6830.
Now apply Newton's method:
x1=0.5−−2.68300.0403≈0.5+0.0150≈0.5150.
Therefore, the new approximation to two decimal places is x≈0.52.
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 Theorem, the general term in the expansion is:
Tk=(kn)(3x)n−k(−4−x8)k, where n=8.
For the coefficient of x2, we need the power of x in Tk to equal 2:
The contribution from (3x)n−k is 3n−kxn−k.
The contribution from (−4−x8)k gives (−4)p⋅(−8)q with terms balancing so that −q+(n−k)=2.
By setting k=3:
T3=(38)(3x)5(−4−x8)3.
Evaluating gives:
(38)35(−4)3(−8)3 yields the coefficient of x2.
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 function f(x)=2extcos−1(x) has a domain where x lies in the interval [-1, 1]. The range is thus determined by evaluating:
At x=−1, f(−1)=2π.
At x=1, f(1)=0.
The graph is a decreasing function.
The domain is:
Domain:x∈[−1,1].
The range is:
Range:f(x)∈[0,2π].
Sketch: Outline the said points and connect them with a smooth curve.
Step 5
Alex's playlist consists of 40 different songs that can be arranged in any order. (i) How many arrangements are there for the 40 songs?
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Answer
The number of arrangements of n items is given by n!.
Thus for 40 songs, it is:
40!=40×39×38×...×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
When Alex plays her three favourite songs first, they can be arranged in 3! ways. The remaining 37 songs can then be arranged in 37! ways.
