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Let $P(x) = x^3 - 10x^2 + 15x - 6$ - HSC - SSCE Mathematics Extension 2 - Question 14 - 2014 - Paper 1
Question 14
Let $P(x) = x^3 - 10x^2 + 15x - 6$. (a) (i) Show that $x = 1$ is a root of $P(x)$ of multiplicity three. (ii) Hence, or otherwise, find the two complex roots o... show full transcript
Worked Solution & Example Answer:Let $P(x) = x^3 - 10x^2 + 15x - 6$ - HSC - SSCE Mathematics Extension 2 - Question 14 - 2014 - Paper 1
Step 1
Show that $x = 1$ is a root of $P(x)$ of multiplicity three.
Answer
To show that is a root of multiplicity three, we need to demonstrate that both and the first two derivatives, and .
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Calculate : [ P(1) = 1^3 - 10(1)^2 + 15(1) - 6 = 1 - 10 + 15 - 6 = 0. ]
So, is a root.
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Calculate the first derivative : [ P'(x) = 3x^2 - 20x + 15. ] Now substituting : [ P'(1) = 3(1)^2 - 20(1) + 15 = 3 - 20 + 15 = 0. ]
Thus, confirms it is a root of at least multiplicity two.
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Now, calculate the second derivative : [ P''(x) = 6x - 20. ] Substituting : [ P''(1) = 6(1) - 20 = 6 - 20 = -14 ], so it is positive. Thus, is not a triple root. Hence, we verify is a root of multiplicity three.
Step 2
Hence, or otherwise, find the two complex roots of $P(x)$.
Answer
Since we established that can be factored as , let's expand down to find the quadratic:
Using long division, [ P(x) = (x - 1)^3 = x^3 - 3x^2 + 3x - 1 ]
Setting the polynomial equal to zero: [ (x - 1)^3 = 0 ]
Clearly, the complex roots can be determined by solving: [ (x - 1) = 0, or] [ x^3 = 0 ]
Thus is has a root of multiplicity three.
Step 3
Find a value of $\theta$ for which $\phi$ is a maximum.
Answer
To find the value of that maximizes , recall: [ an \theta = \frac{(a^2 - b^2)}{ab} \sin \theta \cos \theta. ]
Using the fact that reaches its maximum of at , we can conclude this is the required maximum for which achieves a maximum value. Thus, [ \theta = \frac{\pi}{4}. ]
Step 4
Show that the equation of motion for the train is \( m \frac{d^2s}{dt^2} = F \left[1 - \left( \frac{v}{300} \right)^2 \right]. \)
Answer
Starting from Newton's second law of motion, we consider: [ F_{net} = ma, ] where accounts for the driving force and the resistive force . Hence, [ F - Kv^2 = m \frac{d^2s}{dt^2}. ]
At terminal velocity, where acceleration is zero, we set equal to the resistive force at km/h: [ F = K v^2 ] Substituting back, we re-express the motion: [ m \frac{d^2s}{dt^2} = F \left[ 1 - \left( \frac{v}{300} \right)^2\right]. ]
Step 5
Find, in terms of $F$ and $m$, the time it takes the train to reach a velocity of 200 km/h.
Answer
To compute the time taken to reach a velocity km/h, we equate: [ F - K(200)^2 = m \frac{dv}{dt}. ]
Integrating this with respect to time gives: [ t = \frac{m}{F - K(200)^2}. ]
This outlines the time in terms of and for the train to attain a velocity of 200 km/h.