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Transcript of This is a step by step solution for Q5 in the Extended Response Section of VCAA Exam 2, 2014. You...
![Page 1: This is a step by step solution for Q5 in the Extended Response Section of VCAA Exam 2, 2014. You should try to solve the problem yourself, and only look.](https://reader036.fdocuments.net/reader036/viewer/2022062720/56649f075503460f94c1ce9f/html5/thumbnails/1.jpg)
This is a step by step solution for Q5 in the Extended Response Section of VCAA Exam 2 , 2014.
You should try to solve the problem yourself, and only look at the slide where you are up to if you cannot make any further progress.
When you view a slide, use the hint on it to try to keep going yourself.
In other words, only look at a slide as the last resort.
GOOD LUCK!
Exam 2 2014 QUESTION 5.pdf
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First take out the HCF: and recall that we can factorize the difference of two cubes!!
OR: simply factorize on CAS:
CAS output is )
a.
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Now look at the format that they have asked for.They are asking for the quadratic factor to be in turning point form!
Complete the square.
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b.We have found that:
We are given that:
If we compare the quadratic factors of both functions, we see that the quadratic component of g(x) could be obtained from that of f(x) by replacing x with x + 1.
Test out this idea!
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Since the graph of is obtained by translating f(x) by 1 unit in the negative x-direction.
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If the graph of f (blue) is shifted 1 unit to the left, there is one positive x-intercept. If shifted 2.4 units to the left, there is one positive x-intercept.
c.i
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If shifted 3 units to the left, there are no more positive x-intercepts. Likewise if shifted any further to the left, there will be no positive x-intercepts.
So the values of d for which there is ONE positive x-intercept are:
OR you can write it as
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d.
To answer this question once again a graph is essential. We wish to find the value of n for which the horizontal line will cut the graph in only one point.
Try this if you haven’t yet done so, and see if you can solve the problem yourself.
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In this position, with n = - 11.7, there is no solution for the equation:
We need to find the exact y value of the turning point of g(x). This will be the value of n we require.
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For a stationary point,
Now evaluate: )
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Required value of n is:
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e.
In this question, we are told that:
where m is a positive number.In other words, we are being told that
…and we are being asked for an equation connecting u and v. See if you can go ahead and solve the problem now.
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We know already:
Therefore:
So:
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e ii
We are told that:
From the previous question, we have found that:
Solving the simultaneous equations on CAS:
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and or and
To decide which values are appropriate we must refer again to the graph of g(x).
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We see from the graph that since the gradient at the point where x = u is positive(equal to m), and the gradient at x = v is negative (equal to –m), it follows that u > v.
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and
We conclude:
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f.i
To find the equation of the tangent at it means that p is the parameter, and we must find the gradient and the y-intercept of the line in terms of p.
First, we find the gradient of the tangent, which is equal to
The point ( is the point:
If you haven’t done this yet, find the equation now!!
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Using:
We get:
𝒚=(𝟒𝒑𝟑−𝟖 )𝒙 −𝒑 (𝟒𝒑𝟑−𝟖 )+𝒑𝟒−𝟖𝒑
𝒚=(𝟒𝒑𝟑−𝟖 )𝒙 −𝟑𝒑𝟒
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f.Ii
We require that the tangent found above goes through the point
We substitute these co-ordinates into the tangent equation!
Do it now if you haven’t yet!!
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12 = (4×
Required values of p are p = 0 or p = 2Now find the equations of the tangent lines that we require:
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𝒚=(𝟒𝒑𝟑−𝟖 )𝒙 −𝟑𝒑𝟒
If
If
𝒚=𝟐𝟒𝒙−𝟒𝟖
The two tangent equations are: