VCE Methods Question Thread!

[Robert123]

could someone please explain to me why the answer is 'b' and not 'e'?? Cheers

Question 22
The graph of a differentiable function f has a local maximum at (a, b), where a < 0 and b > 0, and a local
minimum at (c, d), where c > 0 and d < 0.
The graph of y = –|f(x – 2)| has
A. a local minimum at (a – 2, –b) and a local maximum at (c – 2, d)
B. local minima at (a + 2, –b) and (c + 2, d)
C. local maxima at (a + 2, b) and (c + 2, –d )
D. a local minimum at (a – 2, –b) and a local maximum at (a – 2, – d)
E. local minima at (c + 2, –d) and (a + 2, –b)

Ok, this is fairly hard to follow on your head, the best way to approach it would be to graph off.
Now dealing with that pesky d, first off the translation of two has no effect on it as it is in the x direction. Now here's the tricky thing, the magnitude of d actually changes it to -d since d is less than 0. Then you apply the reflection in the x axis to go back to what you started with d.
Does that clarify when your thought pattern went wrong? And remember, when in doubt, draw a graph!

what happened between these steps? i.e. How did it become 0.4Pr(B)?

Think of Pr(B) as a variable like x.
X-0.6x=0.4x

[Blondie21]

How come you got an extra -k? I tried with my calculator and got the same answer as the report.  I think they did area of triangle minus area under curve since  area under tangent is a triangle.

ye I did it again and it worked..

idk what's wrong with my calc sometimes.. 

[RKTR]

what happened between these steps? i.e. How did it become 0.4Pr(B)?

1-0.6 =0.4

[bovawatkins]

Ok, this is fairly hard to follow on your head, the best way to approach it would be to graph off.
Now dealing with that pesky d, first off the translation of two has no effect on it as it is in the x direction. Now here's the tricky thing, the magnitude of d actually changes it to -d since d is less than 0. Then you apply the reflection in the x axis to go back to what you started with d.

yeh i kinda see what you mean, i just cant get my head around how there is are two minimums that arent both below the x axis

[Edward Elric]

Is it just me or was the 2012 exam 2 for methods a lot harder than the 2013 one? Does anyone know the cut off for an A+ during that year?

[silverpixeli]

Is it just me or was the 2012 exam 2 for methods a lot harder than the 2013 one? Does anyone know the cut off for an A+ during that year?

61.5/80 for A+ on exam 2 2012, lowest in a few years, not sure about 2013 (but like I sat the 2013 paper and imo it was harder)

[Reus]

VCAA exam 2 2013 q3di
How on earth?@?@

[psyxwar]

If the gradient of a function f(x) is always increasing, it implies that f'(x) is strictly increasing, right?

A tad confused because Kilbaha 2014 solutions to a question that asks why a graph does not satisfy a criteria which states that 'the gradient of the function is always increasing' says that we require f'(x)>0. Have I misunderstood something?

[RKTR]

VCAA exam 2 2013 q3di
How on earth?@?@

EF is a vertical line. They have same x coordinate. Distance = y of E - y of F

[Reus]

EF is a vertical line. They have same x coordinate. Distance = y of E - y of F

Thanks heaps!!

Also anyone know how to do VCAA exam 2 2013 q3f? Seems so simple but idk.

[Jason12]

Thanks heaps!!

Also anyone know how to do VCAA exam 2 2013 q3f? Seems so simple but idk.

I think you let v'(x) = 0 and then sub in k = 8m

[Reus]

I think you let v'(x) = 0 and then sub in k = 8m

I did :/ can't seem to get an answer

[kx4y]

Hey guys,

Is there any way to plot a vertical line (eg. x=4) on the Ti-nspire CAS?

[Jason12]

I did :/ can't seem to get an answer

I defined v(x) as 8mx^1/2 -mx^2 then derived it and made it equal to zero. Then use cas solve and you should get x = 2^(2/3). You have to include the domain restriction as well.

[Jason12]

how to do these types of MCQ questions?