2020 hsc (1 Viewer)

[=)(=]

does my method work for iii?

 

[member 6003]

yeah it works for ii)

Probably use a different variable to replace y like theta or something for clarity.

 

[Luukas.2]

You haven't accounted for the case when .

 

[member 6003]

You haven't accounted for the case when .

where x<0 y=arccosx has pi/2<y<pi therefore f(x)=tany<0 which matches with g(x). Does this case really need to be considered? x<0 is accounted for its just misleading that an acute angle is drawn.

 

[Luukas.2]

where x<0 y=arccosx has pi/2<y<pi therefore f(x)=tany<0 which matches with g(x). Does this case really need to be considered? x<0 is accounted for its just misleading that an acute angle is drawn.

If pi/2 < y < pi, then it is an obtuse angle, and so the relationship to tan y can't rest on applying Pythagoras' Theorem from when y is acute.

The method will work for both x > 0 and x < 0 cases, but not without justification / explanation.

After all, the question is looking for a solution based on:

​

and the value of C must be shown in each of the two unconnected parts of the domain.

 

[member 6003]

If pi/2 < y < pi, then it is an obtuse angle, and so the relationship to tan y can't rest on applying Pythagoras' Theorem from when y is acute.

The method will work for both x > 0 and x < 0 cases, but not without justification / explanation.

After all, the question is looking for a solution based on:

​

and the value of C must be shown in each of the two unconnected parts of the domain.

The solution considers x<0 and x>0 to ensure that the constant value is the same as the function is not continuous. It doesn't imply you need to consider these cases separately for an alternate proof.

If you look at other proofs of tan(arccosx) they don't consider the x<0 case because its included with the right-angle triangle diagram. There is no assumption with the diagram that either x>0 or that the angle is acute. See wikipedia or any other site also worked solutions for the cambridge year 11 textbook. Maybe specifying that the angle is between 0 and pi could make the solution better but you don't need to consider cases or draw different diagrams or anything. Either way, since most other proofs done online don't consider the cases separately, there's no way you'd be marked down for it.

 

[scaryshark09]

How do you do this?

 

[member 6003]

View attachment 40063
How do you do this?

its monic therefore it should be going upwards on each side therefore either c or d. since its divisible by x^2+x+1 you can think of it as p(x)=(x^2+x+1)(x-a)^2 so only the double root exists so it cant be D therefore C

 

[scaryshark09]

this is question 13cii, the original one on this thread

i did it the other way, but with the method they provide, how do they determine that f(x) - g(x) is a constant?

 

[yanujw]

how do they determine that f(x) - g(x) is a constant?

Integrate both sides with respect to x, and apply

 

[member 6003]

View attachment 40064
this is question 13cii, the original one on this thread

i did it the other way, but with the method they provide, how do they determine that f(x) - g(x) is a constant?

lets say y=f(x)-g(x) we have dy/dx=0 that means at every point the function doesn't change and therefore is a constant function. You can check this by integrating dy/dx=0. The only problem with this is that if the function is discontinuous, the value can jump without changing, think |x|/x. that's why we check x<0 and x>0

 

[scaryshark09]

lets say y=f(x)-g(x) we have dy/dx=0 that means at every point the function doesn't change and therefore is a constant function. You can check this by integrating dy/dx=0. The only problem with this is that if the function is discontinuous, the value can jump without changing, think |x|/x. that's why we check x<0 and x>0

why does dy/dx = 0

 

[member 6003]

why does dy/dx = 0

dy/dx= f'(x) - g'(x) = 0

 

[Luukas.2]

why does dy/dx = 0

Looked at another way:

​

Now, and have the same domain, , which has a discontinuity at . Thus, the value of must be separately determined in each part of the domain.

​

Thus, for all .

 

[Average Boreduser]

why does dy/dx = 0

You could also look at it as a dummy variable/function. that way it's easier to conceptualise.