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3 Unit Maths HSC Exam Revision (1 Viewer)

random-1006

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If so, give a pair of values that satisfies the condition. If not, explain why not

I made this one up myself, its amazing the sort of questions you can make up if you really think about the content.
 
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hscishard

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If so, give a pair of values that satisfies the condition. If not, explain why not

I made this one up myself, its amazing the sort of questions you can make up if you really think about the content.
No, the graph is a negative definite, thus the definite integral is always negative as b is greater than a

Oh sht it's the reciprocal.
Still..the graph is still negative :D
 
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random-1006

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No, the graph is a negative definite, thus the definite integral is always negative as b is greater than a

Oh sht it's the reciprocal
lol, your lucky, it doesnt make a difference, didnt you notice that first time, 1/ negative is still negative
 

Trebla

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Hopefully part d) will make you think lol.

 

cutemouse

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Hopefully part d) will make you think lol.

For part (d) it's something to do with the fact that you need to do 2pi - pi/4 for some reason... I think ... Will have a look later.

Maybe because in actual fact the result in (a) should be stated as arctana + arctanb=arctan[(a+b)/(1-ab)] +2*pi*k (for appropriate integral value of k) due to some trigonometric property...?
 
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random-1006

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when we are proving the formula part 1, we are assuming x is acute

A= arctanx, B = arctany
tanA=x, tanB= y
etc.

the above step is assuming acute x, and the limits of integration are 3 and -2 RADIANS ( which is larger than pi/2, ie larger than 90 degrees, not an acute angle.

either that or something about the range of arctan is : -pi/2 < y < pi/2
 
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Trebla

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Maybe because in actual fact the result in (a) should be stated as arctana + arctanb=arctan[(a+b)/(1-ab)] +2*pi*k (for appropriate integral value of k) due to some trigonometric property...?
Note quite. If you remember the way that the y = tan-1x was defined, it requires that for each x there must be a unique value of y. The addition of 2kπ seems to imply non-uniqueness.

when we are proving the formula part 1, we are assuming x is acute

A= arctanx, B = arctany
tanA=x, tanB= y
etc.

the above step is assuming acute x, and the limits of integration are 3 and -2 RADIANS ( which is larger than pi/2, ie larger than 90 degrees, not an acute angle.
x and y are not "angles". The inverse tangent of them are "angles". So from your labelling, A and B are the angles, not x and y. Even with that correction, tan-14 and tan-12 both give acute angles, so that doesn't quite explain the flaw.
 

cutemouse

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There is a discontinuity at x=4.
Well technically speaking you should say that there's a gap in the domain. Talking about continuity at a point not in the domain of the function doesn't make sense.

Although at HSC level your reasoning should suffice.
 

cutemouse

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I know...



Implies that the LHS and RHS must both be within the (-pi/2, pi/2)... if the LHS is outside this range then the equation is not defined/meaningless, which is the case if the LHS = arctan3+arctan2...
 
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