Mathematics: Post your doubts here!

[princeali97]

LOL I am not even going to bother explaining this now, just re read what I explained, it is very clear. I'm talking about two different points here.

Ok nevermind then.Im sorry if i had said anything wrong against anybody , though i dont mean to.Thanks anyways.

 

[abruzzi]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w10_qp_33.pdf
Someone please explain qn 8ii (b) :/

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_13.pdf

Q1 (ii) and Q4(ii)

ASAP please someone help !

 

[Thought blocker]

tan^2 theta - sin^2 theta ; Always > zero for values of zero to 90
so,
tan^2 theta - sin^2 theta > 0
tan^2 theta > sin^2 theta
tan theta > sin theta

 

[Talha Khatri]

i have nearly done all the questions and only this one bothered me :/ why 360?

Pretty simple actually, I prefer making a diagram like this:

As you can see the letters, the S T and C stand for sin cos tan, these functions are positive in these ranges of angles while in A, all three are positive. When you look at your question, you have tan@ = - 7
I prefer to use this trick, take the inverse of positive 7 instead of negative seven and then look at the angle that you get. Like here you get: 81.67 degrees
Now use the diagram above and see in which ranges will tan@ have negative values, so you will see that happens in the 90 to 180 interval and the 270 to 360 interval.
Now, just use the formulas I have provided in the brackets and you will get your answer
Memorise this, it'll be extremely helpful

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_13.pdf

Q 1 (ii) and Q4 (ii) ..

 

[Talha Khatri]

You technically copied what was written in the mark scheme. I would write something like this:
tan^2@ - sin^2@ is always greater than zero for the values of theta between 0 to 90. Hence,
tan^2@ - sin^2@ > 0
tan^2@ > sin^2@
tan@ > sin@
Now that I think about it, you did write everything that is required plus it is just worth one mark so you aren't expected to do much working. So that should be enough

Haya Ahmed Here's your question 1 part 2

and as for question 4, just look at the range, it is from 0 to 1080. So just understand that the wave cycle keeps repeating itself so how many answers did you get for 0 to 360, multiply it by how many cycles further the range has been extended and you can find that out by dividing 1080 by 360 that'll give you, 3. So 3 into the number of answers you had for 0 to 360. That will be your final answer.

 

[Haya Ahmed]

Haya Ahmed Here's your question 1 part 2

and as for question 4, just look at the range, it is from 0 to 1080. So just understand that the wave cycle keeps repeating itself so how many answers did you get for 0 to 360, multiply it by how many cycles further the range has been extended and you can find that out by dividing 1080 by 360 that'll give you, 3. So 3 into the number of answers you had for 0 to 360. That will be your final answer.

OMG .. thanks a lot bro you're awesome .. THANKS AGAIN !! ^_^

 

[Talha Khatri]

OMG .. thanks a lot bro you're awesome .. THANKS AGAIN !! ^_^

Thanks! Just pray for me and the rest of us here that we all ace it inshaAllah

 

[ZaqZainab]

Pretty simple actually, I prefer making a diagram like this:
View attachment 40958
As you can see the letters, the S T and C stand for sin cos tan, these functions are positive in these ranges of angles while in A, all three are positive. When you look at your question, you have tan@ = - 7
I prefer to use this trick, take the inverse of positive 7 instead of negative seven and then look at the angle that you get. Like here you get: 81.67 degrees
Now use the diagram above and see in which ranges will tan@ have negative values, so you will see that happens in the 90 to 180 interval and the 270 to 360 interval.
Now, just use the formulas I have provided in the brackets and you will get your answer
Memorise this, it'll be extremely helpful

I use the same thing actually but i didn't know to use this for negative values
Thanks

 

[Thought blocker]

Pretty simple actually, I prefer making a diagram like this:
View attachment 40958
As you can see the letters, the S T and C stand for sin cos tan, these functions are positive in these ranges of angles while in A, all three are positive. When you look at your question, you have tan@ = - 7
I prefer to use this trick, take the inverse of positive 7 instead of negative seven and then look at the angle that you get. Like here you get: 81.67 degrees
Now use the diagram above and see in which ranges will tan@ have negative values, so you will see that happens in the 90 to 180 interval and the 270 to 360 interval.
Now, just use the formulas I have provided in the brackets and you will get your answer
Memorise this, it'll be extremely helpful

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_13.pdf
question 4i) how to do this, by your trick ?

 

[princeali97]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_13.pdf
question 4i) how to do this, by your trick ?

Sin2x + 3cos2x=0
sin2x=-3cos2x
sin2x/cos2x= - 3
tan2x= - 3
now tan-1 (3) = 71.56....Since tanx is -ve , x lies in either 2nd or 4th quadrant.Secondly,you see that range is from 0 to 360 and you are finding for 2x.So,
2x=(180-71.56) , (360-71.56) ,(540-71.56) , (720-71.56) as u know: 2nd quad is (180-x) and 4th is (360-x). Then u can see that x has values in 2nd revolution.So,2nd quad is (540-x) and 4th is (720-x)
Then u can easily find x by dividing all values of 2x by 2.Show all values to 2 dp.I hope i was helpful :/

 

[Talha Khatri]

Sin2x + 3cos2x=0
sin2x=-3cos2x
sin2x/cos2x= - 3
tan2x= - 3
now tan-1 (3) = 71.56....Since tanx is -ve , x lies in either 2nd or 4th quadrant.Secondly,you see that range is from 0 to 360 and you are finding for 2x.So,
2x=(180-71.56) , (360-71.56) ,(540-71.56) , (720-71.56) as u know: 2nd quad is (180-x) and 4th is (360-x). Then u can see that x has values in 2nd revolution.So,2nd quad is (540-x) and 4th is (720-x)
Then u can easily find x by dividing all values of 2x by 2.Show all values to 2 dp.I hope i was helpful :/

Yep! That's it although I have to say, this is not at all my trick (I ain't that clever ) All credit goes to my genius math teacher in school mashaAllah

 

[Thought blocker]

Sin2x + 3cos2x=0
sin2x=-3cos2x
sin2x/cos2x= - 3
tan2x= - 3
now tan-1 (3) = 71.56....Since tanx is -ve , x lies in either 2nd or 4th quadrant.Secondly,you see that range is from 0 to 360 and you are finding for 2x.So,
2x=(180-71.56) , (360-71.56) ,(540-71.56) , (720-71.56) as u know: 2nd quad is (180-x) and 4th is (360-x). Then u can see that x has values in 2nd revolution.So,2nd quad is (540-x) and 4th is (720-x)
Then u can easily find x by dividing all values of 2x by 2.Show all values to 2 dp.I hope i was helpful :/

Then, it does not matter at all if there is - sign or +ve sign ?

 

[Talha Khatri]

Then, it does not matter at all if there is - sign or +ve sign ?

No it kinda does but you always have to remove the negative sign and take the inverse and then look for the interval in which the function gives negative values so it matters in that way but you always take the inverse of a positive value with this method.

 

[Thought blocker]

No it kinda does but you always have to remove the negative sign and take the inverse and then look for the interval in which the function gives negative values so it matters in that way but you always take the inverse of a positive value with this method.

got it ty

 

[princeali97]

Then, it does not matter at all if there is - sign or +ve sign ?

It does!.Look if any trigonometrical function(for e.g tan) has a negative value for e.g tanx=-ve ,it means that x lies in a quadrant where tanx is negative (in this case 2nd and 4th quadrant).So,what i do is tan-1 the value without the -ve sign and then find x.

 

[Thought blocker]

It does!.Look if any trigonometrical function(for e.g tan) has a negative value for e.g tanx=-ve ,it means that x lies in a quadrant where tanx is negative (in this case 2nd and 4th quadrant).So,what i do is tan-1 the value without the -ve sign and then find x.

ty

 

[princeali97]

last question last part.I didnt understand.Pls explain if possible.http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_13.pdf

 

[Thought blocker]

last question last part.I didnt understand.Pls explain if possible.http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_13.pdf

Put y as zero, so ; 0 = x(x-2)^2 => 0 = x^3 - 4x^2 + 4x + 0, you will find that you get two roots by solving this, one root will be 2 and second will be zero, and from graph, clearly it is seen x is not zero so x is 2, hence a = 2