Mathematics: Post your doubts here!

[mrpaudel]

june 2011 paper 31 please help??

 

[mrpaudel]

another one from same year.. please help.

 

[XPFMember]

Assalamoalaikum!!

I need help with Q:9 (i) of http://www.xtremepapers.com/CIE/Interna ... 8_qp_3.pdf

A bit messy...but just used the product rule to differentiate.....and..equate to zero..and got the value

Thank you so much!!

I was going wrong somewhere...
I got it now!! Thanks a lot!

 

[mrpaudel]

here's the missin part

 

[mrpaudel]

Solve thz Identity:

( 1/sinx - 1/tanx )^2 = 1-cosx/1+cosx

 

[mrpaudel]

I hope i answered all of the queries...!! if not, sorry..i have my paper tomorrow..though m helping u...!! time to go for me now...bye bye guys..!!

 

[IgoforA]

mrpaudel.. i got 3√2 or 3 root 2.. where is another 1/2?? for p 31 complex part 2

 

[XPFMember]

Assalamoalaikum!

June:2010 P33 Q:6 (i) >>>I've found the eqn. given...but i dont know how to show that x lies between 3 and 4. Can someone please help!

JazakAllah Khair!

 

[kamina1]

Part.ii and part.iii

 

[mrpaudel]

mrpaudel.. i got 3√2 or 3 root 2.. where is another 1/2?? for p 31 complex part 2

I didnot get what u are saying

 

[IgoforA]

can you post your answer in an image form?? maybe by that, i'll understand.. tq

 

[mrpaudel]

Assalamoalaikum!

June:2010 P33 Q:6 (i) >>>I've found the eqn. given...but i dont know how to show that x lies between 3 and 4. Can someone please help!

JazakAllah Khair!

X+1/lnx -x ...put 3 and 4 in this one...and look at the sign....the sign will change means...!! once it was decreasing and...other time its increasing...thus u have some stationary pt...!!

 

[mrpaudel]

can you post your answer in an image form?? maybe by that, i'll understand.. tq

which one??? i didnt understand which question are u talking bout

 

[IgoforA]

i mean....
arg (z - u)= π/4
arg z - arg u= π/4
let, arg u = -π/2
arg z -(-π/2) = π/4
arg z = π/4 - π/2
arg z = -π/4

and i'm stuck until here..

 

[mrpaudel]

Part.ii and part.iii

m writing A for theta..!!

x=sinA
or, dx/dA=CosA
or, dx=CosA.dA-----------1)


now...substitute dx as CosA.dA.....and put SinA in x......then u will get the result.. put y=0 to find the limit...u will get x=0 and x=1...so put x=0 and 1..in X=sinA....to get A=0 and pi/2............

Sin2A=1-2sin^2A....so in case of Sin4A=SIn2.2A=1-2Sin^2 (2A) which means..Sin^2 (2A)=1-Sin4A/2

 

[mrpaudel]

i mean....
arg (z - u)= π/4
arg z - arg u= π/4
let, arg u = -π/2
arg z -(-π/2) = π/4
arg z = π/4 - π/2
arg z = -π/4

and i'm stuck until here..

well dude, m out somewhere..so no scanner here....!! so paper form not possible for me...!!

well if u ask me..i always do these types of question graphically...so i dont know how to do the way u are doing.....!! so...u better ask others to make clear bout that way....!! if u want my way....this is how i do...!!
make a graph for that argument..!! and least possible value means....distance perpendicular from (0,0) to that line...and..apply ur geometrical knowledge to solve it..!!

 

[IgoforA]

to mrpaudel, can you please edit the image...if i'm wrong.. is that the way to represents the complex points?? and solve it for me. tq :sorry:

 

[mrpaudel]

for least |Z| ....the line starting from (0,0) and..perpendicular to the red line will b the least one.....!! and..now since...green line makes π/4 with the vertical line..i.e Y axis..means..the line has the equation y=-x...and therefore.....the pt at which green line and red line interest is (1,-1) or (2,-2)...u will know after u see the fair sketch....!! then.....the distance between (0,0) and ( the pt ) can b calculated by distance formula..!!

 

[vanithavesna]

how to solve this question..my problem is i don't know how to draw a quadratic graph ( 4x^2-1 )
on a trigonometric scale.

 

[mrpaudel]

how to solve this question..my problem is i don't know how to draw a quadratic graph ( 4x^2-1 )
on a trigonometric scale.

For quadratic graphs...always find the vertex first....vertex can be calculated by the completed square form...or this formula

vertex =(-b/2a , (4ac-b^2)/4a ) ...where ax^2+bx+c=0 is the quadratic equation...

then....make a table....suppose the x coordinate of the vertex is 2.....put others x values in such a way that those values are right and left of 2...like...u can put 1 0 -1....and..3 4 5..this way....!!

and..for cotx....put the x as 0 pi/6 pi/4 pi/3 pi/2 ...and look at the values of y u get...and plot it...!!