Mathematics: Post your doubts here!

[Rutzaba]

View attachment 42532

part i

i am getting
'1' instead of 3 .... -.-

a to rha hai three

 

[Rutzaba]

when u substitute u not only substitute x
but u aso find out a value for dx
and u also change the limits

 

[Rutzaba]

let u^2 =x
that means dx/du =2u that means differentiating u^2 woth respect to u
dx= 2u du
next thing is to change limits...
if u^2 =x
when upper limit of x=p^2 then replace x with p^2
u^2 = p^2
new upper limit =p
and since o will remain o
the new limits are p and o

 

[Rutzaba]

agey karun?

 

[daredevil]

let u^2 =x
that means dx/du =2u that means differentiating u^2 woth respect to u
dx= 2u du
next thing is to change limits...
if u^2 =x
when upper limit of x=p^2 then replace x with p^2
u^2 = p^2
new upper limit =p
and since o will remain o
the new limits are p and o

yeah done.... sorry it was the father of all silly mistakes that i did here... -.-

 

[Rutzaba]

no need to be srry

 

[daredevil]

I'm in such a bad place ryt now....
cant even get a simple partial fractions question ryt...

Why have they taken a constant here too? in the ms?

 

[A star]

cause its an improper fraction

 

[daredevil]

cause its an improper fraction

abhi samajh nahi aaye.... paanch daffa aur likho!!

 

[AmeeraKhan]

How do you form the differential equation from that long paragraph they usually give at the end of the P3 exam? Any advice or notes on that would be helpful.

 

[A star]

lol sorry my net is having a coffee break i guess

 

[A star]

How do you form the differential equation from that long paragraph they usually give at the end of the P3 exam? Any advice or notes on that would be helpful.

usually read the whole paragraph once or twice and concentrate on where ever rate of change or numbers apear. whenever they use the word proportional write it down , what ever written and voila it forms :3

 

[Rutzaba]

View attachment 42537
I'm in such a bad place ryt now....
cant even get a simple partial fractions question ryt...

Why have they taken a constant here too? in the ms?

chek the vector thread jub tak mai ye solve krti hun

 

[daredevil]

chek the vector thread jub tak mai ye solve krti hun

ohh i didnt even know there was a vector thread

 

[A*(a*)]

OK you know tat if 1+2i is a factor of this equation that means if this equation is divided by this factor the remainder wud be zero
place the (1+2i in place of z in the eq
(1+2i)^3 -a(1+2i) +10=0
(1+4i+4i²)(1+2i) -a-2ai+10=0
1+4i+4i² +2i+ 8i² +8i³-a-2ai+10=0
now u know that i²= -1
1+4i+4(-1)+2i+ 8(-1) +8(-1)(i)-a-2ai+10 Notice that 8i³ was turned into 8i²(i) the i² became -1 and the i remained.
-1-2i-2ai-a =0 reduced form.
-(1+2i)= 2ai+a
-(1+2i)= a(1+2i)
a=-1

If 1+2i is a root, that means substituting this into the equation will give you 0.

(1+2i)^3 - a(1+2i) +10

The best way to solve (1+2i)^3 is to square it first, then multiply by (1+2i).
(1+2i)^2 = (1)^2 + 4i + (2i)^2
i^2 = -1
=> 4i - 3
(4i-3)(1+2i) = 4i + 8(i)^2 - 3 - 6i
again, i^2 = -1
=> -11 - 2i

-11 - 2i - a - 2ai + 10 = 0
-2i - a - 2ai = 1
-2i - a(1 + 2i) = 1
-a = (1 + 2i)/(1+2i)
a = -1

Thankyou both of you but,
The value of a=1 and roots are 1+2i , 1-2i and -2.

 

[Rutzaba]

Thankyou both of you but,
The value of a=1 and roots are 1+2i , 1-2i and -2.

yar i really dunno complex numbers

 

[Rutzaba]

abhi samajh nahi aaye.... paanch daffa aur likho!!

"http://i1275.photobucket.com/albums/y444/Rutzaba/Picture2_zpsb769909d.png"
iske bad bache ga 2 +(-5x+5)/(x+1)(2x-3)

 

[Rutzaba]

for it to be proper the power of x in the numerator must be smaller than the power of x in the denominator
by the method abv caled long division we have converted into a form of proper fraction

 

[A*(a*)]

yar i really dunno complex numbers

Thankyou, your solution was still helpful

 

[Rutzaba]

Thankyou, your solution was still helpful

i am srry