Answers for Maths P32

[HarisShahzad]

might be out of order a little
1) x>-5/3 or x<-5
2) x=3.68
3) theta = 180 or 28.9 degrees
4) binomial
i) -2/(1+x) + (x+4)/[2+(x^2)]
ii) 5x/2 - 3(x^2) + 7(x^3)/4
5) iteration x=1.35
6) diff where x =ln(tant)
Answer:
i) dy/dx = 2sin^2 t cos^2 t
ii) x-2y = -1
7) i) k=2 then prove
Diff. equation
ii) draw y=2e^(x^2 -1)
8 ) i) 5a/a^2 +4 + i(-10/a^2 +4)
ii)a=-2
iii)dotted circle with center O and radius = 2 units
And perp. bisector of 2 + 2i
And shaded region is circle below the perp. bisector
9) vector question
i) angle = 79.7
ii) (i+3j) + m(-8i +7j+3k)
10) i) area = 2 - 17/e^3
ii) x coordinate of M = 2
iii) x coordinate of P = 1

Got this in a mass forwarded message. decided to share incase anyone hasnt got it yet

 

[HarisShahzad]

anyone knw the marks of binomial question without partial fraction

 

[rocker]

i think five are u sure about first answer?

 

[nour1234]

About the parametric equation, are you sure it's right?
I remember the question: x=ln(tan(t)) and y = sin^2(t)

dy/dx= dy/dt * dt/dx


y=(sin(t))^2 therefore dy/dt= 2sin(t).cos(t)

dx/dt= sec^2(t)/tan(t) therefore dt/dx = 1/dx/dt therefore dt/dx= tan(t)/sec^2(t)

therefore dy/dx= 2sin(t).cos(t) * sin(t).(1)/cos(t).cos^2(t) = 2tan^2(t)



I'm sure that is the right answer, if not can someone please tell me what's wrong with what I did? Any comment appreciated..

 

[nour1234]

Also I think the answer for the vector equation of line is wrong

 

[ahmed t]

well u changed tan(t)/sec^2(t) to sin(t)/cost cos^2t
thats wrong cos sec is 1 /cos
so it will be sint/cost(1/ cos^2t)
so it will be
sint *cos^2 t/cos t

 

[nour1234]

ooooh dammit, thanks for the reply.. So I'll probably lose one mark maximum right?

 

[ahmed t]

for the dy/dx yes probably
and another mark for the line,
who knows?

 

[leebux101]

does anyone rememeber how many marks that equation was for, the one where t was involved. :unknown: :%)

 

[billikidum]

4 marks i guess... and then 2 marks for finding the equation

 

[mhasanriaz]

The answers of Q8 r wrong....it was given that u* has argument 3/4pi.....which means u* wub be in the 2nd quadrant....n the coordinates wud be (-x,+y)....
therefore the answer of u wud be (-x,-y) in the 3rd quadrant i.e (-5a/a^2+4, -10/a^2+4)
answer for part (ii) wud be tan(pi-3/4pi)=10/a^2+4 upon 5a/a^2+4 [DRAW ARGAND DIAGRAM 2 CLEARIFY WHY I HAVE TAKEN pi-3/4pi]
tan 1/4pi=10/5a(a^2+4 wud be cancelled)
1=2/a
a=2 ANSWER!!

BETTER LUCK NEXT TIME!!

 

[ahmed t]

its -2
u cant change x to -x
wat u can do though is change 3/4 pi
to 1/4 pi and find the critical value of a which will be 2
then u can verify that it is in the second quad and therefore it will be -2

 

[mhasanriaz]

maybe people wud be confused why i hav ignored (-) sign in tan(pi-3/4pi)=10/a^2+4 upon 5a/a^2+4 [LENGTHS CAN NOT BE IN NEGATIVE]
Note : the argument coulb be negative in case when a variable lies in the 3rd or 4th quadrant....as u r going clock-wise frm 0 to -pi

 

[WellWIshER]

yo it was not -3/4 pi was it!!!????

i think it was 3/4 pi

 

[mhasanriaz]

no iav ignored negative sign in the lengths which i hav taken of y/x

 

[WellWIshER]

DA ANS is PURE 100% -2

 

[mhasanriaz]

its -2
u cant change x to -x
wat u can do though is change 3/4 pi
to 1/4 pi and find the critical value of a which will be 2
then u can verify that it is in the second quad and therefore it will be -2

man i hav explained frm the facts given in the question.....think abt it!

 

[ahmed t]

its -2
i cant be bothered arguing and can clearly see where u made ur mistake but im goin to sleep now so tell u tomz about dat

 

[ahmed t]

ok here the real part of the complex was
5a/a^2+4
now if a was 2 it would be positive there for not in the secon quad
if it was -2 then it would be negative so in the second quad

 

[mhasanriaz]

anyways gud luck to all including me 8)