Mathematics: Post your doubts here!

[Aahliya]

hey anyone who can solve this second part(ii) ... ??

M/J 2010 (Paper 22)
7 (i) By sketching a suitable pair of graphs, show that the equation
e2x = 2 − x
has only one root. [2]
(ii) Verify by calculation that this root lies between x = 0 and x = 0.5. [2]

 

[reina81]

MJ08, P3, Q 5 and Q6, Q8
Yes it's a lot i know
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_qp_3.pdf

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_ms_3.pdf

 

[Nikesh]

hey anyone who can solve this second part(ii) ... ??

M/J 2010 (Paper 22)
7 (i) By sketching a suitable pair of graphs, show that the equation
e2x = 2 − x
has only one root. [2]
(ii) Verify by calculation that this root lies between x = 0 and x = 0.5. [2]

for ii) u are given two values of x
keep them in the given equation (i.e. y = e^2x +x -2 ) and check the result i.e. value of y
if you get results with different signs i.e. positive and negative numbers or lets say if your two values are 1) > 0, 2) < 0 then the root lies between given two values of x. It is because for any values of x if it is the root of the equation final result is always 0.
hope u understood it

 

[Aahliya]

for ii) u are given two values of x
keep them in the given equation (i.e. y = e^2x +x -2 ) and check the result i.e. value of y
if you get results with different signs i.e. positive and negative numbers or lets say if your two values are 1) > 0, 2) < 0 then the root lies between given two values of x. It is because for any values of x if it is the root of the equation final result is always 0.
hope u understood it

Yup i did understand.. nyz thanks Nikesh ..

 

[Hammad Siddiqi]

Hey, plz solve june 09 p3 Q9 (ii)

 

[Wanzi21]

salam
how do you do the question 4 of nov/2007?
i need ur help..
thanks ^^

http://www.xtremepapers.com/CIE/ind...el/9709 - Mathematics/&file=9709_w07_ms_1.pdf

 

[2pac]

Hey, plz solve june 09 p3 Q9 (ii)View attachment 7957

Lol,I had asked help for the same question but sadly haven't gotten any response.If anyone does reply to your question please do forward me the answer as well.
Thanks

 

[2pac]

hey anyone who can solve this second part(ii) ... ??

M/J 2010 (Paper 22)
7 (i) By sketching a suitable pair of graphs, show that the equation
e2x = 2 − x
has only one root. [2]
(ii) Verify by calculation that this root lies between x = 0 and x = 0.5. [2]

For the second part bring all the values to the left side.
so it'll be e2x+x-2=0.
Now substitute the 0 in place of x,answer will come to -1.
Now substitute 1 in place of x,answer will come to some positive value i assume.
Point to remember is that whenever the sign of two corresponding x value changes,the iterations is said to be converging.
In this case and most normal cases,the sign has changed from a negative to a positive value and hence it's converging.
write this or anything equivalent and you will get the 2 marks allocated.

 

[TSZ]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s10_qp_12.pdf
Hey can u plzz tell me how to do Q.11 (v)
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w10_qp_12.pdf
q.7 part (iii)

 

[anonymous123]

you dont believe me ??
there you go...Proof!
View attachment 7943

c here means the vertex ryt?

 

[leadingguy]

MJ08, P3, Q 5 and Q6, Q8
Yes it's a lot i know
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_qp_3.pdf

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_ms_3.pdf

qstn 5
z = 2cosθ + i(1-2sinθ)
we have to find modulas of |z-i| fr that frst we need to find z-i so

z = 2cosθ + i(1-2sinθ)
i = 0 + i now z-i = z = 2cosθ + i(1-2sinθ) - (0+i) now subtract real part by real and imaginary by imaginary

i.e z-i = 2cosθ - 0 +(1-2sinθ)i - i

= 2cosθ + i (1-2sinθ -1) w have taken i as comman here for the subtraction of imaginary part

= 2cosθ + i(-2sinθ) thsi is the espression fr z-i now take the modulas by the formula (a^2 + b^2)^1/2

((2cosθ)^2+ (-2sinθ)^2)^1/2
= (4cosθ^2 + 4sinθ^2)^1/2
= (4(cosθ^2 + sinθ^2))^1/2
= (4(1))^1/2
= 2 Ans proven

 

[Nikesh]

MJ08, P3, Q 5 and Q6, Q8
Yes it's a lot i know
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_qp_3.pdf

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_ms_3.pdf

Q5 needs more time to think so i wait for other's response
Q6. Given that
-->xy(x + y) = 2a^3
--> use derivative on both sides
--> d [xy(x + y)]/dx = d 2a^3/dx
--> use product rule
--> (x + y) dxy/dx + (xy) d(x +y)/dx = 0 (it is because 2a^3 is a constant so its derivative is 0)...........(3)
--> (x +y) [ x dy/dx + y dx/dx ] + xy [ dx/dx + dy/dx] = 0
--> (x + y) (x* dy/dx + y*1) + xy (1 + dy/dx ) = 0
--> (x + y) (x * 0 + y) + xy ( 1 + 0) = 0 (Since question says tangent is parallel to x-axis, slope is 0)
--> y(x+y) + xy = 0
--> xy + y2 + xy = 0
-->2xy + y2 = 0
--> y(2x + y) = 0
now take the second equation i.e.
2x + y = 0, i.e. Since this equation has a single solution curve has a single point for the above condition.
To find co-ordinates of point solve this equation with the given one you will get values of x and y in terms of a which is your final answer.
Note for colored steps: Use again product rule for dxy/dx
Open bracket of d(x + y)/dx as dx/dx + dy/dx


Q8. you r given that Area of triangle PTN = tanx for 0<x<1/2pie
i.e. 1/2 * PN * TN = tanx
--> 1/2 * TN * y = tanx (PN = y)
--->TN = 2tanx/ y .......(i)
again,
given that
dy/dx = PN/TN
--> dy/dx = y / (2tanx/y)
i.e. dy/dx = 1/2 y^2 cotx......proved
for 2nd part
dy/dx = 1/2 y^2cotx
--> 2/y^2 dy = cotx dx (arrange the equation in such a way that like terms get together)
-->Integrate the above equation you will get your answer

 

[Aahliya]

For the second part bring all the values to the left side.
so it'll be e2x+x-2=0.
Now substitute the 0 in place of x,answer will come to -1.
Now substitute 1 in place of x,answer will come to some positive value i assume.
Point to remember is that whenever the sign of two corresponding x value changes,the iterations is said to be converging.
In this case and most normal cases,the sign has changed from a negative to a positive value and hence it's converging.
write this or anything equivalent and you will get the 2 marks allocated.

Thanks for ur kind help..

 

[leadingguy]

MJ08, P3, Q 5 and Q6, Q8
Yes it's a lot i know
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_qp_3.pdf

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_ms_3.pdf

fr part 2
we have arrived to the expresion of z-i in the frst part which is ............ 2cosθ - 2sinθi

now we jxt need to add 2 in this expression i.e 2cosθ -2isinθ +2
now our expression fr z-i +2 is 2cosθ +2 -2sinθi

now fr solvving 1/(z-i +2 )
substitute the eq. obtained obove ..... 1/( 2cosθ +2 -2sinθi )
its in a fraction form and to obtain it in the form of x + yi we need to multiply the numenator and denominator with the conjugate of of the denominator

1* (2cosθ +2 +2sinθi ) /( 2cosθ +2 -2sinθi )*(2cosθ +2 +2sinθi )

now solve it and u wil get the real part

now put value of theta in the real part the ans each time wil be same

i didnot show working jxt told u the seps as it become too complicated to write here


hope u gt what to do??

 

[smzimran]

Hey, plz solve june 09 p3 Q9 (ii)View attachment 7957

AoA,
(i)
You can form one equation by substituting the coordinate of a point on the line (4 , 2 , -1) in the equation of plane, you will get one equation in terms of 'b' and 'c'. Name it eq(1)
Secondly, we know that line l lies in the plane so its direction vector must be perpendicular to the direction vector 'n' of plane
There dot product should be zero
so n . d = 0
From this equation you will one more equation in terms of 'b' and 'c'
Name this one eq(2)
Solve eq(1) and eq(2) simultaneously to get 'b' and 'c'

(ii)
Name Q as the point where P meets line 'l' PERPENDICULARLY.
Now form a set of coordinates of Q using equation of 'l' because Q lies on 'l'
OQ = (4+2t , 2-t , -1-2t)
OP is given
Use the distance formula to prove length of PQ .

 

[leadingguy]

Hey, plz solve june 09 p3 Q9 (ii)View attachment 7957

for part two just suppose that a point q is on the line l so find the coordinates of that point
i.e x = 4 +2t
y = 2=t
z = -1-2t these are the coordinates of any point online l (obtained throgh the eq. of line)

now considering this point Q , make an eq. of line PQ. u already know the coordinate of p and coordinates of Q are now here in terms of "t"

the eq. of line PQ wil be r= ( 0 2 4 ) +s(4+2t -t -5 -2t)

noe take the direction vector of this line and the drection vector of the line l obtain vector product of them equating it to zero as they are perpendicular! (stated in question)

U wil have the value of t = -2

substitute the value in the cordinates of q obtained at the begining ( u wil have exact coordinates of Q )

subtract p - q and then calculate the magnitude

 

[Esme]

c here means the vertex ryt?

yep !

 

[Esme]

ummmm..... ok lemme try again

fx is (x-2)^2+3. and gx is (x-2). so in wat function will u subsitute (x-2) in order to get(x-2)^2+3
? that will be x^2+3 which is hx. now subsitute (x-2) in the funstion x^2+3 and you will get (x-2)^2+3. therefore your fx=h(gx).
sorry its a little confusing.

ChantooPantoo....... i hope you got this as well.

 

[anonymous123]

yes i did thx alot!!

 

[allysaleemally]

Could some explain me how to do question 9 part ii? i dont understand why do we first have find the volume of the cylinder (as given in the marksheet)?


http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s10_qp_13.pdf

Marksheet: http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s10_ms_13.pdf

Thanks alot!