Mathematics: Post your doubts here!

[soumayya]

Cannot understand the 4 step clearly di,

it should be '4x+ln3' why is it '4x-ln3'

Thanks a lot again!

sorry...c is -1/4 ln 3
Hence it's (x - 1/4 ln 3)..the rest should B ok...

 

[littlecloud11]

Winter 5
Q 5, Q 6, Q10
Please help!!

Thanks a lot!

6i) x = sin^2 θ (Differentiate the right hand side with respect to θ)
dx = 2 Sin θ * Cos θ dθ
Now substitute x with Sin^2θ and dx with dθ in the left hand side of the eq. in the question and you get
∫√( Sin^2θ/ 1-Sin^2θ) * 2 Sin θ * Cos θ dθ
∫√(Sin^2θ/ Cos^2θ )* 2Sinθ * Cos θ dθ [1-Sin^2θ = Cos^2θ]
∫(Sinθ/Cosθ)* 2Sinθ * Cos θ dθ
Cancel out Cosθ and you get ∫2Sin^2θ dθ

ii) Cos2θ = 1- 2Sin^2θ
so re-write 2Sin^2θ as 1 -Cos2θ
You also have to change the limits. x = sin^2 θ so when x= 1/4, θ= π/6. When x= 0, θ =0
So, ∫2Sin^2θ dθ = ∫1- Cos2θ dθ
= [ θ - Sin2θ/2 ] upper limit π/6 and lower limit 0
Substitute the theta values and you get [{π/6 -Sin(π/3)/2 } - {0- Sin0}]
So the answer is π/6 - √3/4

 

[soumayya]

Winter 5
Q 5, Q 6, Q10
Please help!!

Thanks a lot!

 

[tanmaydube]

Thanks a lot soumayya, Minato112, littlecloud11 !!!!

 

[tanmaydube]

Minato112 bhai you will have to do the vector question (q 10) please!!!!!

 

[Nagaanusan]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_31.pdf
someone please help me with no 8
thanks in advance

i want 2012 november pls

 

[Nagaanusan]

you're welcome

can u pls give me 2012 nov cie as and alevel maths p1 p4 s1 s2 pls

 

[Dug]

can u pls give me 2012 nov cie as and alevel maths p1 p4 s1 s2 pls

Here

 

[josephsai]

http://www.mediafire.com/?jegev3920g9rnhf and http://www.mediafire.com/?f437z2633wnygl4

 

[ni2005]

assalamualaikum! i have a question, it's a very old one (june 89), please help me!
The equation y= ax^2 - 2bx + c , where a,b and c are constants, with a>0.
a) Find, in terms of a,b and c the coordinates of the turning point on the curve.
b) Given that the turning point of the curve lies on the line y=x, find an expression for c in terms of a and b. Show that, in this case, whatever the value of b, c> -(1/4a).
it's the "show that" that i'm getting difficulty
jazakallah!

 

[Dug]

assalamualaikum! i have a question, it's a very old one (june 89), please help me!
The equation y= ax^2 - 2bx + c , where a,b and c are constants, with a>0.
a) Find, in terms of a,b and c the coordinates of the turning point on the curve.
b) Given that the turning point of the curve lies on the line y=x, find an expression for c in terms of a and b. Show that, in this case, whatever the value of b, c> -(1/4a).
it's the "show that" that i'm getting difficulty
jazakallah!

Walaikum AsSalam Warahmatullahi Wabarakatohu

a) dy/dx = 2ax - 2b
2ax - 2b = 0
ax - b = 0
x = b/a
Put this in eq and get the y-coordinate:
y = a(b/a)^2 - 2b(b/a) + c
y = b^2/a - 2b^2/a + c
y = -b^2/a + c

b) Vertex lies on the line y = x, so we can form an equation using the coordinates from (a)
b/a = -b^2/a + c
c = (b^2 + b)/a
Now you have to treat this equation as an entirely different function. The methods available are completing square or finding discriminant.

Completing square:
c = 1/a (b^2 + b + (1/2)^2 - (1/2)^2)
c = [1/a (b + 1/2)^2] - 1/4a

The y-coordinate of the vertex is 1/4a. We know that the graph is a U-shaped parabola so the range is c >= -1/4a.

Using discriminant:
b^2 + b - ac = 0
D = 1 - 4(1)(-ac)

For real roots, D>=0
1 + 4ac >=0
c >= -1/4a

My answer includes the the equality symbol but i am sure you made a typo there.

 

[ni2005]

Walaikum AsSalam Warahmatullahi Wabarakatohu

a) dy/dx = 2ax - 2b
2ax - 2b = 0
ax - b = 0
x = b/a
Put this in eq and get the y-coordinate:
y = a(b/a)^2 - 2b(b/a) + c
y = b^2/a - 2b^2/a + c
y = -b^2/a + c

b) Vertex lies on the line y = x, so we can form an equation using the coordinates from (a)
b/a = -b^2/a + c
c = (b^2 + b)/a
Now you have to treat this equation as an entirely different function. The methods available are completing square or finding discriminant.

Completing square:
c = 1/a (b^2 + b + (1/2)^2 - (1/2)^2)
c = [1/a (b + 1/2)^2] - 1/4a

The y-coordinate of the vertex is 1/4a. We know that the graph is a U-shaped parabola so the range is c >= -1/4a.

Using discriminant:
b^2 + b - ac = 0
D = 1 - 4(1)(-ac)

For real roots, D>=0
1 + 4ac >=0
c >= -1/4a

My answer includes the the equality symbol but i am sure you made a typo there.

jazakallahu khairan!!!!
but could u please elaborate a little more...? i didnt really understand how u did the discriminant part.

 

[snowbrood]

two decorations are suspended between the wall and ceiling using light strings AP PQ and QC the strings AP QP and CQare respectively at 30 60 and theta to the vertical the decorations hanging in equilibrium from p and q have weight 4N and 3N respectively.
(a)by considering the equilibrium of forces at p show that the tension in the string pq is 2N
(b) calculate the tension in the string CQ and angle theta.
I HAVE SOLVED A USING LAMI'S THEOREM BUT I CANT SOLVE PART B
ANSWERS FOR PART B 4.36N ,23.4 DEGREES

 

[snowbrood]

assalamualaikum! i have a question, it's a very old one (june 89), please help me!
The equation y= ax^2 - 2bx + c , where a,b and c are constants, with a>0.
a) Find, in terms of a,b and c the coordinates of the turning point on the curve.
b) Given that the turning point of the curve lies on the line y=x, find an expression for c in terms of a and b. Show that, in this case, whatever the value of b, c> -(1/4a).
it's the "show that" that i'm getting difficulty
jazakallah!

IS THIS IN AS? OR IS IT A2

 

[dhahir23]

I need help with Q 10(v) 9709_w06_qp_1

 

[ni2005]

IS THIS IN AS? OR IS IT A2

it is maths A-level

 

[littlecloud11]

I need help with Q 10(v) 9709_w06_qp_1

You solve this like any other quadratic equation.
g(x) = 10
x - 3√x = 10
x- 3√x - 10 =0
√x = -b ±√ (b^2 -4ac)/ 2a
√x = 3 ±√ (9 +10*4) / 2
√x = 3 ± 7 / 2
√x = -2 or 5
As x is greater than or equal to 0
x= 5^2 =25 (Ans)

Hope this helps!

 

[Dug]

jazakallahu khairan!!!!
but could u please elaborate a little more...? i didnt really understand how u did the discriminant part.

Wa iyyakum!!

Following the question's statement, i got the coordinates in terms of a, b and c.
x = b/a and y = -b^2/a + c
We equated them because they lie on the line y = x and we obtained a quadratic in 'b' : b^2 + b - ac = 0
The discriminant of this quadratic cannot be less than 0. Why? Because, if it was less than 0, it would mean that 'b' has imaginary roots. That is not possible since we are already told in the question that a POI does exist. Introducing an imaginary number into the equation will negate the statement.

Consider a curve y = ax^2 + bx + c and the line y = x. If we're told that the line touches the curve at exactly one point, then we know that after setting them equal to one another, D = 0. This is exactly what i did with my discriminant. Hope you got it. But if you're still confused, do let me know.

 

[tanmaydube]

Help please!
Summer 2008 Question 3 4 6 8
little help would also count...

Thank You

 

[~`Heba`~ :)]

Ok, you have to get that A keeps moving up for a certain length of time even after B hits the ground owing to it's velocity at the time B comes to rest (Newton's first law). This is when the string gets slack. A then drops down from the max. height for the same length of time until the string gets taught again and stops it's movement.
So, first you have to find the velocity with which B hit the ground. Use the eq. v= u +at, initial velocity of B was 0 as it was at rest, the acceleration is the same as you calculated in part i and the time is 1.6 sec. The velocity of A is the same as B at the instant B hit the ground.
Now, you have to calculate the time it takes for A to come to rest. Use the same eq as before but take acceleration as -9.8 because gravity is the only force acting on it now. Final velocity is 0 and initial is the one you calculated. Then double this time to find your answer.

Hope this Helps!

thanks a lot!