Mathematics: Post your doubts here!

[chishtyguy]

ok i got the roots -7/3 and -3 but how did u conclude that x is greater than -7/3 and less than -3

 

[yousef]

ok i got the roots -7/3 and -3 but how did u conclude that x is greater than -7/3 and less than -3

If you draw this on the number line //
you will find that the two roots go away from each other .
for eg when you get your answer as x>-2 and x<4
your answer will be -2>x>4 since when you draw the two roots above they coincide

 

[Obsidian Fl1ght]

Sup guys?
If anyone can solve this part, from RHS to the LHS, I'd find it epically awesome. Been trying to solve it (Right to Left) since ages and it's drivin' me mad!
So give it a shot. Every shot is appreciated.

 

[salvatore]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_31.pdf
Please help me with qn no. 9(i).. I have no idea of what to do. I'll be grateful for any help provided
Thanks

 

[yousef]

View attachment 36253

Sup guys?
If anyone can solve this part, from RHS to the LHS, I'd find it epically awesome. Been trying to solve it (Right to Left) since ages and it's drivin' me mad!
So give it a shot. Every shot is appreciated.

replace cotx with cosx/sinx ie ( 1/ tanx )
and cot2x with cos2x / sin2x

and use the identities sin2x =2sinxcosx
and cos2x =2cos^2 x - 1
to get the answer

 

[Obsidian Fl1ght]

Thanx but I got the answer this way.
I want to solve the identity from RHS to LHS, but I keep getting stuck.

 

[Obsidian Fl1ght]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_31.pdf
Please help me with qn no. 9(i).. I have no idea of what to do. I'll be grateful for any help provided
Thanks

Man you've chosen some question...


(1-x)/(1+x)

dy/dx = Use the quotient rule: u/v : dy/dx = (u’.v – v’.u)/v^2

So u get: [(-1)(1-x) – (1)(1+x)]/(1+x)^2

=[-1+x – (1+x)]/(1+x)^2

=[-1+x-1-x]/(1+x)^2

=(-2)/(1+x)^2

Thus you’ve got the dy/dx for (1-x)/(1=x)

Now the bad boy question: u = [(1-x)/(1+x)]^1/2

Use the formula: (x)^n, dy/dx = x’.n.(x)^(n-1) aka just differentiate it, taking the whole thing as one.

Like this:

dy/dx = [(-2)/(1+x)^2].(1/2).[(1-x)/(1+x)]^-1/2

Simplifying it a bit, [(-1).(1+x)^1/2]/[(1+x)^2.(1-x)^(1/2)]

Now this is the dy/dx for the graph, meaning for the normal to graph’s gradient, you gotta take the negative reciprocal.

You get: grad(normal) = [(1+x)^2.(1-x)^(1/2)]/(1+x)^(1/2)

Since we’ve got (1+x) both in the numerator+denominator, so simplify that. (Subtract powers etc)

You should now get, (1+x)^(2-0.5).(1-x)^(1/2) = (1+x)^(1.5).(1-x)^(1/2)

Look at (1+x)^(1.5) now. Power of 1.5 means power of ONE and HALF. Half aka 0.5 aka ½ (power) represents square root. So… (1+x)^(1.5) = (1+x).(1+x)(1/2)

Back to the gradient:

(1+x)^(1.5).(1-x)^(1/2) is now thus: (1+x).(1+x)^(1/2).(1-x)^(1/2)

Notice that the two terms with sq. root are (a+b) and (a-b). What’s the formula involving this?

Yeah, (a+b).(a-b) = (a^2 – b^2)

Thus, (1+x)^(1/2).(1-x)^(1/2) becomes: [(1-x^2)]^(1/2), taking the square roots common.

Finally… (1+x).(1+x)^(1/2).(1-x)^(1/2) is converted to the final answer aka (1+x).[(1-x^2)]^(1/2)

It all looks wickedly messed up here... I'd advise you to grab a pen and jot down the answer as you read it... like the fractions.

 

[ahmed abdulla]

Thanx but I got the answer this way.
I want to solve the identity from RHS to LHS, but I keep getting stuck.

 

[ahmed abdulla]

Trapezium rule ,
i continously get it as 1.15 but the answer is 0.98
any help?

 

[Obsidian Fl1ght]

Hey thanx for the effort bro, but I've no prob in solving it from LHS to RHS... I wanna solve it from RHS ie cosec2x to the LHS.

 

[Obsidian Fl1ght]

Trapezium rule ,
i continously get it as 1.15 but the answer is 0.98
any help?

So you gotta divide the graph into three (equal) intervals, from x = 0 to π/4.

Then just figure out the area for each interval (trapezium). A = 1/2.(a+b).h

The 'h' (aka heights) will be the interval on x axis. Each height will thus be π/12 yeah? (Coz we divide π/4 into three equal bits yadda yadda)


For values of ‘breadths’ aka parallel sides, put x into equation.

Here are the corresponding values of y I got:

For:

X= 0, y = 1

X = π/12, y = 1.069

X = π/6, y = (15)^0.5/3

X = π/4, y = (3)^0.5


Interval 1: x = 0 to π/12

A = (1/2).(π/12)(1+1.069)


Interval 2: π/12 to π/6

A = (1/2).(π/12)(1.069+(15)^0.5/3)


Interval 3: x = π/6 to π/4

A = (1/2).(π/12)((15)^0.5/3+(3)^0.5)


Add ‘em up… (0.2708+0.3089+0.3957 = 0.9754 rounded off to 0.98 (2dp)

 

[Obsidian Fl1ght]

4 (ii) If anyone could give a sketch of what's to be done...

 

[Obsidian Fl1ght]

5(iii). HOW come the answer is ONE???

 

[ahmed abdulla]

Hey thanx for the effort bro, but I've no prob in solving it from LHS to RHS... I wanna solve it from RHS ie cosec2x to the LHS.

i tried , but i didnt get it
Dont worry , doing in both ways will gain full marks .
and i dont think this complicated questions will be asked

 

[ahmed abdulla]

View attachment 36282

5(iii). HOW come the answer is ONE???

from part(i) ..Rsin(A+b)
we know that greatest ( maximum ) value of sin function is 1 which happens when teta is 90 degree

 

[Obsidian Fl1ght]

from part(i) ..Rsin(A+b)
we know that greatest ( maximum ) value of sin function is 1 which happens when teta is 90 degree

Ah got it now. Thanx loads!

And also, did you get the trapezium q. ?

 

[ahmed abdulla]

View attachment 36280
4 (ii) If anyone could give a sketch of what's to be done...

The gradient of this function can be found by the first derivative.
Find the vertex of this function by taking the second derivative, and setting it equal to zero.
Fill in this value ( x value of vertex ) into the first derivative to find the y-value of the vertex.
Take the third derivative to find out if that vertex is a minimum or a maximum.
The third derivative will be always positive which shows it will be never negative

i am not sure correct me if i am wrong

 

[Obsidian Fl1ght]

The gradient of this function can be found by the first derivative.
Find the vertex of this function by taking the second derivative, and setting it equal to zero.
Fill in this value ( x value of vertex ) into the first derivative to find the y-value of the vertex.
Take the third derivative to find out if that vertex is a minimum or a maximum.
The third derivative will be always positive which shows it will be never negative

i am not sure correct me if i am wrong

Hmm... k you've slightly confused me. I get what you mean but... doesn't setting the first derivative to zero give us the x-coordinate of vertex? And putting this value into the second der. show nature of the vertex? Uhh...
Anyway, after this - finding nature of vertex (Assuming the point is a minimum)- what do we do?

 

[snowbrood]

help bros answer is x is less than or equal to -1

 

[ahmed abdulla]

Hmm... k you've slightly confused me. I get what you mean but... doesn't setting the first derivative to zero give us the x-coordinate of vertex? And putting this value into the second der. show nature of the vertex? Uhh...
Anyway, after this - finding nature of vertex (Assuming the point is a minimum)- what do we do?

if this way confuses you ,
find dy/dx =0
you will get x values ,
plug this x values in the second derivative .. and they must be positive which shows its minimum

it you draw the graph , it will be parabola , and its least value ( vertex) will be ABOVE x -axis , which shows its positive and cant be negative