Mathematics: Post your doubts here!

[Obsidian Fl1ght]

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

Hey thanx.

Thanks , i completely got it ..

but i think there is a formula for trapezium rule which is much faster and saves time .. if you can do it that way

Hmm don't know abt it. Dyu have ny idea what this formula is?

 

[ahmed abdulla]

Hey thanx.


Hmm don't know abt it. Dyu have ny idea what this formula is?

how crazy i am , hhhh
this formula , just takes the 1/2* pi/12 out .. and the rest is the same

 

[Obsidian Fl1ght]

how crazy i am , hhhh
this formula , just takes the 1/2* pi/12 out .. and the rest is the same

Oh wow! Man that's some top secret formula!! Never would've thought of it... NEVER
No offense.

 

[ahmed abdulla]

Oh wow! Man that's some top secret formula!! Never would've thought of it... NEVER
No offense.

 

[Obsidian Fl1ght]

p(x) = (x-2)(2x^3+x+2)
What's the justification for (iii)?
Ans is x>2
ER says:


Someone explain the 'justification' aka justify it. Thx in advance

 

[Igcse stuff]

I hope it's readable.

Thanks a lot! Zodiac

 

[Igcse stuff]

can you guys help me with Q4 from o/n 2013 P3

 

[salvatore]

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.

Thanks a tonne bro.. I appreciate it.
Would you mind helping me with part ii of the same question? I'm getting 1.5 as the answer (The ms answer is 0.5).
Sorry for bothering man

 

[Obsidian Fl1ght]

Thanks a tonne bro.. I appreciate it.
Would you mind helping me with part ii of the same question? I'm getting 1.5 as the answer (The ms answer is 0.5).
Sorry for bothering man

Hey,, no worries. Glad 2 help.

So, 9(ii):
I guess you know what we gotta do in this - we treat the gradient of normal as a line. To find the stationary (max) point on it, we find the dy/dx of this 'line'. (Technically we're findin the d^2y/dx^2)

From (i), we know dy/dx is (1+x).[(1-x^2)]^(1/2)___________ [Here, (1+x) is u and v is [(1-x^2)]^(1/2)]
Apply (u.v)' = u'.v+u.v'
d^2y/dx^2 = (1).[(1-x^2)]^(1/2) + (1+x).(1/2)[(1-x^2)]^(-1/2)(-2x)
= [(1-x^2)]^(1/2) + (-x)(1+x)[(1-x^2)]^(-1/2)
Now equate this to zero:
[(1-x^2)]^(1/2) + (-x)(1+x)[(1-x^2)]^(-1/2) = 0
I then got: (1-x^2) - x(1+x) = 0 ___________ (Coz [(1-x^2)]^(1/2).[(1-x^2)]^(1/2) = (1-x^2) )
Then: 1 - x^2 - x - x^2 = 0
-2x^2 - x + 1 = 0 (multiply with -1 in both sides)
2x^2 + x - 1 = 0
Solving this you get two sol., x = 1/2 (ANS.) or x = -1
From graph, we see dat point P ain't on x = -1
Ans. is thus x = 1/2

 

[Tee Kay]

I like the doubt it made me think for a while so here's what i did
for 8) i found the sum of n=1,n=2 and n=3 i could keep going just to be more sure but 3 are enough
i found sum of 1st n terms is -1
sum of 2nd n terms is 4
sum of 3rd n terms is 15
as we know sum of 1st term is -1 that makes it obvious first term is -1
so we have -1,y,z
y and z are 2nd and the 3rd term respectively
as we know sum of 2nd n terms is 4 that means -1+y=4
and so for the 3rd -1+y+z=15
we will get y=5 and z=11
we have -1,5,11
now lets see if its AP or GP
if its GP then un=an^(n-1) lets try that out
5=-1*2^(2-1)
5=2 that is wrong And by now you will know its not GP
lets see if its AP then un=a+(n-1)d (d is the difference between the terms here ist 5-(-1)=6)
5=-1+(2-1)6
5=5 hence proved
you can try for the 3rd which will be
11=-1+(3-1)6
11=11
For the next question you asked you should do the same but if you want me to do it i will do it by the way the first term is 1 and the common difference is 4
I hope i helped and if i did you always free to ask more questions

Thank you and sorry for not replying earlier my net went down and got fixed today
anyway i actually did it the way you did and it's wrong the proof has to be by induction

 

[ZaqZainab]

Thank you and sorry for not replying earlier my net went down and got fixed today
anyway i actually did it the way you did and it's wrong the proof has to be by induction

But i got my answer right this way can i know what i did wrong and what is the right method

 

[Suchal Riaz]

can you guys help me with Q4 from o/n 2013 P3

Sorry for being late.

 

[David Hussey]

Sorry for being late. View attachment 36419

my goodness...

 

[Tee Kay]

But i got my answer right this way can i know what i did wrong and what is the right method

that solution only proves it for the first 3 terms or so, it is required to be proved for all terms, like:
Sn = [a+(n-1)d] + [a+(n-1)d] + [a+(n-1)d] ........ [a+(n-2)d] + [a+(n-1)d] = n(3n-4) and i don't know how to continue from here

 

[ZaqZainab]

that solution only proves it for the first 3 terms or so, it is required to be proved for all terms, like:
Sn = [a+(n-1)d] + [a+(n-1)d] + [a+(n-1)d] ........ [a+(n-2)d] + [a+(n-1)d] = n(3n-4) and i don't know how to continue from here

I believe CIE examiners whould never give this question but i will ask my teacher and let you know

 

[TheStallion-Reborn]

im sorry i cannot provide the link because the 2013 past papers aren't yet uploaded. The question staes, y=mx+14 is tangent to the curve y=12/x+2 at point p. what are the cordinates of point p?

 

[David Hussey]

im sorry i cannot provide the link because the 2013 past papers aren't yet uploaded. The question staes, y=mx+14 is tangent to the curve y=12/x+2 at point p. what are the cordinates of point p?

2013 papers... http://maxpapers.com/category/quali...l/cambridge/a-level/a-level-mathematics-9709/

 

[TheStallion-Reborn]

2013 papers... http://maxpapers.com/category/quali...l/cambridge/a-level/a-level-mathematics-9709/

thankyou! any clue of the question?

 

[Igcse stuff]

Thanks a lot for the detailed explantions!!! ishaallah

 

[David Hussey]

thankyou! any clue of the question?

can u give me the link to the question?