Mathematics: Post your doubts here!

[F.Z.M. 7]

What part of it do you not understand? Its simple
Get an admath book, it covers this chapter in detail

 

[Awesome12]

Can someone help with complete Q11,and the 2 circled parts on the other page?
F.Z.M. 7
Awesome12
MarcoReus
M.Omar

Sorry dude I haven't covered graphs yet in school. However, do mention anything regarding the 'Binomial Theorem' and the beginning of 'Logs'

 

[asadalam]

What part of it do you not understand? Its simple
Get an admath book, it covers this chapter in detail

It would be easier if you could solve this on a copy and write the explanation on the side,would help me understand it.

Sorry dude I haven't covered graphs yet in school. However, do mention anything regarding the 'Binomial Theorem' and the beginning of 'Logs'

I dont know anything about that,havent studied it yet.

 

[MarcoReus]

A certain someone asked a certain question above.

 

[F.Z.M. 7]

Which L.Bostock book covers Binomial Expansion?

Or should I refer to the O Level Add Maths book?

olevel one is basic, get the red one I thimk aur kia F math kee lainee hai ?

 

[MarcoReus]

olevel one is basic, get the red one I thimk aur kia F math kee lainee hai ?

Wo laal wali le li hai (Core Mathematics). Couldn't find it there.

 

[F.Z.M. 7]

It would be easier if you could solve this on a copy and write the explanation on the side,would help me understand it.

I dont know anything about that,havent studied it yet.

it will bee to time consuming, tell me the concept you dont understand and I will explain
or type out your working and I will correct the mistake if there is any

 

[F.Z.M. 7]

Wo laal wali le li hai (Core Mathematics). Couldn't find it there.

Just do the questions sir gave, no need to get so hyper excited

 

[MarcoReus]

it will bee to time consuming, tell me the concept you dont understand and I will explain
or type out your working and I will correct the mistake if there is any

Haha - I understand every bit of it.

I just need to refer back to the book and formulas etc so that takes time. Previously - O Level D1/2/3/4 reading solved it. :3

 

[MarcoReus]

Just do the questions sir gave, no need to get so hyper excited

Tu krle bhai. 2 pages for one question.

 

[F.Z.M. 7]

Tu krle bhai. 2 pages for one question.

Aint nobody got time for that

 

[asadalam]

it will bee to time consuming, tell me the concept you dont understand and I will explain
or type out your working and I will correct the mistake if there is any

Yar apne bhai k liay itna sa bhi time nai nikal sktay?

 

[F.Z.M. 7]

Yar apne bhai k liay itna sa bhi time nai nikal sktay?

nikal bhee loon to I cant give the image, my mobile was stolen and now I have 2 megapixel one now
How can I show it to you ?

I can just give you explanation of concepts wo sun lo

 

[M.Omar]

Can someone help with complete Q11,and the 2 circled parts on the other page?
F.Z.M. 7
Awesome12
MarcoReus
M.Omar

q11(i)
for a quadratic graph of function ..we know tht is either in the shape of U of inverse U depending on the coefficient of x^2
the form of the question is :a(x-h)^2 +c ...c=8 a=-1 and h=2
a<0 so the curve will be of shape inverse U thus it will have a maximum value at the stationary point
we know tht since (x-h)^2 is always positive regardless of any value of x and with a negative coefficient(i.e -1) it will thus always be negative..so for the maximum value we need to make the expression a(x-h)^2 =0 so tht it will not decrease the value of y=8-(x-2)^2 and thus give us the maximum ..comparing with the values given in the question... x shud be 2 to givea -(x-2)^2 =0 and thus give a max value...for this value of x ...find the value of y coordinate which is quite evidently 8
the nature is tht this stationary point is a maximum turning effect

 

[F.Z.M. 7]

q11(i)
for a quadratic graph of function ..we know tht is either in the shape of U of inverse U depending on the coefficient of x^2
the form of the question is :a(x-h)^2 +c ...c=8 a=-1 and h=2
a<0 so the curve will be of shape inverse U thus it will have a maximum value at the stationary point
we know tht since (x-h)^2 is always positive regardless of any value of x and with a negative coefficient(i.e -1) it will thus always be negative..so for the maximum value we need to make the expression a(x-h)^2 =0 so tht it will not decrease the value of y=8-(x-2)^2 and thus give us the maximum ..comparing with the values given in the question... x shud be 2 to givea -(x-2)^2 =0 and thus give a max value...for this value of x ...find the value of y coordinate which is quite evidently 8
the nature is tht this stationary point is a maximum turning effect

Now I was halfway through typing myself -______-
Pahlay kidhr tha ?

 

[M.Omar]

Now I was halfway through typing myself -______-
Pahlay kidhr tha ?

haha...I thought this might happen

 

[M.Omar]

Can someone help with complete Q11,and the 2 circled parts on the other page?
F.Z.M. 7
Awesome12
MarcoReus
M.Omar

now for the other pg..since we,re asked to find the value of k for which the curve and line r tangent to each other
we first equate them y=6x+k and y=7x^1/2
this becomes 6x+k=7x^1/2
now this gets a bit trickier and u shd take the under root x as only one unit on either sides of equation(i.e alone) and now square the expression i.e (6x+k)^2=(7sqtrtx)^2...this gives
36x^2 +12kx +k^2=49x
u,ll end up with a quadratic equation... and now simply for tangent b^2-4ac=0
and u wil find probably two values of k or if the k^2 terms cancel each other u,ll find a single value
for the other part f(x)= (x-2)^2 -4+k .to find the inverse
let f(x)inverse=y
such tht fy)=x
(y-2)^2-4+k=x
(y-2)^2=x+4-k
y-2=+-(under root(x+4-k))
y=2+-(under root(x+4-k))

 

[asadalam]

q11(i)
for a quadratic graph of function ..we know tht is either in the shape of U of inverse U depending on the coefficient of x^2
the form of the question is :a(x-h)^2 +c ...c=8 a=-1 and h=2
a<0 so the curve will be of shape inverse U thus it will have a maximum value at the stationary point
we know tht since (x-h)^2 is always positive regardless of any value of x and with a negative coefficient(i.e -1) it will thus always be negative..so for the maximum value we need to make the expression a(x-h)^2 =0 so tht it will not decrease the value of y=8-(x-2)^2 and thus give us the maximum ..comparing with the values given in the question... x shud be 2 to givea -(x-2)^2 =0 and thus give a max value...for this value of x ...find the value of y coordinate which is quite evidently 8
the nature is tht this stationary point is a maximum turning effect

Well bro, i got most of that figured out myself, h and k are turning points and they're 2,8 so i got that right.My main problem was in the g(x) functions.Could you solve that out on a page and post?

nikal bhee loon to I cant give the image, my mobile was stolen and now I have 2 megapixel one now
How can I show it to you ?

I can just give you explanation of concepts wo sun lo

Allah khair,tha konsa mobile

 

[asadalam]

now for the other pg..since we,re asked to find the value of k for which the curve and line r tangent to each other
we first equate them y=6x+k and y=7x^1/2
this becomes 6x+k=7x^1/2
now this gets a bit trickier and u shd take the under root x as only one unit on either sides of equation(i.e alone) and now square the expression
u,ll end up with a quadratic equation... and now simply for tangent b^2-4ac=0
and u wil find probably two values of k or if the k^2 terms cancel each other u,ll find a single value
for the other part f(x)= (x-2)^2 -4+k .to find the inverse
let f(x)inverse=y
such tht fy)=x
(y-2)^2-4+k=x
(y-2)^2=x+4-k
y-2=+-(under root(x+4-k))
y=2+-(under root(x+4-k))

Could you please please please solve it on a paper?My eyes bleed trying to read out this =P
Thanks for this explanation though.

 

[MarcoReus]

Allah khair,tha konsa mobile

It would be be best that the mobile and the story be best left untold.