Mathematics: Post your doubts here!

[GorgeousEyes]

Please jun012 no.3 v.13 .. jun012 no.7 v.12 .. jun012 v.11 9)iii . Pure Math As . thanks

 

[PhyZac]

When an equation of a plane and equation of a line is given, how do we determine whether the line intersects the plane or not? I know how to find the point of intersection, but if we're asked to tell whether they intersect or not?

When there is a line and plane there are two possibilities,

either, intersection or parallel

If parallel, then the normal and line have 0 scalar product! If not then intersecting!

 

[syed1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_13.pdf

Help please.

Q10)
iii) (did ii)
iv)
v)

iii .. from ii you get... 2(x-2)^2 + 6

meaning minimum point is at (2,6)

so x is 2 when y is minimum..

f(2)= 2(2^2) -8(2) + 14
f(x)≥6 (Since it's the minimum value)


iv.. the smallest value of A for which G has an inverse.. would be the x-cordinate of the minimum point.. A= 2

v)

g^-1(x)= 2(x-2)^2 + 6
y= 2(x-2)^2 + 6
y-6 = 2(x-2)^2
(y-6)/2 = (x-2)^2
√[(y-6)/2] = (x-2)
2+ √[(y-6)/2] = x

g^-1(x) = 2+ √[(x-6)/2]

 

[PANDA-]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_13.pdf

Help please.

Q10)
iii) (did ii)
iv)
v)

iii) Since you did ii, the vertex of the graph is (-b, c) as in the equation a(x+b)+k... and since a>0, then the graph is opening upwards, so y is >= (greater or equal) c.
iv) I'm gonna guess that this is the x value of the vertex, because at the vertex, a horizontal line would intersect it only at one point. The inverse of f is a function only when it is a one to one function.
v) Meh dunno really.

 

[Dudu]

iii .. from ii you get... 2(x-2)^2 + 6

meaning minimum point is at (2,6)

so x is 2 when y is minimum..

f(2)= 2(2^2) -8(2) + 14
f(x)≥6 (Since it's the minimum value)


iv.. the smallest value of A for which G has an inverse.. would be the x-cordinate of the minimum point.. A= 2

v)

g^-1(x)= 2(x-2)^2 + 6
y= 2(x-2)^2 + 6
y-6 = 2(x-2)^2
(y-6)/2 = (x-2)^2
√[(y-6)/2] = (x-2)
2+ √[(y-6)/2] = x

g^-1(x) = 2+ √[(x-6)/2]

Brilliant. Much Appreciated Mate

 

[Esme]

Are the graphs of sin inverse, cos inverse and tan inverse a part of the syllabus for Paper 3? Should we know how to make these graphs?

 

[PhyZac]

Are the graphs of sin inverse, cos inverse and tan inverse a part of the syllabus for Paper 3? Should we know how to make these graphs?

Yes.

 

[gary221]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s07_qp_3.pdf
q 7 the whole.
plz..

 

[PhyZac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_3.pdf
q 7 the whole.
plz..

check this
https://www.xtremepapers.com/commun...st-your-doubts-here.9599/page-269#post-430384

 

[Esme]

Yes.

Ok thanks. Do you have any notes or anything that'll help me learn to make those graphs?

 

[PhyZac]

Ok thanks. Do you have any notes or anything that'll help me learn to make those graphs?

http://www.examsolutions.net/maths-...gonometry/sec-cosec-cot/graphs/tutorial-1.php

 

[Esme]

http://www.examsolutions.net/maths-...gonometry/sec-cosec-cot/graphs/tutorial-1.php

Thanks a lot May Allah grant you success Ameen.

 

[sagar65265]

View attachment 24573

hence the final answer is 2.3 radians....
hope it helps!

It helps a load, that's an awesome diagram! Thank you so, so much!

Good Luck for all your exams!

 

[Soldier313]

It helps a load, that's an awesome diagram! Thanks you so, so much!

Good Luck for all your exams!

You're welcome anytime.....
All the best to you too......

 

[Maz]

Please jun012 no.3 v.13 .. jun012 no.7 v.12 .. jun012 v.11 9)iii . Pure Math As . thanks

falcon678
I'll try to answer the others later, InshaAllah

 

[falcon678]

thanks a lot man.....

 

[GorgeousEyes]

View attachment 24622
falcon678
I'll try to answer the others later, InshaAllah

I am very grateful , thanks

 

[ahmed abdulla]

When an equation of a plane and equation of a line is given, how do we determine whether the line intersects the plane or not? I know how to find the point of intersection, but if we're asked to tell whether they intersect or not?

by using the discriminant rule
ie) b^2-4ac
if b^2-4ac >0 (2 point intersection)
if b^2-4ac=0 (1 point intersection)
if b^2-4ac<0 (no intersection) thats for P1 and i dont now which paper ur talking about

 

[Asad]

Please help me in this question, here's a link to it's mark scheme: http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_ms_33.pdf

 

[Maz]

Please jun012 no.3 v.13 .. jun012 no.7 v.12 .. jun012 v.11 9)iii . Pure Math As . thanks

June-12/12
7) Using the formula given first find the sum of the first 1 term. (sounds funny)
Sn= 1^2 + 8(1)
= 9
Then find of sum of the first two terms
Sn= 2^2 + 8(2)
= 20
subtract the two terms to get the second term
20-9 = 11
and then 11-9 = 2 is the common difference

[i cant seem to be able to solve the second part]

June-12/13
9(iii) Since BC = AD [since its a square], the translation of D from A is the same that of C from B.
i.e. the y-step is 6 and x-step is 2
D= (-3+2, 2+6)
= (-1, 8)

|AD|= use the formula that's all. (x2-x1)^2 - (y2-y1)^2 The whole square root. [i dont know how to type that in]

Hope that helps.