Mathematics: Post your doubts here!

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PhyZac

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Esme said:
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?
Click to expand...
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

syed1995

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Dudu said:
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)
Click to expand...

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-

PANDA-

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Dudu said:
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)
Click to expand...

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

Dudu

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syed1995 said:
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]
Click to expand...

Brilliant. Much Appreciated Mate :)
 
Esme

Esme

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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?
 
ahmed abdulla

ahmed abdulla

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Esme said:
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?
Click to expand...
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
 
Maz

Maz

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