Mathematics: Post your doubts here!

[qwertypoiu]

View attachment 59739 help in part iv

f(x) = x(x-2)^2
f'(x) = (x-2)(3x-2) [or equivalent]

You want to find the minimum value for this gradient function. Simple! Just find it as if it were any other function. (ie differentiate it, find when derivative of this is zero, etc)

So:
f''(x) = 6x - 8 [or equivalent??]
f''(X) = 0
x = 8/6 = 4/3

substitute into gradient function:

f'(4/3) = -4/3

 

[mohmed ahmed soliman]

helpView attachment 59741

qwertypoiu
thank may u help in this

 

[qwertypoiu]

helpView attachment 59741

A decreasing function is always decreasing, so its gradient is always negative. Therefore, you need to differentiate the function and prove that the expression obtained is always negative.

f(x) = x^-3 - x^3
f'(x) = -3/(x^4) - 3x^2

Now 1/x^4 and x^2 are two terms that are always positive no matter which value of x you substitute. Since you have negative coefficients for both of these, the gradient function will always output a negative value. Therefore, the function is a decreasing function.

It is interesting to note that f(-1.5) = -3.1, whereas f(0.5) = +7.9. It seems that it isn't decreasing here. I'll leave you to figure out why .

PS This is why the question specified x>0.

PPS Has something to do with continuity.

 

[Copy Cat]

 

[mohmed ahmed soliman]

HELP IN III

 

[mohmed ahmed soliman]

HELP IN
I PART B
AND II OF M/J/7 Q5

 

[mohmed ahmed soliman]

MAJ JUNE 2008
help in I

 

[mohmed ahmed soliman]

View attachment 59754

WHICH PAPER GIVE ADDRESS

 

[Copy Cat]

 

[darks]

Unable to do these. Please help. thanks.

 

[qwertypoiu]

View attachment 59754

We basically need to find the area of the shaded region first in terms of h, then get the volume by multiplying 40.

There are several ways of getting the area of shaded region. You can split it into two triangles like shown above. (I only showed one). Then:

tan30 = h/w
1/√3 = h/w
w = h*√3

So area of one triangle = 1/2 * w * h = 1/2 * h*√3*h = (√3/2)*h^2
Area of shaded region (which is two triangles) = 2 * Ans = √3 * h^2
Volume of thingy = area * length = √3 * h^2 * 40 = (40√3)*h^2

 

[qwertypoiu]

View attachment 59755 HELP IN III

There are 6 men and 4 women.

Combinations of selecting 3 men and 2 women = 6C3 * 4C2 = 120 ways.
Now let's say the particular man and woman are preselected and chosen to be in the group together:
5 men and 3 women left to be chosen from, and 2 men and 1 woman is to be selected:
5C2 * 3C1 = 30 ways

But they refused these combinations. So:

120 - 30 = 90 other ways of making this work.

 

[qwertypoiu]

View attachment 59757
HELP IN
I PART B
AND II OF M/J/7 Q5

REFRIGERATOR

If the R's are all together:
[RRRR]EFIGEATO

Basically you consider the [RRRR] to be one unit. So now you can permute the above:
Total number of (movable) elements = 9 (we include the [RRRR] unit)
Any repeating element? Yes, there are 2 E's. So:

Permutations = 9! / 2! = 181440

 

[qwertypoiu]

View attachment 59757
HELP IN
I PART B
AND II OF M/J/7 Q5

Sorry I forgot the second part.

They want you to select four but two have already been selected. (the two E's):
EE**
There is no R you can choose (and no E either, cuz they said TWO E's specifically)

So the letters left to you are the following six:

F, I, G, A, T, O

From these you'll choose two. Order doesn't matter here:

6C2 = 15.

 

[qwertypoiu]

View attachment 59759

If ABC is really a straight line, the points A, B, and C are said to be collinear.

If ABC is really a straight line, the vector AB and the vector BC must be in the same direction. (parallel). So:

AB = OB - OA = [2 3 1]
BC = OC - OB = [1 , p-5 , q+2]

If two vectors are parallel, a common factor can set them equal to one another:

AB = kBC
[2 3 1] = k*[1, p-5, q+2]

It is clear than k=2 here. (By equating the i component of the vector equation)
So now we can equate the other components (j and k):
3 = 2(p - 5)
p = 6.5
1 = 2(q + 2)
q = -1.5

 

[qwertypoiu]

View attachment 59760 View attachment 59761
Unable to do these. Please help. thanks.

The general coordinates of any point lying on the two lines may be expressed as:

If these two lines are to join, these coordinates need to become equal to each other at some value of λ and μ.
So:

1. 3 - λ = 4 + aμ
2. -2 + 2λ = 4 + bμ
3. 1+λ = 2 - μ

You'll notice the last equation doesn't contain a or b. It can be used to eliminate either λ or μ from the first two equation:
1+ λ = 2 - μ
λ= 1 - μ

So equation 1 becomes (by substituting λ = 1 - μ into it):
3 - 1 + μ = 4 + aμ
μ = 2 + aμ
(1-a)μ = 2
μ = 2/(1-a)

The reason we made μ the subject here will become clear later.

Equation 2 becomes (again after substituting λ = 1 - μ into it):
-2 + 2(1 - μ) = 4 + bμ
-2 + 2 - 2μ = 4 + bμ
(-b-2)μ = 4

Now we can substitute μ from above into here:

(-b-2)*( 2/(1-a) ) = 4
-2b - 4 = 4 - 4a
4a - 2b = 8
2a - b = 4

 

[qwertypoiu]

View attachment 59760 View attachment 59761
Unable to do these. Please help. thanks.

Part (ii):

If two lines are perpendicular, the dot product of their direction vectors = 0
So:
[-1 2 1] ⋅ [a b -1] = 0
-a + 2b -1 = 0
-a + 2b = 1

Solve this simultaneously with 2a- b = 4 and you'll get:
a = 3, b = 2

Now for the location of their intersection, you just have to substitute these values and do it the usual way:


Solve for λ and μ from any two of these three components (check for consistency by substituting into the third component), then obtain the coordinate by substituting either λ or μ into either of the line equations.
You'll get λ = 2, and μ = -1

So the coordinates will be (1,2,3)

 

[qwertypoiu]

View attachment 59760 View attachment 59761
Unable to do these. Please help. thanks.

Sorry forgot your last question:
For these types of question just apply log to both sides so that they become like linear equations:

5^(2x-1) = 2(3^x)

ln(5^(2x-1)) = ln(2(3^x))
(2x-1)*ln5 = ln2 + x*ln3

(2ln5)x - ln5 = ln2 + (ln3)x
(2ln5 - ln3)x = ln2 + ln5

Do the rest using calculator:
x = 1.085990045

 

[qwertypoiu]

View attachment 59758
MAJ JUNE 2008
help in I

6 pop, 3 jazz, 2 classical.
So we group jazz together:

PPPPPP[JJJ]CC

The pops and classics can move around but jazz has to stick together (but can move as a group).
So there are 9 elements all together (including Jazz block). (these can be arranged 9! ways)
Note that each of the J's are different from one another. (same goes for P's and C's)
So within the [JJJ] block, the three CD's can move around to form different arrangements. These can permute 3! ways.
So answer = 9! * 3! = 2,177,280

 

[Copy Cat]

If ABC is really a straight line, the points A, B, and C are said to be collinear.

If ABC is really a straight line, the vector AB and the vector BC must be in the same direction. (parallel). So:

AB = OB - OA = [2 3 1]
BC = OC - OB = [1 , p-5 , q+2]

If two vectors are parallel, a common factor can set them equal to one another:

AB = kBC
[2 3 1] = k*[1, p-5, q+2]

It is clear than k=2 here. (By equating the i component of the vector equation)
So now we can equate the other components (j and k):
3 = 2(p - 5)
p = 6.5
1 = 2(q + 2)
q = -1.5

Thanks....