Mathematics: Post your doubts here!

[mimiswift]

The contents of this moduli function is -4x+12. Depending on whether -4x+12 is positive or negative, we will proceed to redefine things sans
the modulus brackets.

Therefore, it simply means |-4x+12| =

(i) -4x+12 for -4x+12 ≥0 ====>x ≤3 (simply remove the modulus sign since the contents are positive)

(ii) -(-4x+12)= 4x-12 for -4x+12 <0 =====>x >3 (append a negative sign to ensure the contents are now positive)

(shown)

Essentially y=|-4x+12| is interpreted as two separate functions, one for x ≤3 and another for x>3.

Hope this helps. Peace.

how do we know it's <= and >and not >= and <.
I kind of got the idea how to use modulus. a positive and a negative function comes out of it, right?

 

[KJS98097090]

I need URGENT help, have a test tommorow
Can someone help me out on the loci of complex no?

 

[anonymous123]

I need URGENT help, have a test tommorow
Can someone help me out on the loci of complex no?

yes...ask asap..

 

[KJS98097090]

yes...ask asap..

Can u help me on the theory of loci of complex no?

 

[anonymous123]

Can u help me on the theory of loci of complex no?

yes dude....just ask!!

 

[KJS98097090]

yes dude....just ask!!

arg(z-2)=pi/3
how will we do this?

 

[ASstudent]

How to show that .... d/dx(ln cosec x) = -cot x .. Thank you !!
Please show in steps ....

 

[anonymous123]

arg(z-2)=pi/3
how will we do this?

arg(z-2)=pi/3
how will we do this?

either u posted half the question or this is not CIE stuff...u will need a protractor to draw the angle

 

[whitecorp]

How to show that .... d/dx(ln cosec x) = -cot x .. Thank you !!
Please show in steps ....

Generally speaking, d/dx { ln[f(x)] } = f ' (x)/ f(x)

d/dx(ln cosec x) = (-cosec x cot x)/ cosec x = -cot x (shown)

(Note that in this instance f(x) = cosec x and f '(x) =- cosec x cot x )

Hope this helps. Peace.

 

[whitecorp]

how do we know it's <= and >and not >= and <.
I kind of got the idea how to use modulus. a positive and a negative function comes out of it, right?

It is fine if you choose the alternate set of inequalities, because at the point for x=3, both -4x+12 and 4x-12 will give a y value of zero. The modulus function is piecewise continuous.

Hope this clarifies. Peace.

 

[zain786]

i need help in this questions
VECTORS A2

Q)
FIND THE DISTANCE OF THE POINT (1,1,4) FROM THE LINE (r= i-2j+k+ t(-2j+j+2k)
its easy but i havnt done these types so plx kindly explain it thanks

 

[Alice123]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w10_qp_61.pdf
please explain no 2

 

[Iffat]

Can aum1plz help me wif d foll question:
Given that h:x→x^2+2, stats the domain and range of h (x)

 

[Dug]

Can aum1plz help me wif d foll question:
Given that h:x→x^2+2, stats the domain and range of h (x)

Domain: All Real Numbers.
Range: y≥2

 

[Iffat]

Domain: All Real Numbers.
Range: y≥2

thanx but how do u kno dat domain is all real numbers and what are real numbers

 

[Silver Wing]

thanx but how do u kno dat domain is all real numbers and what are real numbers

The real numbers are all the numbers including rational and irrational ones, however they do not include imaginary numbers. These are square root of negative numbers, they include the symbol "i" equivalent to square root of negative one, they do not exist as cartseian coordinates and a graph cannot contain them, they are only plotted as points on an argand diagram, between if you are not an A2 student, then you do not have to know about imaginary numbers. In addition, the domain is all real numbers (negative infinity < x < positive infinity) because if you substitute any value of x, then the graph would give a valid value for y.

Hope I have helped you

 

[Dug]

thanx but how do u kno dat domain is all real numbers and what are real numbers

Because there is no real value of 'x' for which the function cannot produce a corresponding real value of 'y'.
The set of real numbers is a union of all rational and irrational numbers.

If the function was something like f(x) = x^2/(x+1), then the domain would've been as follows:

x + 1 = 0
x = -1

Domain: All real numbers except -1. f(-1) is undefined since division by zero is attempted. You can observe this phenomenon in tan curves too.

 

[Iffat]

The real numbers are all the numbers including rational and irrational ones, however they do not include imaginary numbers. These are square root of negative numbers, they include the symbol "i" equivalent to square root of negative one, they do not exist as cartseian coordinates and a graph cannot contain them, they are only plotted as points on an argand diagram, between if you are not an A2 student, then you do not have to know about imaginary numbers. In addition, the domain is all real numbers (negative infinity < x < positive infinity) because if you substitute any value of x, then the graph would give a valid value for y.

Hope I have helped you

thanx

 

[Iffat]

Because there is no real value of 'x' for which the function cannot produce a corresponding real value of 'y'.
The set of real numbers is a union of all rational and irrational numbers.

If the function was something like f(x) = x^2/(x+1), then the domain would've been as follows:

x + 1 = 0
x = -1

Domain: All real numbers except -1. f(-1) is undefined since division by zero is attempted. You can observe this phenomenon in tan curves too.

thanx

 

[7devilsallaround]

taking sub u=2x+3:
du/dx = 2
dx=du/2

x=(u-3)/2

insert values in original eq:
=> ⌡{u-3)/2}/u du/2
take 1/4 outside: 1/4 ⌡(u-3)/u du
simplify: 1/4 ⌡1 - (3/u) du
integrate: 1/4 [u - 3lnu] + k
replace u by x: 1/4[(2x+3) - 3ln|2x+3|] +k
simplify: x/2 + 3/4 -3/4 ln|2x+3| + k

mine includes one more term too..are u sure that answer is correct??

That's exactly how I did it!
Guess its a misprint or something. The books can be wrong sometimes. //shrugs
Thanks!