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Mathematics: Post your doubts here!

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Walaikum AsSalam Warahmatullahi Wabarakatohu

i am assuming you already know how |z-3i|=<2 is a circle
Centre = 0,3
Radius = 2

27zj2ap.jpg


In exam you will have a compass and a graph paper so everything will be accurate..the inequality says the circle is less than the radius so you have to shade inside....i forgot to do so :p

To find the maximum arg..draw a tangent such that u create the largest angle as done in the picture....now u have a triangle in the 2nd quadrant and so u can find arg z easily
Jazaka Allah Khairan........Thank you so much !!! Really helpful, thanks for the graph bit !
May Allah reward you for your help Ameen, And In Shaa Allah your exams shall be easy and you score the highest grades in this life and hereafter Ameen!!!!!!!!!!!!!!!!
 
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Jazaka Allah Khairan........Thank you so much !!! Really helpful, thanks for the graph bit !
May Allah reward you for your help Ameen, And In Shaa Allah your exams shall be easy and you score the highest grades in this life and hereafter Ameen!!!!!!!!!!!!!!!!
wow thanks for the prayer!!!! really needed it :D May Allah give you success in this world and the Hereafter and make these exams easy for you...Ameen ;)
 
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ok. so my question is not from any papers or anything. But my teacher gave it to me. And I have no idea how to solve it.
write y=f(x)= modulus of -4x+12 as a piecewise function

how do I do it?:eek:
 
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ok. so my question is not from any papers or anything. But my teacher gave it to me. And I have no idea how to solve it.
write y=f(x)= modulus of -4x+12 as a piecewise function

how do I do it?:eek:

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