• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Mathematics: Post your doubts here!

shingiechingz@8

shingiechingz@8

Messages
112
Reaction score
24
Points
28
guys how do you solve this one?
when a polynomial p(x) is divided by x-1 the remainder is 5 and when p(x) is divided by x-2 the remainder is 7. Find the remainder when p(x) is divided by (x-1)(x-2)
 
Thought blocker

Thought blocker

Messages
8,477
Reaction score
34,837
Points
698
shingiechingz@8 said:
how did you find it..................................:)
Click to expand...
Its simple.
We know that when we will divide p(x) with (x-1)(x-2) the remainder will be in Ax + B form. Yes?
Thus,
p(x) = Quotient x divisor + Remainder
p(1) = 5 and p(2) = 7
p(1) = Quotient x (1-1)(1-2) + A + B = 5
so, our 1st equation is A + B = 5
similarly, 2nd equation will be 2A + B = 7

Then use substitution method to find A and B , A = 2 and B = 3 thus remainder is 2x + 3
 
Thought blocker

Thought blocker

Messages
8,477
Reaction score
34,837
Points
698
areeba240 said:
View attachment 61503
can anyone help me with qtn 8(iii)
thank u
Click to expand...
I get -pi/2 and 0.464(3sf) as answer.
the equation of line : y = mx + c
the equation of circle : (x - x')^2 + (y - y')^2 = r^2
m = 1/1 = 1
c = -1
eqn of line is y = x - 1
x' = 0 and y' =1
eqn of circle is x^2 + (y - 1)^2 = 4
Now equate both and solve for y and x and then use normal procedure to find arguments that is using arctan (y/x)
and you will get answers as mine.
 
Last edited:
DragonCub

DragonCub

Messages
260
Reaction score
104
Points
53
areeba240 said:
View attachment 61503
can anyone help me with qtn 8(iii)
thank u
Click to expand...

If you calculated correctly in part (ii), the loci should be:
y = x - 1
x^2 + (y - 1)^2 = 4
Where x is the real part and y is the imaginary part.

Now, to find the intersection points of the loci is the same as to find sets of x and y that satisfy both of the equations.

To do that, simply plug in y = x - 1 into x^2 + (y - 1)^2 = 4:
x^2 + (x - 1 - 1)^2 = 4
x^2 + (x - 2)^2 = 4
x^2 + x^2 - 4x + 4 = 4
2 x^2 - 4 x = 0
x (x - 2) = 0
Solve to get x = 0 , x = 2.

Then, use y = x - 1 to get corresponding y: y = -1 , y = 1 .
So the points of intersection should be (0 , -1) and (2 , 1).

Now, to obtain the argument, use arctan (y/x).

- For (0 , -1):
arctan (-1/0) = -pi/2 rad

- For (2 , 1):
arctan (1/2) = 0.464 rad

------------------------------------------------
Final answer:
-pi/2 rad , 0.464 rad
 
Last edited:
Thought blocker

Thought blocker

Messages
8,477
Reaction score
34,837
Points
698
areeba240 said:
View attachment 61509
plzz help me with this qtn 10(ii) and (iii)
thank u
Click to expand...
There is nothing like (iii) I guess u asking for hole part b. If so,
(i) Draw a circle with center at (-1,3) with radius of 1. We chose (-1,3) as a center because |z + 1 - 3i| <= 1 can be return as |z - (-1 + 3i)| <= 1. Draw a straight line parallel to x axis that is on imaginary axis at 3. And shade the upper half of the circle as that is the common region.

(ii) if u have plot the diagram perfectly, you can see that least value of arg(z) is pi/2 and the greatest value of arg(z) is arctan(2/3) + pi/2. Arctan(2/3) because we draw a line from origin to the other part of the circle forming a diameter between lines of least and great arguments. You will get a visual idea once you plot it perfectly.
Difference will be arctan(2/3) and that is your answer.
 
You must log in or register to reply here.
Top