Maths p1 urgent help (part 2)

[Hasnain1sds]

Guys more problems:

1- M/J 2010 P11 Q1

2- M/J 2010 P12 Q9

 

[arlery]

M/J 2010 P11 Q 1

(i) tan(pi − x) = -k

because it will fall under the second quadrant, where tan is negative.

(ii) tan(0.5 pi -x) = 1/k

the value of tan (0.5pi -x) = cos x/sin x
= 1/tan x
= 1/k

(iii) sin x = k/square root (1+k^2 )

this is because: tan x = k
therefore we can assume tan x = k/1
k = opposite, adjacent = 1

so hypotenuse = square root(1^2 + k^2)
= square root (1 +k^2)

sin x = opposite/hypotenuse
k/square root (1 + k^2)

 

[arlery]

M/J 2010 P12 Q 9.

It's simple, really.
First find the co ordinates A & B.

First equate the equation of the line to make y the subject so that now it is y= 7-2x
since both y's are the subject so now solve it this way:

y =y
7-2x = x^2 -4x+4
x^2 -2x -3 = 0
x^2 -3x +x -3 = 0
x(x-3)+1(x-3)= 0
(x-3)(x+1) = 0
x= 3 & x= -1
y = 7-2(3) & y = 7-2(-1)
= 1 = 9
B(3,1) A(-1,9)

Now first integrate the equation of the curve by using these 2 values of x
the answer you'll get will be 28/3

Then integrate the equation of the line using the same x co-ordinates
the answer will 20.

Then subtract 28/3 from 20
20 - 28/3 = 32/3
= 10 (2/3)

 

[Hasnain1sds]

Lol M/J 2010 p12 Q9 was very easy actually i was ridiculously double integrating equation of a line. thanx anyways

 

[arlery]

You're welcome. Yeah I did that the first time I tried it too. lol