MATHS ........THX!!!

9. g(x) = a(x - 1)(x + 1) - b( x + 1)^2
g(1) = a(1 - 1)(1 + 1) - b(1 + 1)^2 = - 4b = 4, so b = - 1.
g(0) = a(0 -1)(0 + 1) - (-1)(0 + 1)^2 = - a+ 1 = -3, so a = 4
so g(x) = 4(x - 1)(x + 1) + (x + 1)^2 = (x + 1)[(4(x - 1) + (x + 1)] = (x + 1)(4x - 4 + x + 1) = (x + 1)(5x - 3)
g(x) = 0 means (x + 1)(5x - 3) = 0, so x = - 1 or 3/5.
10.
F(-1) = (-1)^2 + k(-1) + 7 = 1 - k + 7 = 8 - k = 1, so k = 7.
f(2p) = (2p) + 3
F(p -2) = (p-2)^2 + 7(p-2) + 7 = p^2 - 4p + 4 + 7p - 14 + 7 = p^2 + 3p - 3
so 2p + 3 = p^2 + 3p - 3
p^2 + p - 6 = 0
(p + 3)(p - 2) = 0
p = - 3 or 2.
12.
slope of L1 = sqrt 3
L2 : (2 sqrt 3) x -2y + 5 = 0,
2y = (2 sqrt 3) x + 5
y = (sqrt 3) x + 5/2, so slope = sqrt 3, so L1//L2.
slope = sqrt 3, that is tan a = sqrt 3, a = arctan sqrt 3 = 60 degree
Put y = 0,
(2 sqrt 3) x - 2(0) + 5 = 0
x = -5/(2 sqrt 3) = x - intercept of L2.
Let distance between 2 lines = p
so p/(x - intercept ) = sin a
p = sin a ( x - intercept) = sin 60 (-5/2 sqrt 3) = (sqrt 3)/2 (-5/2 sqrt 3) = -5/4
= 5/4 because distance is positive.
14.
Using formula of distance between 2 points, equation of L(=AB) is
y - 2.5/[x - (-1)] = (- 0.5 - 2.5)/[5 - (-1)]
(y - 2.5)/(x + 1) = -3/6 = -1/2
2y - 5 = - x - 1
2y = - x + 4
y = -x/2 + 2
Put x = 0, y = 2, so A is (0,2)
Put y = 0, x = 4, so B is (4,0)
AB is diameter of circle, so mid-point is center of circle P, so P is
[(0 + 4)/2, (2 + 0)/2] = (2,1)
By Pythagoras theorem, OA^2 + OB^2 = diameter^2 = (2r)^2
2^2 + 4^2 = 4r^2
4 + 16 = 4r^2 , so r^2 = 5
so area of circle = 5(pi).
11.
Slope of OA = (3 - 0)/(-4 - 0) =-3/4
slope of OA x slope of AB = - 1
so slope of AB = (-1)/(-3/4) = 4/3
Using slope - point form, equation of AB is
y - 3 = (4/3)( x + 4)
3y - 9 = 4x + 16
3y = 4x + 25
Put x = 0, y = 25/3, so OB = 25/3.
Area of triangle = OB x ( x - coordinate of A)/2 = (25/3)(4)/2 = 50/3.
[Note : x - coordinate of A is the height of the triangle]