1)Solve for x in the equation 2sin^2x-1=cos^2x+sin x for 0°<or=x<or=360°

Question

2)An object A of area 10cm^2 is mapped onto its image B of area 60cm^2 by a transformation whose matrix is given by P=([x. 4]
[3x. x+3] )

3)The position vectors of points A and B are 5i+4j-6k and 2i-2j respectively. A point X divides AB in the ratio -3:5.Find the coordinates of X.

4)A closed box has a square base of side X metres and its height h metres.The total surface area of the box is 24cm^2.
a)Find the expression of h in terms of x
b)Hence find the volume of x that would make the volume of the box maximum

5)The image of a point A,under the transformation represented by the matrix T=([1. -1]
[0. 2]) is A^1 (-2,4).Find the coordinates of A

Answer 1

1) Solving the equation 2sin^2x - 1 = cos^2x + sinx, we get:
2(1-cos^2x) - 1 = (1-sin^2x) + sinx
2 - 2cos^2x - 1 = 1 - sin^2x + sinx
1 - 2cos^2x = 1 - sin^2x + sinx
2cos^2x - sin^2x - sinx = 0
2cos^2x - 2sin^2x + sinx = 0
2(cosx - sinx)(cosx + sinx) + sinx = 0
(cosx - sinx)(2cosx + 2sinx + 1) = 0

Setting each factor to zero:
cosx = sinx or 2cosx + 2sinx + 1 = 0

For cosx = sinx:
cosx = √(1-sin^2x)
sinx = √(1-cos^2x)
sinx = √(1-sin^2x)
sin^2x = 1 - sin^2x
2sin^2x = 1
sin^2x = 1/2
sinx = ± (√2)/2

Therefore, x = 45° or 135°

For 2cosx + 2sinx + 1 = 0:
cosx + sinx = -1/2
sinx = -1 - cosx

Substitute sinx = -(1+cosx) into sin^2x = 1-cos^2x:
(1+cosx)^2 = 1 - cos^2x
1 + 2cosx + cos^2x = 1 - cos^2x
2cos^x + cosx = 0
cosx(2cosx + 1) = 0
cosx = 0 or -1/2

Therefore, x = 90° or 135°

2) We are given the transformation matrix P=([x, 4] [3x, x+3]).
Since the area of the object A is 10cm^2 and the area of the image B is 60cm^2, we can set up the transformation equation using the determinants of the transformation matrix:
Area B/Area A = |det(P)|
Area B/Area A = |x(x+3) - 4(3x)|
60/10 = |x^2 + 3x - 12x|
6 = |x^2 - 9x|
6 = |x(x-9)|
x(x-9) = 6
x^2 - 9x - 6 = 0
(x-3)(x+2) = 0
x = 3 or x = -2

3) Given position vectors of points A and B:
A = 5i + 4j - 6k
B = 2i - 2j

Let the position vector of X be the point dividing AB in a ratio of -3:5, then we have:
X = (5B + (-3)A)/(5 + (-3))
X = (5(2i - 2j) + (-3)(5i + 4j - 6k))/(5 - 3)
X = (10i - 10j - 15i - 12j + 18k)/2
X = (-5i - 22j + 18k)/2
X = -5/2 i - 11 j + 9 k

Therefore, the coordinates of X are (-5/2, -11, 9).

4) Given a closed box with a square base of side x meters and height h meters, and the total surface area of the box is 24cm^2.
a) The total surface area of the box can be written as:
2(x^2) + 4(xh) = 24
2x^2 + 4xh = 24
x^2 + 2xh = 12
h = (12 - x^2) / 2

b) To find the maximum volume of the box, we need to maximize the volume V = x^2h:
V = x^2(12 - x^2) / 2
To maximize V, we will take the derivative of V with respect to x and set it to zero:
dV/dx = 2x(12 - x^2)/2 - x^2 = 0
12 - x^2 - x^2 = 0
12 - 2x^2 = 0
2x^2 = 12
x^2 = 6
x = √6

Therefore, the volume will be maximum when x = √6.

5) Given the transformation matrix T=([1, -1] [0, 2]) and image of point A as A^1(-2, 4).
Let the coordinates of point A be (x, y).
Then, by applying the transformation matrix T to point A, we get the image A^1 as:
A^1 = T * A
(-2, 4) = ([1, -1] [0, 2]) * (x, y)
-2 = x - y
4 = 2y
y = 2

Substitute y = 2 into -2 = x - y:
-2 = x - 2
x = 0

Therefore, the coordinates of point A are (0, 2).