ce 1991 maths mc

7 out of syllabus in dse. But I can tell you the answer is B.method of bisection is like predicting the answer by guessing some values. When f(x)=0, y=0 and x is a certain value. When x is slightly larger or smaller than the answer, x is positive and there is no change in sign. If we are predicting and do not know the answer, there are no ways that we can know the answer. However, the other graphs has change in sign when x is slightly smaller and then larger than the answer. B

12 for the shaded trapezium, when AD is the base, height=(a-b)/2
Please be noted that 4 trapeziums are identical.
Area of shaded trapezium=(b+a)(a-b)/2/2
=(a^2-b^2)/4+b^2
=(a^2+3b^2)/4 A

39. (X-2)(x-3)=(a-2)(a-3)
By looking at the equation, x= a is clearly a solution.
By thinking twice, x=5-a is also a solution.
Sub x=5-a into LHS, (5-a-2)(5-a-3)=(3-a)(2-a)=(a-3)((a-2)= RHS
Therefore, answer is E

44. By looking at the first three circles, a triangle can be formed by linking the three centers together. It is a isosceles triangle with 2 sides equal to 2cm. As the width of the rectangle is 3cm, the height of the triangle is 3-1-1=1 cm
By Pythagoras theorem, the third side of the triangle is root (4-1)x2=2 root3
As the length of rectangle is 40 cm, (40-2)/(root3)=21.9
We look at the horizontal length 40cm, if we neglect the 1st half of the 1st circle and the last half of the last one, the remaining length should be 38 cm. 2 root3/2 is the horizontal space occupied by 1 circle, so there is the formua above.
So the answer should be 21.9+1=22.9-> 22 C