# Tag Archives: H2 Math

## CJC/II/Q4

The diagram shows the region R bounded by the two parabolas

Y = x2 and x = (y – 2)2 – 2  and the y – axis. Find the points indicated A and B in the diagram.

(a)Find the area of the region R

(b)Find the volume formed when R is rotated 2π radian about the y – axis

Solutions:  A(-1 , 1), B(0 , 2 – 2√2)

(a) area = 0.448

(b) volume = 1.10

Sub Y = x2 into x = (y – 2)2 – 2

y = 2 ±√x+2

x = (x2 – 2)2 – 2

x = x4 – 4x2 + 4 – 2

0 = x4 – 4x2 – x + 2 = (x2 – x – 2)(x2 + x – 1)

X = 2, -1, -1.618, 0.6180

By long division,  Quadratic factor = (x – 2)(x + 1) = x2 – x – 2

A(-1 , 1)    B( (-1 + √5)/2 , [(-1 + √5)/2]2)

(a)

(b)

## VJC/2013/Prelim/P1/Q1

(a)Without using a calculator, solve the inequality

(b) Deduce the range of values of x that satisfies.

Solutions: (a): x≤-6 or 1≤x<3 or x>3

(b): 0<x≤e-6 or e≤x<e3 or x>e3

(a)

(b)

lnx ≤ -6

x ≤ e-6

1≤ lnx <3

e≤x<e3

lnx>3

x>e3

## Vertical Line Test

The use of the vertical line test determines if a line equation is a function.

If the vertical line intersects only 1 point, the line equation is a function. If the vertical line intersects more than 1 point, the line equation is not a function.

## Circular Permutation 1

At the table, 5 boys and 4 girls are to be seated on a round table. How many ways can this be done if

(i) there is no restrictions

(ii) there is no restrictions and the seats are numbered

(iii) 4 girls must sit together

(iv) 3 particular boys cannot sit together

Solution

(i) 40320

(9-1)! = 40320

(ii) 362880

(iii) 2880

(6-1)! x 4! = 2880

Note: (6-1)! is arranging the number of entities in a circle, not 6!

(iv) 14400

(6-1)! x 6C3 x 3! = 14400

note:

3 particular boys are seperated

use ‘slotting’ method

6C3 is the number of ways the 3 PARTICULAR boys can be ‘slotted’