Mathematics 2005 Jamb Examination

If the zeros are -2, -1, 3, the polynomial is:

\[(x+2)(x+1)(x-3) = x^3 - 7x - 6\]

Because: Product of the factors
(x+2)(x+1)(x-3) = 0
(x² + 3x + 2)(x-3)
x(x² + 3x + 2) -3(x² + 3x + 2) = 0
x³ + 3x² + 2x - 3x² - 9x - 6 = 0
x³ - 7x - 6 = 0

Let the number of men required be \(30 \times 6 = x \times 4 \quad \Rightarrow \quad x = 45\)

Alternatively:
Let
t = time taken and N = number of men
t ∝ ¹/_N
t = ^K/_N
K = Nt
K = 30 * 6
K = 180
∴t = ¹⁸⁰/_N
4 = ¹⁸⁰/_N
4N = 180
N = ¹⁸⁰/₄
45 men

Since \(W \propto L d^2\), \(W = k L d^2\).

When \(W = 140, L = 54, d = 42/3\), solve for k and rearrange to express \(d\) in terms of \(W\) and \(L\):

\[d = \sqrt{\dfrac{42 W}{5 L}}\]

7x - 3 > 25 + 3x
7x - 3x > 25 + 3
4x > 28
x > ²⁸/₄
x > 7

From the graph, \(f(x) \le 0\) for \(x \le -2\) and \(0 \le x \le 2\)

Let the first term be \(a\) and common difference \(d\).

7th term: \(a + 6d\), 3rd term: \(a + 2d\)

Given: \(a + 6d = 2(a + 2d) \implies a + 6d = 2a + 4d \implies a = 2d\)

Sum of first 4 terms:
S₄ = 4/2 (2a + 3d)
2(2a + 3d) = 4d + 6d
10d = 42
d = 42/10 = 4.2

S₄ = 2(2a + 3d)
2(4d + 3d) =
14d = 42
d = 3

AP with first term \(a=8\), common difference \(d=4\).

Sum of n terms: \(S_n = n/2 [2a + (n-1)d]\)

\(S_{20} = 20/2 [2*8 + 19*4] = 10 [16 + 76] = 10*92 = 920\)

Option D: 920

Let identity element be \(e\). Then \(a * e = a\) for all \(a\).

\(a e + 2(a + e +1) = a \)
\(\implies ae + 2a + 2e + 2 = a \)
\(\implies ae + a + 2e +2 = 0 \)
\(\implies e(a+2) + (a+2) =0 \)
\(\implies e+1 =0 \implies e = -1\)

Using properties of intersecting lines and vertical angles:

\(\angle EOG = 180 - (52 + 85) = 43^\circ\)

Recall the formula for the sum of interior angles of a polygon:

\[S = (n-2) \cdot 180°\]

Substitute the given sum:

\[1800° = (n-2) \cdot 180° \]\[\implies n-2 = \dfrac{1800}{180} = 10\]\[ \implies n = 12\]

Find one exterior angle using:

\[\text{Exterior angle} = \dfrac{360°}{n} = \dfrac{360°}{12} = 30°\]