JAMB UTME - Mathematics - 2008

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2x2 - kx - 12 is divisible by x-4
implies x is a factor ∴ x = 4
f(4) implies 2(4)2 - k(4) - 12 = 0
32 - 4k - 12 = 0
-4k + 20 = 0
-4k = -20
k = 5

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x-3, 3x+2,x-1, 4x, 2x-1, x-2, 2x-2, 3x, 3x+1
Range = 3x+2 - (x-3)
= 3x+2 - x - 3
= 2x + 5

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sinθ = 3/5
x2 = 52 - 32

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X = {1,5,10}
Y = {3,4,6}
X∩Y = ∅

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X = 20/4
= 5
mean deviation = 8/4 = 2

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3X + 2Y + 1 = 0
2Y = -3X - 1
$\frac{-3}{2}X-\frac{1}{2}$
Gradient of 3X + 2Y +1 = 0 is -3/2
Gradient of a line perpendicular to 3X + 2Y + 1 = 0
$=-1÷\frac{3}{2}=-1\times \frac{-2}{3}=\frac{2}{3}$

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Area of semicircle + Area of rectangle

A = $\frac{1}{2}\pi r^2$ + LB

A = $\frac{1}{2}\times \frac{22}{77}\times (\frac{5}{2}{)}^2+(15\times 5)$

= $\frac{1}{2}\times \frac{22}{7}\times \frac{25}{4}+75$

A = $\frac{275}{28}+\frac{75}{1}$

$\frac{275+2100}{28}=\frac{2375}{28}$

A = 84.8cm2

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125x = 20
1xX2 + 2xX1 + 5xX0 = 20
X2 + 2X + 5 = 20
X2 + 2X - 15 = 0
(X + 5)( X - 3) = 0
X + 5 implies X = -5
X - 3 implies X = 3
But X cannot be negative
∴X = 3

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n(M) only = 30-10 = 20
n(E) only = 20-10 = 10
n(M∩E) = 10
∴M∪E = 20+10+10
= 40

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(tan $\theta$ = $\frac{opp}{adj}$)
|AB|2 = 52 + 42 $\to$ |AB|2 = 41
$\to$ AB = $\sqrt{41}$ sin2$\theta$ - cos2$\theta$
$\to$ $\frac{5^2}{\sqrt{41}}$ - (${\frac{4}{\sqrt{41}}}^2$) = $\frac{25}{41}$ - $\frac{16}{41}$
= ($\frac{9}{41}$)

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x2 - 5x + 6 = 0
(X-2)(X-3) = 0
X-2 = 0 implies X = 2
X-3 = 0 implies X = 3
∴ x ≤ 2, x ≥ 3

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X * Y = 2X - 3Y + 2
2*3 = 2(2) - 3(3) + 2
=4-9+2
= -3
But -3 does not belong to positive integer

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${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}cosxdx=[sinx{]}_{\frac{-\pi }{2}}^{\frac{\pi }{2}}=sin\frac{\pi }{2}-sin\frac{-\pi }{2}$
= sin90 – sin-90
= sin90 – sin270
= 1 – (-1)
= 1+1
= 2

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mΔn = m+n+mn
Let e be the identity element
∴mΔe = eΔm = m
m+e+me = m
e+me = m-m
e+me = 0
e(1+m) = 0
e = 0 / (1+m)
e = 0

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4X - 7 $\le$ 3X and 3X - 4 $\le$ 4X
4X - 3X $\le$ 7 and 3X - 4X $\le$ 4
X $\le$ 7 and -X $\le$ 4 = X $\ge$ -4
Range -4 $\le$ x $\le$ 7

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y = X2 - 3x + 2, $\frac{dy}{dx}$ = 2x - 3
at turning pt, $\frac{dy}{dx}$ = 0
∴ 2x - 3 = 0
∴ x = $\frac{3}{2}$
$\frac{d^{2}y}{d{x}^2}$ = $\frac{d}{dx}$($\frac{d}{dx}$)
= 270
∴ ymin = 2$\frac{3}{2}$ - 3$\frac{3}{2}$ + 2
= $\frac{9}{4}$ - $\frac{9}{2}$ + 2
= -$\frac{1}{4}$

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sin x - x cos x
dy/dx = cos x - [1.cos x + x -sin x]
= co x - [cos x - x sin x]
= cos x - cos x + x sin x
= x sin x

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N85 - N60 = N25 increase
∴ percentage increase =$\frac{25}{60}\times \frac{100}{1}=\frac{125}{3}=41.67\%=42\%$

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${\int }_1^2(6x^2-2x)dx=[\frac{6x^3}{3}-\frac{2x^2}{2}{]}_1^2=[2x^3-x^2{]}_1^2$
= [2(2)3 - (2)2] – [2(1)3 - (1)2]
= [16-4] – [2-1]
= 12 – 1
= 11

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(x – 3) ∝ y2
X-3 = Ky2
K = X-3 / y2
= 5-2/22
= 2/4
= 1/2
∴X-3 = 1/2y2
X-3 = 1/2(6)2
X-3 = 1/2 x 30/1
X-3 = 18
X = 21

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(4x+3y)2 - (3x-2y)2
(4x+3y+3x-2y)(4x+3y-(3x-2y))
(4x+3y+3x-2y)(4x+3y-3x+2y)
(x+5y)(7x+y)

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Total possible outcome
12+18+x+30+2x+45 = 105+3x
∴105+3x = 150
3x = 150-105
3x = 45
x = 15
P(obtaining 5) =$\frac{2x}{(105+3x)}Butx=15=\frac{2\left(15\right)}{(105+3(15\left)\right)}=\frac{30}{(105+45)}=\frac{30}{150}=\frac{1}{5}$

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11012 + 101112 + 1112 = 101011

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Area of the figure = Area of rect + area of semi circle
$=L\times h+\frac{1}{2}\pi r^{2}5\times 15+\frac{1}{2}\times \frac{22}{7}\times {\left(\frac{5}{2}\right)}^2=75+\frac{(22\times 25)}{2\times 7}=75+9\frac{23}{28}=84.8cm$

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U5 = 24, n = 5 and U11 = 96, n = 11
Un = a + (n-1)d
24 = a + (5-1)d imply 24 = a+4d .....eqn1
96 = a + (11-1)d imply 96 = a+10d ...eqn2
eqn1 - eqn2 -72 = -6d
d = 72/6 = 12
but 24 = a+4d
24 = a + 4(12)
24 = a + 48
a = 24-48
a = -24

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Ratio of least to most = 3:12
= 3/12
= 1/4

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a = a(base ∠s of Iss Δ)
∴ a+a+74 = 180
2a + 74 = 180
2a = 180-74
2a = 106
a = 53
∴∠OPQ = 53${}^{\circ }$

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V = πr2h
1m = 100cm
14cm = 1400cm
$\therefore V=\frac{22}{7}\times \frac{100x100x1400}{1000}=44,000liters$

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$\begin{array}{ccccccc}Number& 1& 2& 3& 4& 5& 6\\ Frequency& 12& 18& x& 30& 2x& 45\end{array}$
12 + 18 + x + 30 + 2x + 45 = 150
3x + 105 = 150
3x = 150 - 105
3x = 45
x = $\frac{45}{3}$
x = 15
probability of 5 = $\frac{30}{150}$
= $\frac{1}{5}$

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2 x s = 280o(Angle at centre = 2 x < at circum)

S = $\frac{{280}^o}{2}$

= 140

< O = 360 - 280 = 80o

60 + 80 + 140 + a = 360o

(< in a quad); 280 = a = 360

a = 360 - 280

a = 80o

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(x2 + x - 12) $\ge$ 0 , (x - 3)(x + 4) $\ge$ 0
For the condition to hold, each of (x - 3) and (x + 4) must be of the same sign
.i.e. x - 3 $\ge$ 0 and x + 4 $\ge$ 0
or x - 3$\le$ 0 and x + 4 $\le$ 0
when x $\ge$ 3, the condition is satisfied
when x $\ge$ -4, the condition is not satisfied.
when x $\le$ 3, the condition is not satisfied
when x $\le$ -4 , the condition is not satisfied. Thus, the solution of the inequality is x $\ge$ 3 or x $\le$ -4 ,

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P ∝ 1/q
P = k/q
K = q2P
= 428
∴P = 128/q
32 = 128/q
q2 = 128/32
q2 = 4
q = √4 = +/-2

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ACCEPTANCE = 10 Letters
A = 2 letters
C = 3 letters
E = 2 letters
Can be arranged in 10! / (2!3!2!) ways

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OBSTRUCTION
Total possible outcome = 11
Number of chance of getting T = 2
P(picking T) = 2/11

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If logx1/264 = 3
(X 1/2)3 = 64
(X 1/2)3 = 4 3
X 1/2 = 4
X = 42
X = 16

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