JAMB UTME - Mathematics - 2011

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${\int }_0^1$(3 - 2x)dx
[3x - x2]o
[3(1) - (1)2] - [3(0) - (0)2]
(3 - 1) - (0 - 0) = 2 - 0
= 2

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2y = 5x + 4 (4, 2)
y = $\frac{5x}{2}$ + 4 comparing with
y = mx + e
m = $\frac{5}{2}$
Since they are perpendicular
m1m2 = -1
m2 = $\frac{-1}{m_1}$ = -1
$\frac{5}{2}$ = -1 x $\frac{2}{5}$
The equator of the line is thus
y = mn + c (4, 2)
2 = -$\frac{2}{5}$(4) + c
$\frac{2}{1}$ + $\frac{8}{5}$ = c
c = $\frac{18}{5}$
$\frac{10+5}{5}$ = c
y = -$\frac{2}{5}$x + $\frac{18}{5}$
5y = -2x + 18
or 5y + 2x - 18 = 0

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P(x, y) Q(8, 6)
midpoint = (5, 8)
x + 8 = 5
$\frac{y+6}{2}$ = 8
x + 8 = 10
x = 10 - 8 = 2
y + 6 = 16
y + 16 - 6 = 10
therefore, P(2, 10)

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From Pythagoras theorem
|OA|2 = |AN|2 + |ON|2
72 = |AN|2 + (5)2
49 = |AN|2 + 25
|AN|2 = 49 - 25 = 24
|AN| = $\sqrt{24}$
= $\sqrt{4\times 6}$
= 2√6 cm
|AN| = |NB| (A line drawn from the centre of a circle to a chord, divides the chord into two equal parts)
|AN| + |NB| = |AB|
2√6 + 2√6 = |AB|
|AB| = 4√6 cm

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T2 = 4, T4 = 16
Tx = arn-1
T2 = ar2-1 = 4 i.e. ar3 = 16, i.e. ar = 4
T4 = ar4-1
therefore, $\frac{T_4}{T_r}$ = $\frac{a{r}^3}{ar}$ = $\frac{16}{4}$
r2 = 4 and r = 2
but ar = 4
a = $\frac{4}{r}$ = $\frac{4}{2}$
a = 2
Sn = $\frac{a(r^n-1)}{r-1}$
S5 = $\frac{2(2^5-1)}{2-1}$
= $\frac{2(32-1)}{2-1}$
= 2(31)
= 62

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($\sqrt{2}+\frac{1}{\sqrt{3}}\left)\right(\sqrt{2}-\frac{1}{\sqrt{3}}$)
$\sqrt{4}-\frac{\sqrt{2}}{\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{3}}-\frac{1}{\sqrt{9}}$
= 2 - $\frac{1}{3}$
= $\frac{16-1}{3}$
= $\frac{5}{3}$

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y = x3 + x2 - x + 1
$\frac{dy}{dx}$ = $\frac{d\left(x^3\right)}{dx}$ + $\frac{d\left(x^2\right)}{dx}$ - $\frac{d\left(x\right)}{dx}$ + $\frac{d\left(1\right)}{dx}$
$\frac{dy}{dx}$ = 3x2 + 2x - 1 = 0
$\frac{dy}{dx}$ = 3x2 + 2x - 1
At the maximum point $\frac{dy}{dx}$ = 0
3x2 + 2x - 1 = 0
(3x2 + 3x) - (x - 1) = 0
3x(x + 1) -1(x + 1) = 0
(3x - 1)(x + 1) = 0
therefore x = $\frac{1}{3}$ or -1
For the maximum point
$\frac{d^{2}y}{d{x}^2}$ < 0

$\frac{d^{2}y}{d{x}^2}$ 6x + 2
when x = $\frac{1}{3}$
$\frac{d{x}^2}{d{x}^2}$ = 6($\frac{1}{3}$) + 2
= 2 + 2 = 4
$\frac{d^{2}y}{d{x}^2}$ > o which is the minimum point
when x = -1
$\frac{d^{2}y}{d{x}^2}$ = 6(-1) + 2
= -6 + 2 = -4
-4 < 0

therefore, $\frac{d^{2}y}{d{x}^2}$ < 0

the maximum point is -1

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$\begin{array}{cccc}\text{Class Interval}& 3-3& 6-8& 9-11\\ x& 4& 7& 10\\ f& 2& 2& 2\\ f-x& 8& 14& 20\\ |x-\overline{x}{|}^2& 9& 0& 9\\ |x-\overline{x}{|}^2& 180& 18& \end{array}$
$\overline{x}$ = $\frac{\sum fx}{\sum f}$
= $\frac{8+14+20}{2+2+2}$
= $\frac{42}{6}$
$\overline{x}$ = 7
S.D = $\sqrt{\frac{\sum f(x-\overline{x}{)}^2}{\sum f}}$
= $\sqrt{\frac{18+0+18}{6}}$
= $\sqrt{\frac{36}{6}}$
= $\sqrt{6}$

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log${}_3^{18}$ + log${}_3^3$ - log${}_3^x$ = 3
log${}_3^{18}$ + log${}_3^3$ - log${}_3^x$ = 3log33
log${}_3^{18}$ + log${}_3^3$ - log${}_3^x$ = log333
log3($\frac{18\times 3}{X}$) = log333
$\frac{18\times 3}{X}$ = 33
18 x 3 = 27 x X
x = $\frac{18\times 3}{27}$
= 2

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< TUW = 110${}^{\circ }$ = 180${}^{\circ }$ (< s on a straight line)
< TUW = 180${}^{\circ }$ - 110${}^{\circ }$ = 70${}^{\circ }$
In $△$ XTU, < XUT + < TXU = 180${}^{\circ }$
i.e. < YTS + 70${}^{\circ }$ = 180
< XTU = 180 - 110${}^{\circ }$ = 70${}^{\circ }$
Also < YTS + < XTU = 180 (< s on a straight line)
i.e. < YTS + < XTU - 180(< s on straight line)
i.e. < YTS + 70${}^{\circ }$ = 180
< YTS = 180 - 70 = 110${}^{\circ }$
in $△$ SYT + < YST + < YTS = 180${}^{\circ }$(Sum of interior < s)
SYT + 40 + 110 = 180
< SYT = 180 - 150 = 30
< SYT = < XYW (vertically opposite < s)
Also < SYX = < TYW (vertically opposite < s)
but < SYT + < XYW + < SYX + < TYW = 360
i.e. 30 + 30 + < SYX + TYW = 360
but < SYX = < TYW
60 + 2(< TYW) = 360
2(< TYW) = 360${}^{\circ }$ - 60
2(< TYW) = 300${}^{\circ }$
TYW = $\frac{300}{2}$ = 150${}^{\circ }$
< SYT

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$\frac{90}{360}\times 100=\frac{1}{4}\times 100$
$=25\%$

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$\left|\begin{array}{cc}2& 3\\ 5& 3x\end{array}\right|$ = $\left|\begin{array}{cc}4& 1\\ 3& 2x\end{array}\right|$
(2 x 3x) - (5 x 3) = (4 x 2x) - (3 x 1)
6x - 15 = 8x - 3
6x - 8x = 15 - 3
-2x = 12
x = $\frac{12}{-2}$
= -6

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Perimeter of circle = Perimeter of square
28cm = 4L
L = $\frac{28}{4}$ = 7cm
Area of square = L2
= 72
= 49cm2

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N + Y = XY + X + Y
3 + -$\frac{2}{3}$ = 3(- $\frac{2}{3}$) + 3 + (- $\frac{2}{3}$)
= -2 + 3 -$\frac{2}{3}$
= $\frac{1-2}{1-3}$
= $\frac{1}{3}$

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${\int }_0^1$ cos4 x dx
let u = 4x
$\frac{dy}{dx}$ = 4
dx = $\frac{dy}{4}$
${\int }_0^1$cos u. $\frac{dy}{4}$ = $\frac{1}{4}$$\int$cos u du
= $\frac{1}{4}$ sin u + k
= $\frac{1}{4}$ sin4x + k

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Volume of cube = L3
33 = 27cm3
volume of rectangular tank = L x B X h
= 3 x 4 x 5
= 60cm3
volume of H2O the tank can now hold
= volume of rectangular tank - volume of cube
= 60 - 27
= 33cm3

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tan$\theta$ = $\frac{100}{100}$ = 1
$\theta$ = tan-1(1) = 45o
The bearing of x from z is ₦45oE or 135o

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$\frac{\sin \theta }{\cos \theta }$
$\frac{\cos \theta \frac{d\left(\sin \theta \right)}{d\theta }-\sin \theta \frac{d\left(\cos \theta \right)}{d\theta }}{{\cos }^2\theta }$
$\frac{\cos \theta .\cos \theta -\sin \theta (-\sin \theta )}{co{s}^2\theta }$
$\frac{co{s}^2\theta +{\sin }^2\theta }{co{s}^2\theta }$
Recall that sin2 $\theta$ + cos2 $\theta$ = 1
$\frac{1}{{\cos }^2\theta }$ = sec2 $\theta$

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9y2 - 16x2
= 32y2 - 42x2
= (3y - 4x)(3y +4x)

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N = [2 3]
N-1 = $\frac{adjN}{|N|}$
adj N = $\left|\begin{array}{cc}4& -3\\ -1& 2\end{array}\right|$
|N| = (2 x4) - (1 x 3)
= 8 - 3
=5
N-1 = $\frac{1}{5}$ $\left|\begin{array}{cc}4& -3\\ -1& 2\end{array}\right|$

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$(\frac{16}{81}{)}^{\frac{1}{4}}÷(\frac{9}{16}{)}^{-\frac{1}{2}}$
$(\frac{16}{81}{)}^{\frac{1}{4}}÷(\frac{16}{9}{)}^{\frac{1}{2}}$
$(\frac{2^4}{3^4}{)}^{\frac{1}{4}}÷(\frac{4^2}{3^2}{)}^{\frac{1}{2}}$
$\frac{2^{4\times \frac{1}{4}}}{3^{4\times \frac{1}{4}}}÷\frac{4^{2\times \frac{1}{2}}}{3^{2\times \frac{1}{2}}}$
$\frac{2}{3}÷\frac{4}{3}$
$\frac{2}{3}\times \frac{3}{4}$
$\frac{2}{4}$
$\frac{1}{2}$

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3, 6, 9,..., 36.
a = 3, d = 3, i = 36, n = 18
Sn = $\frac{n}{2}$ [2a + (n - 1)d
S18 = $\frac{18}{2}$ [2 x 3 + (18 - 1)3]
= 9[6 + (17 x 3)]
= 9 [6 + 51] = 9(57)
= 513

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$\frac{2-\sqrt{5}}{3-\sqrt{5}}$ x $\frac{3+\sqrt{5}}{3+\sqrt{5}}$
$\frac{(2-\sqrt{5})(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}$ = $\frac{6+2\sqrt{5}-3\sqrt{5}-\sqrt{2}5}{9+3\sqrt{5}-3\sqrt{5}-\sqrt{2}5}$
= $\frac{6-\sqrt{5}-5}{9-5}$
= $\frac{1-\sqrt{5}}{4}$

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The first person will sit down and the remaining will join.
i.e. (n - 1)!
= (5 - 1)! = 4!
= 24 ways

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Mode = L1 + ($\frac{D_1}{D_1+D_2}$)C
D1 = frequency of modal class - frequency of the class before it
D1 = 5 - 2 = 3
D2 = frequency of modal class - frequency of the class that offers it
D2 = 5 - 3 = 2
L1 = lower class boundary of the modal class
L1 = 5 - 5
C is the class width = 8 - 5.5 = 3
Mode = L1 + ($\frac{D_1}{D_1+D_2}$)C
= 5.5 + $\frac{3}{2+3}$C
= 5.5 + $\frac{3}{5}$ x 3
= 5.5 + $\frac{9}{5}$
= 5.5 + 1.8
= 7.3 $\approx$ = 7

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The first poster has 7 ways to be arranges, the second poster can be arranged in 6 ways and the third poster in 5 ways.
= 7 x 6 x 5
= 210 ways
or $\frac{7}{P_3}$ = $\frac{7!}{(7-3)!}$ = $\frac{7!}{4!}$
= $\frac{7\times 6\times 5\times 4!}{4!}$
= 210 ways

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x2 + 2x > 15
x2 + 2x - 15 > 0
(x2 + 5x) - (3x - 15) > 0
x(x + 5) - 3(x + 5) >0
(x - 3)(x + 5) > 0
therefore, x = 3 or -5
then x < -5 or x > 3
i.e. x< 3 or x < -5

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Let the numbers be a, a + 1, a + 2, a + 3
a + a + 1 + a + 2 + a + 3 = 34
4a = 34 - 6
4a = 28
a = $\frac{28}{4}$
= 7
The least of these numbers is a = 7

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$\left|\begin{array}{ccc}4& 2& -1\\ 2& 3& -1\\ -1& 1& 3\end{array}\right|$
4 $\left|\begin{array}{cc}3& -1\\ 1& 3\end{array}\right|$ -2 $\left|\begin{array}{cc}2& -1\\ -1& 3\end{array}\right|$ -1 $\left|\begin{array}{cc}2& 3\\ -1& 1\end{array}\right|$
4[(3 x 3) - (-1 x 1)] -2 [(2x 3) - (-1 x -1)] -1 [(2 x 1) - (-1 x 3)]
= 4[9 + 1] -2 [6 - 1] -1 [2 + 3]
= 4(10) - 2(5) - 1(5)
= 40 - 10 - 5
= 25

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T $\alpha \frac{1}{R^3}$
T = $\frac{k}{R^3}$
k = TR3
= $\frac{2}{81}$ x 33
= $\frac{2}{81}$ x 27
dividing 81 by 27
k = $\frac{2}{2}$
therefore, T = $\frac{2}{3}$ x $\frac{1}{R^3}$
When R = 2
T = $\frac{2}{3}$ x $\frac{1}{2^3}$ = $\frac{2}{3}$ x $\frac{1}{8}$
= $\frac{1}{12}$

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-2x -5y = 3
x + 3y = 0
x = -3y
-2 (-3y) - 5y = -3
6y - 5y = 3
y = 3
but, x = -3y
x = -3(3)
x = -9
therefore, x = -9, y = 3

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X2 + 1 $\frac{X-2}{\sqrt{X^3-2X^2+3n-3}}$
= $\frac{-6X^3+n}{-2X^2+2X-3}$
= $\frac{(-2X^2-2)}{2X-1}$
Remainder is 2X - 1

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x $\alpha \sqrt{y}$
x = k$\sqrt{y}$
81 = k$\sqrt{9}$
k = $\frac{81}{3}$
= 27
therefore, x = 27$\sqrt{y}$
y = 1$\frac{7}{9}$ = $\frac{16}{9}$
x = 27 x $\sqrt{\frac{16}{9}}$
= 27 x $\frac{4}{3}$
dividing 27 by 3
= 9 x 4
= 36

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tan$\theta$ = $\frac{3}{4}$
from Pythagoras tippet, the hypotenus is T
i.e. 3, 4, 5.
then sin $\theta$ = $\frac{3}{5}$ and cos$\theta$ = $\frac{4}{3}$
cos$\theta$ - sin$\theta$
$\frac{4}{5}$ - $\frac{3}{5}$ = $\frac{1}{5}$

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ELATION
Since there are 7 letters. The first letter can be arranged in 7 ways, , the second letter in 6 ways, the third letter in 5 ways, the 4th letter in four ways, the 3rd letter in three ways, the 2nd letter in 2 ways and the last in one way.
therefore, 7 x 6 x 5 x 4 x 3 x 2 x 1 = 7! ways

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S.I. = $\frac{P\times R\times T}{100}$
If T = 9 months, it is equivalent to $\frac{9}{12}$ years
S.I. = $\frac{5000\times 4\times 9}{100\times 12}$
S.I. = ₦150

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(2x + 1)(3x + 1) IS
2x + 1 $\frac{d(3x+1)}{d}$ + (3x + 1) $\frac{d(2x+1)}{d}$
2x + 1 (3) + (3x + 1) (2)
6x + 3 + 6x + 2 = 12x + 5

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2q35 = 778
2 x 52 + q x 51 + 3 x 50 = 7 x 81 + 7 x 80
2 x 25 + q x 5 + 3 x 1 = 7 x 8 + 7 x 1
50 + 5q + 3 = 56 + 7
5q = 63 - 53
q = $\frac{10}{5}$
q = 2

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$\frac{3\frac{2}{3}\times \frac{5}{6}\times \frac{2}{3}}{\frac{11}{15}\times \frac{3}{4}\times \frac{2}{27}}$
$\frac{\frac{11}{3}\times \frac{5}{6}\times \frac{2}{3}}{\frac{11}{15}\times \frac{3}{4}\times \frac{2}{27}}$
$\frac{110}{54}÷\frac{66}{1620}$
50

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M:N:Q == 5:4:3
i.e M = 5, N = 4, Q = 3
Substituting values into equation, we have...
$\frac{2N-Q}{M}$
= $\frac{2\left(4\right)-3}{5}$
= $\frac{8-3}{5}$
= $\frac{5}{5}$
= 1

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Interior angle = (n - 2)180
but, n = 12
= (12 -2)180
= 10 x 180
= 1800
let each interior angle = x
x = $\frac{(n-2)180}{n}$
x = $\frac{1800}{12}$
= 150o

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-6(x + 3) $\le$ 4(x - 2)
-6(x +3) $\le$ 4(x - 2)
-6x -18 $\le$ 4x - 8
-18 + 8 $\le$ 4x +6x
-10x $\le$ 10x
10x $\le$ -10
x $\le$ 1