JAMB UTME - Mathematics - 2011

The perpendicular bisector of a line XY is the locus of a point B. whose distance from Y is always twice its distance from X. C

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A chord of circle of radius 7cm is 5cm from the centre of the maximum possible area of the square?

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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

Find ${\int }_0^1$ cos4 x dx

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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

A man walks 100 m due West from a point X to Y, he then walks 100 m due North to a point Z. Find the bearing of X from Z.

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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

Find the value of x at the minimum point of the curve y = x3 + x2 - x + 1

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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

A man invested ₦5,000 for 9 months at 4%. What is the simple interest?

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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

If log318 + log33 - log3x = 3, Find x.

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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

In the diagram, STUV is a straight line. < TSY = < UXY = 40o and < VUW = 110o. Calculate < TYW

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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

The bar chart above shows the distribution of SS2 students in a school.

Find the total number of students

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

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

In how many ways can five people sit round a circular table?

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

Raial has 7 different posters to be hanged in her bedroom, living room and kitchen. Assuming she has plans to place at least a poster in each of the 3 rooms, how many choices does she have?

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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

If x varies directly as square root of y and x = 81 when y = 9, Find x when y = 1$\frac{7}{9}$

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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

A solid metal cube of side 3 cm is placed in a rectangular tank of dimension 3, 4 and 5 cm. What volume of water can the tank now hold

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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

Which of these angles can be constructed using ruler and a pair of compasses only?

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Find the equation of a line perpendicular to line 2y = 5x + 4 which passes through (4, 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

If | 2 3 | = | 4 1 |. find the value of y. 7

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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

Find the derivative of $\frac{\sin \theta }{\cos \theta }$

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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 $\theta$

Find the remainder when X3 - 2X2 + 3X - 3 is divided by X2 + 1

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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

The pie chart shows the distribution of courses offered by students. What percentage of the students offer English?

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

Solve the inequality x2 + 2x > 15.

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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

Find the sum of the first 18 terms of the series 3, 6, 9,..., 36.

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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

Simplify $\frac{3\frac{2}{3}\times \frac{5}{6}\times \frac{2}{3}}{\frac{11}{15}\times \frac{3}{4}\times \frac{2}{27}}$