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JAMB UTME - Mathematics - 2009

Question 1 Report

Determine the value of x for which (x2 - 1) > 0

x < -1 or x > 1

-1 < x < 1

x > 0

x < -1

Answer Details

To find the value of x for which (x^2 - 1) > 0, we need to factorize the expression. (x^2 - 1) can be written as (x - 1)(x + 1). So now we have: (x - 1)(x + 1) > 0 For the product of two factors to be greater than zero, either both factors must be positive or both factors must be negative. If (x - 1) and (x + 1) are both positive, then x > 1. If (x - 1) and (x + 1) are both negative, then x < -1. Therefore, the solution is: x < -1 or x > 1 which means option (A) is the correct answer.

Read lesson note on Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

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Question 5 Report

No. of days	1	2	3	4	5	6
No. of students	20	\(2x\)	60	40	\(x\)	50

The distribution above shows the number of days a group of 260 students were absent from school in a particular term. How many students were absent for at least four days in the term

180

120

110

40

Answer Details

3x + 170 = 260
3x = 260 - 170
3x = 90

\to

x = 30
therefore, 20 + 2x + 60 + 40
= 180

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Question 6 Report

Simplify \(4\frac{3}{4} - 6\frac{1}{4}\)

-7

\frac{7}{8}

\frac{- 2}{7}

\frac{- 10}{21}

\frac{10}{21}

Answer Details

4

\frac{3}{4}

- 6

\frac{1}{4}

\frac{19}{4}

-

\frac{25}{4}

............(A)

\frac{21}{5}

-

\frac{5}{4}

.............(B)
Now work out the value of A and the value of B and then find the value

\frac{A}{B}

A =

\frac{19}{4}

-

\frac{25}{4}

=

\frac{- 6}{4}

B =

\frac{21}{5}

x

\frac{5}{20}

=

\frac{105}{20}

=

\frac{21}{4}

But then

\frac{A}{B}

=

\frac{- 6}{4}

\frac{21}{4}

=

\frac{- 6}{4}

\div

\frac{21}{4}

=

\frac{- 6}{4}

x

\frac{4}{21}

=

\frac{- 24}{84}

=

\frac{- 2}{7}

View Answer

Question 8 Report

If the hypotenuse of a right-angled isosceles triangle is 2cm , what is the area of the triangle?

\frac{1}{\sqrt{2}}

cm2

1 cm2

\sqrt{2}

cm2

2

s q r t 2

cm2

Answer Details

An isosceles triangle has two equal sides and two equal angles opposite those sides. In a right-angled isosceles triangle, the two equal angles are 45 degrees each, since the sum of all angles in a triangle is 180 degrees and the right angle is 90 degrees. Let's use the Pythagorean theorem to find the length of the legs of the triangle. If the hypotenuse is 2cm, then each leg is x cm. Applying the Pythagorean theorem, we have: x^2 + x^2 = 2^2 2x^2 = 4 x^2 = 2 x = sqrt(2) Therefore, the length of each leg of the triangle is sqrt(2) cm. To find the area of the triangle, we can use the formula: Area = (base * height) / 2 Since the triangle is isosceles, the base and height are equal. Therefore, we can use either leg as the base and the other leg as the height. Area = (sqrt(2) * sqrt(2)) / 2 Area = 2 / 2 Area = 1 cm^2 So, the area of the right-angled isosceles triangle is 1 cm^2.

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Question 10 Report

If \( p = \begin{pmatrix} x + 3 & x + 2 \\ x + 1 & x - 1 \end{pmatrix} \) evaluate x if \(|p| = -10\)

-5

-2

2

5

Answer Details

If p =

(\begin{matrix} x + 3 & x + 2 \\ x + 1 & x - 1 \end{matrix})

evaluate x if |p| = -10
|P| = (x + 3)(x - 1) - (x + 2)(x + 1)
x2 + 2x - 3 - (x2 + 3x + 2)
x2 + 2x - 3 - x2 - 3x - 2
-x - 5 = -10
-x = -10 + 5
x = 5

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Polynomials

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Question 22 Report

If s = (2 + 3t)(5t - 4), find \( \frac{dy}{dx} \) when t = \( \frac{4}{5} \) sec

0 units per sec.

15 units per seconds

22 units per seconds

26 units per seconds

Answer Details

s = (2 + 3t)(5t - 4), find

\frac{d y}{d x}

when t =

\frac{4}{5}

sec
= 10t - 8 + 15t2 - 12t

\frac{d s}{d t}

= 15t2 - 2t - 8
= 30t - 2 ; t =

\frac{4}{5}

= 30t - 2
= 30 x

\frac{4}{5}

- 2

\to

24 - 2
= 22 sec

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Differentiation

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Question 23 Report

Solve 52(x - 1) x 5x + 1 = 0.04

\frac{1}{3}

\frac{1}{4}

-

\frac{1}{5}

-

\frac{1}{3}

Answer Details

52(x - 1) x 5x + 1 = 0.04
52x - 1 x 5x + 1 =

\frac{1}{25}

53x - 1 = 5-2
3x - 1= -2
3x = -1
therefore, x = -

\frac{1}{3}

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Algebraic Expressions

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Question 24 Report

In the figure above, TS//XY and XY = TY,

47

^{\circ}

52

^{\circ}

56

^{\circ}

99

^{\circ}

Answer Details

TXY = XTY = 47

^{\circ}

(Basa angle of an isosceles

△

)

< TYZ = 94 $^{\circ}$

n $^{\circ}$ = 180 - (94 $^{\circ}$ + 34 $^{\circ}$ )

= 180 $^{\circ}$ - 128 $^{\circ}$

= 52 $^{\circ}$

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Triangles And Polygons

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Question 29 Report

Marks	2	3	4	5	6	7	8	9
No. of students	3	4	1	0	4	5	2	1

The table above shows the distribution of marks of students in a test. Find the probability of passing the test if the pass mark is 5.

\frac{3}{5}

\frac{2}{5}

\frac{7}{20}

\frac{1}{5}

Answer Details

p (pass at 5mark) =

\frac{12}{20}

=

\frac{3}{5}

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Probability

Surds (radicals)

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Question 32 Report

Evaluate \( \int \sec^2 \theta \, d\theta \)

sec

θ

tan

θ

+ k

tan

θ

+ k

2 sec

θ

+ k

sec

θ

+ k

Answer Details

\int

sec2

θ

d

θ

= tan

θ

+ k

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Integration

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Question 33 Report

If \(P = \begin{bmatrix} x+3 & x+2 \\ x+1 & x-1 \end{bmatrix}\) evaluate x if \(|P| = -10\)

-5

-2

2

5

Answer Details

P = [\begin{array}{cc} x + 3 & x + 2 \\ x + 1 & x - 1 \end{array}]

evaluate x if |P| = -10
(x+3)(x-1) - {(x+1)(x+2)} = -10
x2 - x + 3x - 3 - {x2 + 2x + x + 2} = -10
x2 + 2x - 3 - {x2 + 3x + 2} = -10
-x - 5 = -10
-5 + 10 = x
5 = x
∴x = 5

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Read lesson note on Algebraic Expressions (WAEC)

Read lesson note on Angles Of Elevation And Depression (WAEC)

Read lesson note on Surds (radicals) (WAEC)

Solution Of Linear Equations

Algebraic Expressions

Angles Of Elevation And Depression

Surds (radicals)

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Question 35 Report

Find the acute angle between the straight lines \(y = x\) and \(y = \sqrt{3x}\)

15o

30o

45o

60o

Answer Details

x

\leq

90o (acute angle)
x2 = 3x
3x = 90o
x =

\frac{90^{o}}{3}

= 30o

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Coordinate Geometry

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Question 36 Report

Find the range of values of \(x\) for which \(3x - 7 \le 0\) and \(x + 5 > 0\)

-5 < x <

\frac{7}{3}

-5

\leq

x

\leq

\frac{7}{3}

-5 < x

\leq

\frac{7}{3}

-5

\leq

x <

\frac{7}{3}

Answer Details

To solve this inequality, we need to work with each inequality separately and then find the values of x that satisfy both. Starting with the first inequality: 3x - 7 ≤ 0 Adding 7 to both sides: 3x ≤ 7 Dividing by 3: x ≤ 7/3 Now, let's work with the second inequality: x + 5 > 0 Subtracting 5 from both sides: x > -5 To satisfy both inequalities, x must be greater than -5 and less than or equal to 7/3. Therefore, the range of values for x is: -5 < x ≤ 7/3

Read lesson note on Linear Inequalities (WAEC)

Read lesson note on Inequalities (JAMB)

Linear Inequalities

Inequalities

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Question 40 Report

A student spent \( \frac{1}{5} \) of his allowances on books, \( \frac{1}{3} \) of the remainder on food and kept the rest for contingences. What fraction was kept?

$\frac{7}{16}$

\frac{8}{15}

$\frac{2}{3}$

$\frac{4}{5}$

Answer Details

Spent

\frac{1}{5}

on books remaining

\frac{4}{5}

spent

\frac{1}{3}

of

\frac{4}{5}

on food

\frac{4}{5}

x

\frac{1}{3}

=

\frac{4}{15}

therefore,

\frac{4}{5}

-

\frac{4}{15}

=

\frac{12 - 4}{15}

=

\frac{8}{15}

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Read lesson note on Number Bases (WAEC)

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Fractions, Decimals And Approximations

Number Bases

Surds (radicals)

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Question 41 Report

If \(m * n = n-(m + 2)\) for any real numbers m and n, find the value of \(3 * (-5)\)

-6

-8

-10

-12

Answer Details

If m x n = n
= n - (m + 2)
3

*

(-5) = -5 - (3 + 2)
= -5 - 5
= -10

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Read lesson note on Differentiation (JAMB)

Read lesson note on Quadratic Equations (WAEC)

Solution Of Linear Equations

Differentiation

Quadratic Equations

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Question 44 Report

Find the value of \( \sin 45^\circ - \cos 30^\circ \)

\frac{2 + \sqrt{3}}{4}

\frac{\sqrt{2} + \sqrt{3}}{4}

\frac{\sqrt{2} + \sqrt{3}}{2}

\frac{\sqrt{2} - \sqrt{3}}{2}

Answer Details

S i n 45 - c o s 30 \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} = \frac{\sqrt{2} - \sqrt{3}}{2}

Read lesson note on Sine, Cosine And Tangent Of An Angle (WAEC)

Read lesson note on Trigonometry (JAMB)

Sine, Cosine And Tangent Of An Angle

Trigonometry

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