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JAMB UTME - Mathematics - 2006

Question 1 Report

If dy/dx = x + cos x, find y

x2 - sin x + c

x2 + sin x + c

x2/2 - sin x + c

x2/2 + sin x + c

Answer Details

We need to integrate the given expression to find the value of y. Integrating both sides with respect to x, we get: ∫dy = ∫(x + cos x)dx y = 1/2 * x^2 + sin x + c where c is the constant of integration. Therefore, the correct answer is: - x^2/2 + sin x + c

Read lesson note on Differentiation (JAMB)

Differentiation

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Question 2 Report

In the diagram above, |QR| is the diameter of the semicircle QR. Find the area of the figure to the nearest whole number. {\( \pi = \frac{22}{7} \)}

90cm2

89cm2

80cm2

70cm2

Answer Details

Area of PQRS = length x breath(7 x 10)cm2 = 70

Area of semi-circle $\frac{1}{2} π r^{2} = \frac{1}{2} \times \frac{22}{7} \times \frac{7}{2}$

= 19.5cm2 = 20cm2(appr.)

Area of the figure = (70 + 20)cm2

= 90cm2

Read lesson note on Areas (WAEC)

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Question 4 Report

If \(x = \begin{vmatrix} 1 & 0 & 1 \\ 2 & -1 & 0 \\ -1 & 0 & 1 \end{vmatrix}\) and \(y = \begin{vmatrix} -1 & 1 & 2 \\ 0 & -1 & -1 \\ 2 & 0 & 1 \end{vmatrix}\)

find \(2x - y\)

| \begin{matrix} 3 & - 1 & 0 \\ 4 & - 3 & - 1 \\ - 4 & 1 & 1 \end{matrix} |

| \begin{matrix} 3 & - 1 & 0 \\ 4 & - 1 & 1 \\ - 4 & 1 & 1 \end{matrix} |

| \begin{matrix} 3 & - 1 & 0 \\ 4 & - 3 & 1 \\ - 4 & 1 & 1 \end{matrix} |

| \begin{matrix} 3 & - 1 & 0 \\ 4 & 1 & 1 \\ - 4 & - 1 & 1 \end{matrix} |

Answer Details

2(10 - (-1) = 3 2(2) - (0) = 4 2(-1) - (2) = -4
2(0) - (1) = -1 2(-1) - (-1) = -1 2(0) - (-1) = 1
2(10 - (20) = 0 2(0) - (-1) = 1 2(1) - (1) = 1

| \begin{matrix} 3 & - 1 & 0 \\ 4 & - 1 & 1 \\ - 4 & 1 & 1 \end{matrix} |

Read lesson note on Linear Inequalities (WAEC)

Read lesson note on Number Bases (JAMB)

Linear Inequalities

Number Bases

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Question 9 Report

Find p, q for which \(\begin{pmatrix}2p & 8 \\ 3 & -5q\end{pmatrix}\) \(\begin{pmatrix}1 \\ 2\end{pmatrix}\) \(\begin{pmatrix}24 \\ -17\end{pmatrix}\)

-4

45

4,2

2

Answer Details

2p x 2 + 8 x 1 = 24

\to

4p = 24 - 8 = 16, p = 4
3 x 1 + -5q x 2 = -17

\to

-10q = -17 - 3
-10q = -20
q = 2

Read lesson note on Solution Of Linear Equations (WAEC)

Read lesson note on Algebraic Expressions (WAEC)

Read lesson note on Differentiation (JAMB)

Solution Of Linear Equations

Algebraic Expressions

Differentiation

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Question 10 Report

If \(y = x^2 - x - 12\), find the range of values of \(x\) for which \(y \ge 0\)

x < -3 0r x > 4

x

\leq

-3 or x

\geq

4

-3 < x

\geq

4

-3

\leq

x

\leq

4

Answer Details

y = x2 - x - 12
= (x - 4)(x - 3)
∴ x = 4 or x = -3
-3 < x < 4

But y $\geq$ 0
∴ -3 < x $\leq$ 4

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Read lesson note on Quadratic Equations (WAEC)

Inequalities

Quadratic Equations

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Question 17 Report

If \(T = 2\pi\sqrt{\frac{L}{g}}\), make \(g\) the subject of the formula

\frac{4 π L^{2}}{T}

\frac{4 π^{2} L^{2}}{T}

\frac{4 π^{2} L}{T}

\frac{2 π \sqrt{T}}{T}

Answer Details

T = 2

π

\sqrt{\frac{L}{g}}

T2 = 4

π^{2} \frac{L}{g}

g =

\frac{4 π^{2} L}{T}

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Read lesson note on Simple Operations On Algebraic Expressions (WAEC)

Change Of Subject Of A Formula/Relation

Simple Operations On Algebraic Expressions

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Question 19 Report

Evaluate \( \frac{2}{6 - 5\sqrt{3}} \)

- (\frac{12}{39} - \frac{10 \sqrt{3}}{39})

\frac{12}{39} - \frac{10 \sqrt{3}}{39}

- (\frac{12}{39} + \frac{10 \sqrt{3}}{39})

\frac{12}{39} + \frac{10 \sqrt{3}}{39}

Answer Details

\frac{2}{6 - 5 \sqrt{3}} = \frac{2}{6 - 5 \sqrt{3}} \times \frac{6 + 5 \sqrt{3}}{6 + 5 \sqrt{3}} = \frac{2 (6 + 5 \sqrt{3})}{(6 - 5 \sqrt{3}) (6 + 5 \sqrt{3})} = \frac{12 + 10 - \sqrt{3}}{36 - 25 (3)} = \frac{12 + 10 - \sqrt{3}}{36 - 75} = \frac{12 + 10 - \sqrt{3}}{- 39} = - (\frac{12}{39} - \frac{10 \sqrt{3}}{39})

Read lesson note on Surds (radicals) (WAEC)

Read lesson note on Application Of Differentiation (JAMB)

Surds (radicals)

Application Of Differentiation

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Question 20 Report

The sum of the first n positive integers is

\frac{1}{2}

(n + 1)

n

\frac{1}{2}

(n + 1)

n(n - 1)

n(n + 1)

Answer Details

Sn =

\frac{n}{2}

(a(n - d)d)

Read lesson note on Sequence And Series (WAEC)

Sequence And Series

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Question 23 Report

A binary operation * on the set of rational numbers is defined as \(x \times y = \frac{x^2-y^2}{2xy}\) find \(-5 \times 3\)

\frac{- 8}{15}

\frac{8}{15}

\frac{17}{15}

\frac{- 17}{15}

Answer Details

To evaluate -5 * 3 using the binary operation * on the set of rational numbers, we substitute x = -5 and y = 3 in the given expression: -5 * 3 = (-5)^2 - 3^2 / 2(-5)(3) Simplifying the numerator, we get: (-5)^2 - 3^2 = 25 - 9 = 16 Simplifying the denominator, we get: 2(-5)(3) = -30 Substituting these values back into the expression, we get: -5 * 3 = 16 / -30 Simplifying the fraction, we get: -5 * 3 = -8/15 Therefore, the value of -5 * 3 using the binary operation * on the set of rational numbers is -8/15.

Read lesson note on Matrices And Determinants (JAMB)

Read lesson note on Binary Operations (JAMB)

Matrices And Determinants

Binary Operations

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Question 27 Report

A binary operation \( \oplus \) on the set of rational numbers is defined as \(x \oplus y = \frac{x^2-y^2}{2xy}\). Find \(-5 \oplus 3\)

\frac{8}{15}

-

\frac{17}{15}

\frac{17}{15}

\frac{8}{- 15}

Answer Details

To find -5 ⊕ 3, we simply substitute x = -5 and y = 3 into the formula: -5 ⊕ 3 = (-5)^2 - 3^2 / (2 * -5 * 3) = 25 - 9 / (-30) = 16 / 30 = 8 / 15 Therefore, the answer is option D: 8-15. To explain how we arrived at this answer, we first plugged in the values of x and y into the formula for the binary operation ⊕. We then simplified the expression using basic arithmetic operations, following the order of operations (parentheses first, then exponents, then multiplication and division from left to right, and finally addition and subtraction from left to right). Finally, we expressed the result as a fraction in lowest terms.

Read lesson note on Binary Operations (JAMB)

Binary Operations

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Question 28 Report

In triangle XYZ, \(\angle XYZ = 15^\circ\), \(\angle XZY = 45^\circ\) and \(lXYl = 7\text{ cm}\). Find \(lYZl\).

14√2 cm

7 (\frac{\sqrt{6}}{2})

7√2 cm

7 cm

Answer Details

∠yxz = 180 - (45 + 15)

= 180 - 60
= 120

^{\circ}

Using sine rule

\frac{x}{s i n x} = \frac{7}{s i n z} \frac{x}{s i n 120} = \frac{7}{s i n 45} x = \frac{7 s i n 120}{s i n 45} x = \frac{7 s i n (180 - 120)}{s i n 45} x = \frac{7 s i n 60}{s i n 45} = (7 (\frac{\sqrt{3}}{2}) \div \frac{1}{\sqrt{2}}) x = (7 (\frac{\sqrt{3}}{2}) \div \frac{1}{\sqrt{2}}) x = 7 (\frac{\sqrt{6}}{2})

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Question 29 Report

In the diagram given PQ = 10cm, PS = 8cm and < PSR is 60 while < SRQ is a right angle. Find SR

10

\sqrt{3}

cm

10cm

45cm

14cm

Answer Details

Draw a line perpendicular to |SR| to form |PT| in

△

PST, cos 60 = \(\frac{|ST|}{8}\(

|PT| = 8 cos 6 = 4cm

Since |TR| = |PQ|

SR = ST + TR

= (4 + 10)cm

= 14cm

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Question 30 Report

he pie chart above shows the expenditure of a family whose income is ₦30,000. If the expenditure on food is twice that on housing and that on school fees is twice that on transport, how much does the family spend on food?

₦28,000

₦25,500

₦15,000

₦12,500

Answer Details

Expenditure on housing = x
Expenditure on food = 2x
Expenditure on school fees = 90

^{\circ}

Expenditure on transport = 45

^{\circ}

x + 2x + 90 + 45 = 360

^{\circ}

3x + 135 = 360

^{\circ}

3x = 360 - 135
3x = 225
x = 225/3 = 75

^{\circ}

∴∠ for food = 2x = 2 * 75
= 150
360

^{\circ}

= N30,000
1

^{\circ}

= ?
1

^{\circ}

= 30,000/360
150

^{\circ}

= (30000/360) * (150/1)
= N12500

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Read lesson note on Financial Arithmetic (WAEC)

Financial Arithmetic

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Question 34 Report

simplify \( \frac{\frac{7}{9}-\frac{2}{3}}{\frac{1}{3}+\frac{\frac{2}{5}}{\frac{4}{5}}} \)

\frac{11}{12}

\frac{5}{6}

\frac{1}{5}

\frac{2}{15}

Answer Details

79−2313+2545=7−6913+2545=1913+(25×54)=1913+12\=192+36=1956=19×65=215

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Number Bases

Number Bases

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Question 35 Report

How many terms of the series 3, -6, +12, - 24, + ..... are needed to make a total of 1-28?

12

10

9

8

Answer Details

3, -6, +12, -24
a = 3, r = -2

8 n = \frac{a (1 - r^{n})}{1 - r} ∴ 1 - 2^{8} = \frac{3 (1 - (- 2^{n - 1}))}{1 - (- 2)} 1 - 2^{8} = \frac{3 (1 - (- 2^{n - 1}))}{3}

1-28 = 1-(-2)n-1
-28 = -2n-1
8 = n-1
n = 9

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Sequence And Series

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Question 38 Report

In a triangle PQT, \(QR = \sqrt{3}\text{ cm}\), PR = 3cm, \(PQ = 2\sqrt{3}\text{ cm}\) and PQR = 30^o. Find angles P and R

P = 60o and R = 90o

P = 30o and R = 120o

P = 90o and R = 60o

P = 60o and R 60o

P = 45o and R = 105o

Answer Details

By using cosine formula, p2 = Q2 + R2 - 2QR cos p
Cos P =

\frac{Q^{2} + R^{2} - p^{2}}{2 Q R}

=

\frac{(3)^{2} + 2 (\sqrt{3})^{2} - 3^{2}}{2 \sqrt{3}}

=

\frac{3 + 12 - 9}{12}

=

\frac{6}{12}

=

\frac{1}{2}

= 0.5
Cos P = 0.5
p = cos-1 0.5 = 60

^{\circ}

= < P = 60

^{\circ}

If < P = 60

^{\circ}

and < Q = 30

< R = 180 $^{\circ}$ - 90 $^{\circ}$
angle P = 60 $^{\circ}$ and angle R is 90 $^{\circ}$

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Question 39 Report

In the parallelogram PQRS given, find angle < SQR

50

^{\circ}

80

^{\circ}

80cm2

100

^{\circ}

Answer Details

< QRS = 30

^{\circ}

(corresponding < ), < QSR = 50

^{\circ}

< SQR = 180 $^{\circ}$ - (30 + 50) $^{\circ}$ (Sum of < S in a $△$ )

180 $^{\circ}$ - 80 $^{\circ}$ = 100 $^{\circ}$

Read lesson note on Angles (WAEC)

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Question 40 Report

The response of 160 pupils in a school asked to indicate their favourite subjects is given in the bar chart above. What percentage of the pupils have English English and Health education as the their favourite subjects?

55%

52%

36%

22%

Answer Details

(

\frac{52}{160} = \frac{36}{160}

)% =

\frac{88}{160}

%

= 55%

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Percentages

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Question 41 Report

The binary operation \( \oplus \) defined on the set of real numbers is such that \( x \oplus y = \frac{xy}{6} \) for all \( x, y \in R \). Find the inverse of 20 under this operation when the identity element is 6

\frac{9}{5}

\frac{1}{20}

\frac{10}{3}

\frac{1}{12}

Answer Details

\frac{\frac{1}{20} \times 6}{6}

=

\frac{6}{20}

x

\frac{1}{6}

=

\frac{1}{20}

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Binary Operations

Progression

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Question 43 Report

The solution set of the shaded area above is

y + x

\leq

4, y

\geq

x

y

\leq

0, y

\leq

x and y + x

\geq

4

y

\geq

x, y + x

\geq

4 and y

\leq

0

y

\geq

x, y + x

\geq

4

Linear Inequalities

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Question 48 Report

If \( \tan \theta = \frac{5}{4} \) find \( \sin^2 \theta - \cos^2 \theta \)

\frac{9}{41}

$\frac{5}{4}$

1

$\frac{9}{1}$

0

Answer Details

(tan

θ

=

\frac{o p p}{a d j}

)
|AB|2 = 52 + 42

\to

|AB|2 = 41

\to

AB =

\sqrt{41}

sin2

θ

- cos2

θ

\to

\frac{5^{2}}{\sqrt{41}}

- (

{\frac{4}{\sqrt{41}}}^{2}

) =

\frac{25}{41}

-

\frac{16}{41}

= (

\frac{9}{41}

)

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Introductory Calculus

Trigonometry

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