JAMB UTME - Mathematics - 2009
Question 1 Report
Find the range of values of \(x\) for which \(3x - 7 \le 0\) and \(x + 5 > 0\)
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Question 2 Report
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Question 3 Report
i. \(S \cap T \cap W = S\)
ii. \(S \cup T \cup W = W\)
ii. \(T \cap W = S\)
If \(S \subset T \subset W\). Which of the above statements are true?
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Question 4 Report
Find the infinity, the sum of the sequence \(1, \frac{9}{10}, \left(\frac{9}{10}\right)^2, \left(\frac{9}{10}\right)^3, ...\)
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Question 5 Report
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Question 6 Report
Find to infinity, the sum of the sequence \(1, \frac{9}{10}, \left(\frac{9}{10}\right)^2, \left(\frac{9}{10}\right)^3, \ldots\)
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Question 7 Report
A binary operation \(plus\) defined on the set of integers is such that m \(plus\) n = n + mn for all integers m and n. Find the inverse of -5 under this operation, if the identity element is 0.
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Question 8 Report
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Question 9 Report
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Question 10 Report
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Question 11 Report
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Question 12 Report
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Question 13 Report
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x2 + 25 = 169
x2 = 144
x = √ 144
= 12
Length of the chord AB = 2x
2*12= 24cm
Question 14 Report
The cumulative frequency curve above shoes the distribution of the scores of 50 students in an examination. Find the 36th percentile scores
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Question 15 Report
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Question 16 Report
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52x - 1 x 5x + 1 = 125
53x - 1 = 5-2
3x - 1= -2
3x = -1
therefore, x = -13
Question 17 Report
A cliff on the bank of a river is 300 meter high. if the angle of depression of a point on the opposite side of the river is \(60^\circ\), find the width of the river.
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We can solve this problem using trigonometry. Let's draw a diagram of the situation:
A /| / | / | h = 300 m / | ----------- x B
Where point A is the top of the cliff, point B is the unknown point on the opposite side of the river, and x is the width of the river.
We know that the angle of depression from A to B is 60 degrees. This means that the angle of elevation from B to A is also 60 degrees.
Using trigonometry, we can set up the following equation:
tan(60) = h / x
where h is the height of the cliff and x is the width of the river. We can solve for x:
x = h / tan(60) x = 300 / √3 x = 100√3 meters
Therefore, the width of the river is 100√3 meters. Answer is correct.
Question 18 Report
If \(y = 3 \cos 4x\), \(\frac{dy}{dx}\) equals
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Question 19 Report
If \(P = \begin{bmatrix} x+3 & x+2 \\ x+1 & x-1 \end{bmatrix}\) evaluate x if \(|P| = -10\)
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(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
Question 20 Report
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= -5 - (3+2)
= -5-5
= -10
Question 21 Report
The probability of a student passing any examination is \( \frac{2}{3} \). If the student takes three examinations, what is the probability that he will not pass any of them?
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Question 22 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?
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spent 13 of 45 on food
45 x 13 = 415
therefore, 45 - 415
= 12−415
= 815
Question 23 Report
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Question 24 Report
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∫(cos)-2 dθ,
let u = cos θ
∴∫u-2 = 1/u + c
∫cos θ = sin θ + c
∫sec-2θ = 1/u sin θ + c
= (sinθ / cosθ) + c
= Tan θ + c
Question 25 Report
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Question 26 Report
If \(m * n = n-(m + 2)\) for any real numbers m and n, find the value of \(3 * (-5)\)
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= n - (m + 2)
3 ∗ (-5) = -5 - (3 + 2)
= -5 - 5
= -10
Question 27 Report
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Question 28 Report
A regular polygon has \(150^\circ\) as the size of each interior angle. How many sides does it have?
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Question 29 Report
Find the value of x in the diagram.
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x = 100 - 30
= 70o
Question 30 Report
In the figure above, TS//XY and XY = TY,
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< TYZ = 94∘
n∘ = 180 - (94∘ + 34∘ )
= 180∘ - 128∘
= 52∘
Question 31 Report
Evaluate \( \int \sec^2 \theta \, d\theta \)
Question 32 Report
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Question 33 Report
I.S∩T∩W=S II. S∪T∪W=S
III. T∩W=S
If S⊂T⊂W, which of the above statements are true?
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S∪T∪W = W (imply W is a universal set for S and T) T∩W = S imply all the elements of S are in T and W)
S ⊂ T ⊂ W
∴ I, II, and III
Question 34 Report
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Let u = 2+3t ∴du/dt = 3
and v = 5t-4 ∴dv/dt = 5
dx/dt = Vdu/dt + Udv/dt
= (5t-4)3 + (2+3t)5
= 15t - 12 + 10 + 15t
= 30t - 2
= 30x4/5 - 2
= 24 - 2
= 22
Question 35 Report
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x2 + 9 = 25
x2 = 25 – 9
x2 = 16
x = √ 16
= 4 cm
∴ The locus is a circle of radius 4 cm with the center 0
Question 36 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.
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= 35
Question 37 Report
| No. of days | 1 | 2 | 3 | 4 | 5 | 6 |
| No. of students | 20 | x | 50 | 40 | 2x | 60 |
The distribution above shows the number of days a group of 260 students were absents from school in a particular term. How many students were absent for at least four days in the term
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3X + 170 = 260
3X = 260 - 170
3x = 90
x = 30
Absent for at least 4 days
| 4 | 5 | 6 |
| 40 | 2x | 60 |
i.e 40 + 2x + 60 = 40 + (2 x 30) + 60
= 40 + 60 + 60
= 160
Question 38 Report
Simplify \( \frac{5+\sqrt{7}}{3+\sqrt{7}} \)
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Question 39 Report
Find the value of \( \sin 45^\circ - \cos 30^\circ \)
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Question 40 Report
The histogram above represents the number of candidates that sat for Mathematics examination in a school. How many candidate scored more than 50 marks?
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Question 41 Report
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Question 42 Report
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Imply x = 4/3 and - 3/5
3x = 4 and 5x = -3
∴3x-4 = 0 and 5x+3 = 0 are factors
(3x-4)(5x+3) = 0 product of the factors
15x2 + 9x – 20x – 12 = 0 By expansion
15x2 - 11x – 12 = 0
Question 43 Report
If \( p = \begin{pmatrix} x + 3 & x + 2 \\ x + 1 & x - 1 \end{pmatrix} \) evaluate x if \(|p| = -10\)
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|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
Question 44 Report
Find the acute angle between the straight lines \(y = x\) and \(y = \sqrt{3x}\)
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x2 = 3x
3x = 90o
x = 90o3 = 30o
Question 45 Report
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Question 46 Report
Evaluate \(\frac{81.81+99.44}{20.09+36.16}\) correct to 3 significant figures.
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Question 47 Report
Simplify \(4\frac{3}{4} - 6\frac{1}{4}\)
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194 - 254 ............(A)
215 - 54 .............(B)
Now work out the value of A and the value of B and then find the value AB
A = 194 - 254
= ?64
B = 215 x 520
= 10520
= 214
But then AB = -64
214 = -64 ÷ 214
= -64 x 421
= -2484
= -27
Question 48 Report
If s = (2 + 3t)(5t - 4), find \( \frac{dy}{dx} \) when t = \( \frac{4}{5} \) sec
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= 10t - 8 + 15t2 - 12t
dsdt = 15t2 - 2t - 8
= 30t - 2 ; t = 45
= 30t - 2
= 30 x 45 - 2 → 24 - 2
= 22 sec
Question 49 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
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3x = 260 - 170
3x = 90 → x = 30
therefore, 20 + 2x + 60 + 40
= 180
Question 50 Report
