JAMB UTME - Mathematics - 2007
Question 1 Report
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Question 2 Report
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Question 3 Report
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f(x) = 6x2 - 2x - 4 at turning point, f1(x) = 0
6x2 - 2x - 4 = 0, 3x2 - x - 2 = 0, 3x2 - 3x + 2x - 2 = 0
(3x + 2)(x - 1) = 0, x = -23 or 1
f11(x) = 12x - 2,
when x = 23 , f11(x) = 12(-23 ) - 2 = -10 < 0
→
f(x) is maximum @ x = -23
when x = 1, f11(x) = 12(1)- 2 = 10 > 0
→
f(x) is maximum @ x = 1
Question 4 Report
The solution of the quadratic inequality \( (x^2 + x - 12) \ge 0 \) is
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For the condition to hold, each of (x - 3) and (x + 4) must be of the same sign
.i.e. x - 3 ≥ 0 and x + 4 ≥ 0
or x - 3≤ 0 and x + 4 ≤ 0
when x ≥ 3, the condition is satisfied
when x ≥ -4, the condition is not satisfied.
when x ≤ 3, the condition is not satisfied
when x ≤ -4 , the condition is not satisfied. Thus, the solution of the inequality is x ≥ 3 or x ≤ -4 ,
Question 5 Report
The table above shows the number of pupils in each age group in a class. What is the probability that a pupil chosen at random is at least 1 years old?
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= 27/40 + 7/40
= 34/40
= 17/20
Question 6 Report
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Question 7 Report
Find the sum to infinity to the following series \(3 + 2 + \frac{4}{3} + \frac{8}{9} + \frac{16}{17} + .....\)
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Question 8 Report
Integrate \( \frac{x^2-\sqrt{x}}{x} \) with respect to x
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∫x2−√xx dx = ∫(x2x−√xx )dx
= ∫(x - x12 )dx
= x22 - x1212 + k
= x22 - 2√x + k
Question 9 Report
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Question 10 Report
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Question 11 Report
The volume of a hemispherical bowl is \(718\frac{2}{3}\,\text{cm}^3\). Find its radius.
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23πr3 = 71823
23 x 227r3
= 21563
r3 = 21563 x 32 x 722
r3 = 49 x 7
r = 3√343
= 7cm3
Question 12 Report
If y = x cosx, find \( \frac{dy}{dx} \).
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Using product rule dydx = cosx - x sinx
Question 13 Report
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Question 14 Report
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5 . 2n-2 - n. To understand why, we can look at how the sequence is generated. The first term is 3/2, the second term is 3, and each subsequent term is generated by doubling the previous term and subtracting its position in the sequence. For example, to get the third term, we double the second term (which is 3) to get 6, and then subtract the position of the term (which is 3) to get 3+3=6. Similarly, to get the fourth term, we double the third term (which is 6) to get 12, and then subtract the position of the term (which is 4) to get 12-4=8. Using this pattern, we can derive the general formula for the nth term: 5 . 2n-2 - n.Question 15 Report
Make L the subjects of the formula if \( \sqrt{\frac{42w}{5l}} \)
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square both side of the equation
d2=(√42W5l)2d2=42W5l5ld2=42Wl=42W5d2
Question 16 Report
The nth term of the sequence \( \frac{3}{2} \), 3, 7, 16, 35, 74, ..., is
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Question 17 Report
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Question 18 Report
Determine the value of \( \displaystyle \int_{0}^{\frac{\pi}{2}}(-2\cos x)\,dx \)
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Question 19 Report
Simplify \( \frac{3}{5} \div \left(\frac{2}{7} \times \frac{4}{3} \div \frac{4}{9}\right) \)
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Question 20 Report
The graph above is represented by
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then the factors; x+2, x+1 and x-1
Product of the factors; (x+2)(x+1)(x-1)
= y = (x + 2)(x2 - x + x - 1)
= y = (x+2)(x2-1)
x3 - x + 2x2 - 2 = y
x3 + 2x2 - x - 2 = y
Question 21 Report
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32x-1 = 162/2
32x-1 = 81
32x-1 = 32
2x - 1 = 4 (equating the indices)
2x = 5
5/2
x =
Question 22 Report
If X = {all the perfect squares less than 40}
Y = {all the odd numbers fro, 1 to 15}. Find X ∩ Y.
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X = {1, 4, 9, 16, 25, 36}
All the odd numbers from 1 to 15
Y = {1, 3, 5, 7, 9, 11, 13, 15}
X ∩ Y = {1, 9}
Question 23 Report
The histogram above represents the weights of students who travelled out to their school for an examination. How many people made the trip.
Question 24 Report
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26% profit = 100 + 26 = 126%
∴ 105% → N60,000
1% → 60000/15
126% = 1000/105 x 126/1
=N72,000
Question 25 Report
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Question 26 Report
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= 34
2x - 1 = 4 , 2x = 5
x = 52
Question 27 Report
If the lines \(2y - kx + 2 = 0\) and \(y + x - \frac{k}{2} = 0\) intersect at (1, 2), find the value of k
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2(-2) - k(1) + 2 = 0
-4 - k + 2 = 0
k = -2
Or y + x - k2 = 0, @ x= 1, y = -2
-2 + 1 - k2 = 0
-1 = k2
k = -1 x 2
= -2
Question 28 Report
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= 125 + 50 + 5 + 4 18410
x = 184
Question 29 Report
| Marks | 3 | 4 | 5 | 6 | 7 | 8 |
| Frequency | 5 | \(y - 1\) | \(y\) | 9 | 4 | 1 |
The table above gives the frequency distribution of marks obtained by a group of students in a test. If the total mark scored is 200, the value of y is
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15 + 4y - 4 + 5y + 54 + 28 + 8 = 200
9y + 105 - 4 = 200
9y + 101 = 200
9y = 200 - 101
9y = 99
y = 11
Question 30 Report
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=7!(7−4)!=7!3!=7×6×5×4×3!3!=840
Question 31 Report
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f’(x) = 6x2 - 2x – 4
As f’(x) = 0
Implies 6x2 - 2x – 4 = 0
3x – x – 2 = 0 (By dividing by 2)
(3x – 2)(x + 1) = 0
3x – 2 = 0 implies x = -2/3
Or x + 1 = 0 implies x = -1
f’(x) = 6x2 - 2x – 4
f’’(x) = 12x – 2
At max point f’’(x) < 0
∴f’’(x) = 12x – 2 at x = -1
= 12(-1) – 2
= -12 – 2 = -14
∴Max at x = 1
Question 32 Report
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Question 33 Report
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Question 34 Report
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Question 35 Report
Find the sum to infinity of the series \(2 + \frac{3}{2} + \frac{9}{8} + \frac{27}{32} + \ldots\)
Question 36 Report
\(W \alpha L^2\) and \(W = 6\) when \(L = 4\). If \(L = \sqrt{17}\), find \(W\).
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642 = k
W(√17)2
W = 17×616
= 518
= 638
Question 37 Report
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Question 38 Report
Integrate \( \frac{x^2-\sqrt{x}}{x} \) with respect to x
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Question 39 Report
The volume of a hemispherical bowl is \(718\frac{2}{3}\). Find its radius .
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Question 40 Report
The graph above is represented by
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y = (x + 2)(x + 1)(x - 1) = (x + 2)(x2 - 1)
= x3 + 2x2 - x - 2
Question 41 Report
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= x10 = 1 x 53 + 2 x 52 + 1 * 51 + 4 x 50
= 1 x 125 + 2 x 25 + 1 x 5 + 4 x 1
= 125 + 50 + 5 + 4
= 184
Question 42 Report
Simplify \( \frac{3}{5} \div \left(\frac{2}{7} x \frac{4}{3} \div \frac{4}{9}\right) \)
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= 35 ÷ 67
= 35 x 76
= 710
Question 43 Report
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(x + 4)(x - 3) ≥ 0
Either x + 4 ≥ 0 implies x ≥ -4
Or x - 3 ≥ 0 implies x ≥ 3
∴ x ≥ 3 or x ≥ -4
Question 44 Report
If \(y = (1 + x)^2\), find \(\frac{dy}{dx}\)
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Question 45 Report
A binary operation \( \oplus \) on real numbers is defined by \( x \oplus y = xy + x + y \) for any two real numbers \( x \) and \( y \). The value of \( \left(-\frac{3}{4}\right) \oplus 6 \) is
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(-34 ) ⊕ 6 = -34.6 - 34 + 6
= -92 - 34 + 6
= -34
Question 46 Report
Find y, if \( \sqrt{12} - \sqrt{147} + y\sqrt{3} = 0 \)
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√4×3−√49×3 + y√3 = 0
y - 5 = 0
y = 5
Question 47 Report
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Question 48 Report
In the diagram P, Q, R, S are points on the circle RQS = 30o. PRS = 50o and PSQ = 20o. What is the value of xo + yo?
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< PQS = < PRS (angle in the sam segment)
< PQS = 50o
Also, < QSR = < QPR(angles in the segment)
< QPR = xo
x + y + 5= = 180(angles in a triangle)
x + y = 180 - 50
x + y = 130o
Question 49 Report
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Question 50 Report
Given
P = {1, 3, 5, 7, 9, 11}
And Q = {2, 4, 6, 8, 1, 12}. Determine the relationship between P and Q
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