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JAMB UTME - Mathematics - 2003

Frage 1 Bericht

Determine the maximum value of y = 3x2 - x3

zero

2

4

6

Antwortdetails

To find the maximum value of y = 3x^2 - x^3, we can take the derivative of y with respect to x, set it equal to zero, and solve for x. First, we find the derivative of y: dy/dx = 6x - 3x^2 Setting this equal to zero gives: 6x - 3x^2 = 0 Factor out x: x(6 - 3x) = 0 Solving for x, we get two solutions: x = 0 or x = 2 To determine which value of x gives the maximum value of y, we can use the second derivative test. Taking the second derivative of y: d^2y/dx^2 = 6 - 6x Plugging in x = 0 and x = 2: d^2y/dx^2 |x=0 = 6 > 0 d^2y/dx^2 |x=2 = -6 < 0 Since the second derivative is positive at x = 0 and negative at x = 2, this means that x = 0 corresponds to a minimum value of y, while x = 2 corresponds to a maximum value of y. Therefore, the maximum value of y = 3x^2 - x^3 occurs when x = 2, and is equal to: y = 3(2)^2 - (2)^3 = 4. So the answer is 4.

Lies die Unterrichtsnotiz zu Quadratic Equations (WAEC)

Quadratic Equations

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Frage 16 Bericht

An arc of a circle subtends an angle of \(30^\circ\) on the circumference of a circle of radius 21cm. Find the length of the arc.

11cm

22cm

44cm

66cm

Antwortdetails

θ = 30 x 2(∠ at center twice ∠ at θ)
∴ θ = 60

^{\circ}

Length of arc =

\frac{θ}{360} \times 2 π r = \frac{60}{360} \times 2 \times \frac{22}{7} \times \frac{21}{1} = 22 c m

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Frage 20 Bericht

Find the range of the value of \(x\) satisfying the inequalities \(5 + x \le 8\) and \(13 + x \ge 7\)

-3 $\leq$ x $\leq$ 3

3 $\leq$ x $\leq$ 6

-6 $\leq$ x $\leq$ 3

-6 $\leq$ x $\leq$ -3

Antwortdetails

5 + x $\leq$ 8 and 13 + x $\geq$ 7
5 + x $\leq$ 8
X $\leq$ 8 – 5
X $\leq$ 3
And 13 + x $\geq$ 7
X $\geq$ 7 – 15
X $\geq$ -7
Combining the two together
-6 $\leq$ x $\leq$ 3

Lies die Unterrichtsnotiz zu Linear Inequalities (WAEC)

Lies die Unterrichtsnotiz zu Inequalities (JAMB)

Linear Inequalities

Inequalities

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Frage 32 Bericht

Find the range of values of \(x\) which satisfy the inequalities \(4x - 7 \le 3x\) and \(3x - 4 \le 4x\)

-4

\leq

x

\leq

7

-7

\leq

x

\leq

4

x

\geq

-7

-7

\leq

x

\leq

6

Antwortdetails

To solve the system of inequalities 4x - 7 ≤ y ≤ 3x and 3x - 4 ≤ y ≤ 4x, we can use the following steps: 1. For the first inequality, isolate y in the middle by adding 7 to both sides: 4x - 7 + 7 ≤ y + 7 ≤ 3x + 7, which simplifies to 4x ≤ y + 7 ≤ 3x + 7. 2. For the second inequality, isolate y in the middle by adding 4 to both sides: 3x - 4 + 4 ≤ y + 4 ≤ 4x + 4, which simplifies to 3x ≤ y + 4 ≤ 4x + 4. 3. We want to find the range of values of x that satisfy both inequalities. To do this, we need to find the intersection of the two intervals [4x, 3x + 7] and [3x, 4x + 4] for y. 4. We start by finding the maximum lower bound of the two intervals. The lower bound of [4x, 3x + 7] is 4x, and the lower bound of [3x, 4x + 4] is 4x. So the maximum lower bound is 4x. 5. We then find the minimum upper bound of the two intervals. The upper bound of [4x, 3x + 7] is 3x + 7, and the upper bound of [3x, 4x + 4] is 4x + 4. So the minimum upper bound is 3x + 7. 6. Therefore, the range of values of x that satisfy both inequalities is 4x ≤ y + 7 ≤ 3x + 7 and 3x - 4 ≤ y ≤ 4x, which simplifies to 4x ≤ y + 7 ≤ 3x + 7 and 3x - 4 ≤ y ≤ 4x. 7. To solve for x, we can use the second inequality: 3x - 4 ≤ y ≤ 4x. We can substitute the upper bound of y with 4x and the lower bound of y with 3x - 4 to get 3x - 4 ≤ 3x - 4 ≤ x ≤ 4x. Therefore, the range of values of x that satisfy both inequalities is -4 ≤ x ≤ 7. Therefore, the correct option is "-4 ≤ x ≤ 7".

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Frage 33 Bericht

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

150o

140o

130o

120o

Antwortdetails

< 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

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Frage 34 Bericht

Triangle OPQ is the solution of the inequalities

x - 1 < 0, y +

\geq

0, y - x

\geq

0

x + 1

\leq

0, y + x

\geq

0, y - x

\leq

0

x - 1

\geq

0, y - x

\geq

0, y + x

\geq

0

y + x

\geq

0, y - x

\leq

0

Linear Inequalities

Inequalities

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Frage 35 Bericht

An aeroplane flies due north from airport P to Q and then flies due east R. If Q is equidistant from P and R, find the bearing of P and R

90

^{\circ}

135

^{\circ}

225

^{\circ}

270

^{\circ}

Antwortdetails

Since PQ = QR, Δ PQR is a right angle isosceles triangle

∴ ∠PQR = 45

^{\circ}

=> The bearing of P from R = 180 + 45
= 225

^{\circ}

Lies die Unterrichtsnotiz zu Bearings (WAEC)

Lies die Unterrichtsnotiz zu Circles (WAEC)

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