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JAMB UTME - Mathematics - 2002

Question 1 Report

Solve for x in the equation x3 - 5x2 - x + 5 = 0

1, - 1, or 5

1, 1, or -5

-1, 1, or -5

1, 1, or 5

Answer Details

To solve this equation, we need to use algebraic methods to isolate the variable x on one side of the equation. First, we can combine like terms on the left-hand side of the equation: x^3 - 5x^2 - x + 5 = 0 Next, we can try to factor the equation, looking for factors that will make the equation equal to zero. By trial and error, we can see that (x-1) is a factor of the equation: (x-1)(x^2 - 4x - 5) = 0 Now we can use the zero product property, which tells us that if the product of two factors is zero, then at least one of the factors must be zero. So we can set each factor equal to zero and solve for x: x-1 = 0 or x^2 - 4x - 5 = 0 Solving for x-1, we get x = 1. Solving for x in the quadratic equation, we can use the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / 2a In this case, a = 1, b = -4, and c = -5. Plugging these values into the quadratic formula, we get: x = (-(-4) ± sqrt((-4)^2 - 4(1)(-5))) / 2(1) Simplifying, we get: x = (4 ± sqrt(36)) / 2 x = (4 ± 6) / 2 x = 5 or x = -1 Therefore, the solutions for x are 1, 5, and -1.

Read lesson note on Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

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Question 2 Report

The triangle PQR above is

an obtuse-angled triangle

a scalene triangle

an isosceles triangle

an equilateral triangle

Answer Details

∠PQR = 180 - 128 (∠s on a straight line)
= 52

^{\circ}

∠QPR + 76 + 52 = 180(extr ∠ = sum of intr. opp ∠s)
∠QPR + 128 = 180
∠QPR = 180 - 128
= 52

^{\circ}

∴ΔPQR is an isosceles triangle

Read lesson note on Circles (WAEC)

Read lesson note on Triangles And Polygons (WAEC)

Triangles And Polygons

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Question 12 Report

Make \(r\) subject of the formula given that \(\frac{x}{r+a}=\frac{a}{r}\)

\frac{a^{2}}{(x - a)}

\frac{a^{2}}{(x + a)}

\frac{a}{x - a}

\frac{a}{x + a}

Introductory Calculus

Change Of Subject Of A Formula/Relation

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Question 31 Report

Using the graph find the values of \(p\) and \(q\) if \(px + qy \ge 4\)

p = 2, q = -1

p = -1, q = 2

p = 2, q = 2

p = 1, q = 2

Answer Details

m =

\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{2 - 0}{0 - (4)} = \frac{2}{4} = \frac{1}{2}

$\frac{y_{2} - y_{1}}{x_{2} - x_{1}} \geq m$

$\frac{y - 0}{x + 4} \geq \frac{1}{2}$

2y $\geq$ x + 4, -x + 2y $\geq$ 4 = px + qy $\geq$ 4

p = -1, q = 2

Read lesson note on Solution Of Linear Equations (WAEC)

Read lesson note on Algebraic Expressions (WAEC)

Read lesson note on Graphs Of Linear And Quadratic Functions (WAEC)

Solution Of Linear Equations

Algebraic Expressions

Graphs Of Linear And Quadratic Functions

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Question 33 Report

In the diagram, a cylinder is surmounted by a hemispherical bowl. Calculate the volume of the solid

198

π c m^{3}

180

π c m^{3}

162

π c m^{3}

216

π c m^{3}

Answer Details

V =

π r^{2} h + \frac{2}{3} π r^{3}

N.B h = r

V = $π x (3)^{2}$ x 20 = $\frac{2}{3} \times π \times (3)^{3}$

180 $π + 18 π = 198 c m^{3}$

Read lesson note on Volumes (WAEC)

Read lesson note on Euclidean Geometry (JAMB)

Euclidean Geometry

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Question 42 Report

In the diagram are two concentric circles of radii r and R respectively with centre O. If \(r = \frac{2}{5}R\), express the area of the shaded portion in terms of \(\pi\) and R

\frac{5}{9} π R^{2}

\frac{21}{25} π R^{2}

\frac{21}{23} π R^{2}

\frac{9}{25} π R^{2}

Answer Details

A

shaded portion

=

π R^{2} - π r^{2}

r =

\frac{2}{5}

R

A $shaded portion$ = $π R^{2} - π (\frac{2}{3} R)^{2} = π R^{2} - \frac{4 π R^{2}}{25}$

= $π R^{2} (1 - \frac{2}{3}) = \frac{21}{25} π R^{2}$

Read lesson note on Circles (WAEC)

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