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JAMB UTME - Mathematics - 2002

Ibeere 1 Ìròyìn

If y = x2 - (1/x). find dy/dx

2x - (1/x2)

2x + x2

2x - x2

2x + (1/x2)

Awọn alaye Idahun

To find dy/dx, we need to differentiate y with respect to x using the rules of differentiation. Let's start by expressing y in terms of x: y = x^2 - (1/x) To differentiate this expression, we need to use the power rule and the quotient rule. The power rule states that if y = x^n, then dy/dx = nx^(n-1). The quotient rule states that if y = u/v, then dy/dx = (v du/dx - u dv/dx) / v^2. Using the power rule, we can find the derivative of x^2: d/dx (x^2) = 2x Using the quotient rule, we can find the derivative of (1/x): d/dx (1/x) = -1/x^2 Now we can use these results to find the derivative of y: dy/dx = d/dx (x^2) - d/dx (1/x) = 2x - (-1/x^2) = 2x + 1/x^2 Therefore, the answer is option D: 2x + (1/x^2).

Ka ìwé ẹ̀kọ́ nípa Differentiation (JAMB)

Differentiation

Ibeere 15 Ìròyìn

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}

Awọn alaye Idahun

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 $π + 18 π = 198 c m^{3}$

Ka ìwé ẹ̀kọ́ nípa Volumes (WAEC)

Ka ìwé ẹ̀kọ́ nípa Euclidean Geometry (JAMB)

Euclidean Geometry

Ibeere 18 Ìròyìn

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}

Awọn alaye Idahun

A

shaded portion

=

π R^{2} - π r^{2}

r =

\frac{2}{5}

R

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

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

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Ibeere 19 Ìròyìn

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

Ibeere 23 Ìròyìn

The triangle PQR above is

an obtuse-angled triangle

a scalene triangle

an isosceles triangle

an equilateral triangle

Awọn alaye Idahun

∠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

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Ka ìwé ẹ̀kọ́ nípa Triangles And Polygons (WAEC)

Triangles And Polygons

Ibeere 45 Ìròyìn

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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Solution Of Linear Equations (WAEC)

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Ka ìwé ẹ̀kọ́ nípa Graphs Of Linear And Quadratic Functions (WAEC)

Solution Of Linear Equations

Algebraic Expressions

Graphs Of Linear And Quadratic Functions

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