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JAMB UTME - Mathematics - 1995

Question 1 Report

For what value of x is the tangent to the curve \(y = x^2 - 4x + 3\) parallel to the x-axis?

3

2

1

0

Answer Details

To find the value of x for which the tangent to the curve y = x^2 - 4x + 3 is parallel to the x-axis, we need to find the point on the curve where the slope of the tangent line is zero. The slope of the tangent to a curve y = f(x) at a point (a, f(a)) is given by f'(a), the derivative of f(x) evaluated at x = a. So, we need to find the derivative of y = x^2 - 4x + 3, which is y' = 2x - 4. Setting y' equal to zero, we get: 2x - 4 = 0 Solving for x, we get: x = 2 Therefore, the value of x for which the tangent to the curve y = x^2 - 4x + 3 is parallel to the x-axis is 2.

Read lesson note on Statistics (WAEC)

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Question 4 Report

Find T in terms of K, Q and S if S = 2r\( \pi \)QT + K)

\frac{S^{2}}{2 π r^{2} Q} - \frac{k}{Q}

\frac{S^{2}}{2 π r^{2} Q}

- k

\frac{S^{2}}{4 π r^{2} Q} - \frac{k}{Q}

\frac{s^{2}}{4 π r^{2} Q}

Answer Details

\frac{s^{2}}{4 r^{2}}

= QT

π

+ KT

\frac{s^{2}}{4 r^{2}}

- k

π

= QT

π

T = $\frac{s^{2}}{4 Q π r^{2}}$ - k

Read lesson note on Quadratic Equations (WAEC)

Quadratic Equations

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Question 5 Report

A school boy lying on the ground 30m away from the foot of a water tank tower observes that the angle of elevation of the top of the tank 60o. Calculate the height of the water tank.

60m

30

\sqrt{3}

m

20

\sqrt{3}

m

10

\sqrt{3}

m

Answer Details

To solve this problem, we can use trigonometry. Let h be the height of the water tank, and d be the horizontal distance between the foot of the water tank tower and the position of the boy. Then, we have: tan(60°) = h / d We know that d = 30m, and tan(60°) = √3, so we can solve for h: h = d * tan(60°) h = 30m * √3 h = 30√3 m Therefore, the height of the water tank is 30√3 meters.

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Question 6 Report

Determine the value of x in the figure.

134

^{\circ}

81

^{\circ}

53

^{\circ}

46

^{\circ}

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Question 8 Report

If S = (x : x2 = 9, x > 4), then S is equal to

4

{0}

\oplus

{

\oplus

}

Algebraic Fractions

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Question 10 Report

Which of the following binary operations is cumulative in the set of integers?

a

*

b = a + 2b

a

*

b = a + b - ab

a

*

b = a2 + b

a

*

b =

\frac{a (b + 1)}{2}

Binary Operations

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Question 14 Report

P is on the locus of points equidiatant from two given points X and Y. UV is a straight line throuh Y parallel to the locus. If < PYU is 40°, find < XPY.

100o

80 o

50 o

40 o

Answer Details

< PYU = < YPQ = 40°(Alternative) = PQ

∥

UV

< XPQ = < YPQ = 40°, Since QX = QY

therefore < XPY = 80°

Read lesson note on Loci (WAEC)

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Question 15 Report

Use the graph of the curve \(y = f(x)\)to solve the inequality \(f(x) \le 0\)

-1

\geq

x

\geq

1, x

\geq

2

x

\leq

-1, 1

\geq

x

\geq

2

x

\geq

-1, 1

\geq

x

\geq

2

x

\geq

2, -1

\geq

x

\geq

1

Answer Details

-1

\geq

x

\geq

1, 1 < x

\geq

2

Combining solutions

= x $\leq$ 1; 1 $\geq$ x $\geq$ 2

Read lesson note on Linear Inequalities (WAEC)

Linear Inequalities

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Question 19 Report

PT is a tangent to the circle TYZX. YT = YX and < PTX = 50o. Calculate < TZY

50o

65o

85o

130o

Answer Details

XTY = 50o = TXY =

\frac{130^{o}}{2}

= 65o

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Question 20 Report

Class Interval	Frequency	Class boundaries	Class Mid-point
\(1.5 - 1.9\)	\(2\)	\(1.45 - 1.95\)	\(1.7\)
\(2.0 - 2.4\)	\(21\)	\(1.95 - 2.45\)	\(2.2\)
\(2.5 - 2.9\)	\(4\)	\(2.45 - 2.95\)	\(2.7\)
\(3.0 - 2.9\)	\(15\)	\(2.95 - 3.45\)	\(3.2\)
\(3.5 - 3.9\)	\(10\)	\(3.45 - 3.95\)	\(3.7\)
\(4.0 - 4.4\)	\(5\)	\(3.95 - 4.45\)	\(4.2\)
\(4.5 - 4.9\)	\(3\)	\(4.45 - 4.95\)	\(4.7\)

The median of the distribution above is

4.0

3.4

3.2

3.0

Answer Details

Median =

\frac{a + b}{f m}

(

\frac{1}{2} \sum f - C F_{b}

)
= 2.95 +

\frac{0.5}{15}

(2.-7)
= 2.95 +

\frac{0.5}{15}

x 13
= 2.95 + 0. 43
= 3.38
= 3.4

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Question 23 Report

Let \(p\) be a probability function on set \(S\), where \(S = (a_1, a_2, a_3, a_4)\). Find \(P(a_1)\) if \(P(a_2) = \frac{1}{3}\), \(p(a_3) = \frac{1}{6}\) and \(p(a_4) = \frac{1}{5}\)

\frac{7}{3}

\frac{2}{3}

\frac{1}{3}

\frac{3}{10}

Probability

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Question 26 Report

Express \( \frac{5x-12}{(x-2)(x-3)} \) in partial fractions

\frac{2}{x + 2} - \frac{3}{x - 3}

\frac{2}{x - 2} + \frac{3}{x - 3}

\frac{2}{x - 3} - \frac{3}{x - 2}

\frac{5}{x - 3} - \frac{4}{x - 2}

Answer Details

$\frac{5 x - 12}{(x - 2) (x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3}$

= $\frac{A (x - 3) + B (x - 2)}{(x - 2) (x - 3)}$

⟹5x−12=Ax−3A+Bx−2B

$A + B = 5 . . . (i)$

−(3A+2B)=−12⟹3A+2B=12...(ii)

From (i), $A = 5 - B$

$3 (5 - B) + 2 B = 12$

15−3B+2B=12⟹B=3

A+3=5⟹A=2

$\frac{5 x - 12}{(x - 2) (x - 3)} = \frac{2}{x - 2} + \frac{3}{x - 3}$

Read lesson note on Algebraic Expressions (WAEC)

Algebraic Expressions

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Question 29 Report

A die has four of it's faces coloured white and the remaining two coloured black. What is the probability that when the die is thrown two consecutive time, the top face will be white in both cases?

\frac{2}{3}

\frac{1}{9}

\frac{4}{9}

\frac{1}{36}

Answer Details

The probability of getting a white face on a single roll of the die is 4/6 or 2/3, since four of the six faces are white. Similarly, the probability of getting a white face on a second roll of the die is also 2/3. To find the probability of both events occurring (getting a white face on both rolls), we multiply the probabilities of each event together. So, the probability of getting a white face on the first roll AND getting a white face on the second roll is (2/3) x (2/3) = 4/9. Therefore, the correct answer is not provided in the options given.

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Read lesson note on Probability (JAMB)

Probability

Probability

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Question 31 Report

class	1−3	4−6	7−9
Frequency	5	8	5

Find the standard deviation of the data using the table above

5

\(\sqrt{6}\0

\frac{5}{3}

\sqrt{5}

Answer Details

\begin{matrix} class intervals & F r e (F) & class-marks(x) & F x & (x - x) & (x - x)^{2} & F (x - x)^{2} \\ 1 - 3 & 5 & 10 & 10 & - 3 & 9 & 90 \\ 4 - 6 & 8 & 40 & 40 & 0 & 0 & 0 \\ 7 - 9 & 5 & 40 & 40 & 3 & 9 & 360 \\ 18 & 90 & 450 \end{matrix}

x =

\frac{\sum f x}{\sum f}

=

\frac{90}{18}

= 5
S.D =

\frac{\sum f (x - x)^{2}}{\sum f}

=

\frac{450}{18}

=

\sqrt{25}

= 5

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Question 34 Report

Simplify \( \frac{x^2 - 1}{x^3 + 2x^2 - x - 2} \)

\frac{1}{x + 2}

\frac{x - 1}{x + 1}

\frac{x - 1}{x + 2}

x−1x+2

\frac{1}{x - 2}

Answer Details

\frac{x^{2} - 1}{(x - 1) (x + 2) (x + 1)}

=

\frac{(x - 1) (x + 1)}{(x - 1) (x + 2) (x + 1)}

=

\frac{1}{x + 2}

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Question 35 Report

Class Interval	Frequency	Class boundaries	Class Mid−point
\(1.5-1.9\)	\(2\)	\(1.45-1.95\)	\(1.7\)
\(2.0-2.4\)	\(21\)	\(1.95-2.45\)	\(2.2\)
\(2.5-2.9\)	\(4\)	\(2.45-2.95\)	\(2.7\)
\(3.0-2.9\)	\(15\)	\(2.95-3.45\)	\(3.2\)
\(3.5-3.9\)	\(10\)	\(3.45-3.95\)	\(3.7\)
\(4.0-4.4\)	\(5\)	\(3.95-4.45\)	\(4.2\)
\(4.5-4.9\)	\(3\)	\(4.45-4.95\)	\(4.7\)

Find the mode of the distribution above to find the mode of the distribution.

3.2

3.3

3.7

4.2

Answer Details

Mode = a + (b - a)(fm - Fb)

2Fm - Fa - Fb

= 3.0 + $\frac{(3.4 - 3) (15 - 4)}{2 (15) - 4 - 10}$

= 3 + $\frac{(6.4) (11)}{30 - 14}$

= 3 + $\frac{4.4}{16}$

= 3 + 0.275

= 3.275

= 3.3cm

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Read lesson note on Measures Of Dispersion (JAMB)

Representation Of Data

Measures Of Dispersion

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Question 40 Report

Find the distance between two towns p(45\(^{o}\)N, 30\(^{o}\)W) and Q(15\(^{o}\)S, 30\(^{o}\)W) if the radius of the earth is 7000km. [\(\pi = \frac{22}{7}\)]

\frac{1100}{3}

km

\frac{2200}{3}

km

\frac{22000}{3}

km

\frac{1100}{3}

km

Coordinate Geometry

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Question 47 Report

Solve the inequality \( (x - 3)(x - 4) \le 0 \)

3

\leq

x

\leq

4

3 < x < 4

3

\leq

x < 4

3 < x

\leq

4

Answer Details

To solve this inequality, we need to find the values of x that make the expression (x - 3)(x - 4) less than or equal to 0. To do this, we can use a method called the "sign chart" or "interval notation". First, we can find the critical values of x by setting the expression equal to 0: (x - 3)(x - 4) = 0. This gives us x = 3 and x = 4. Next, we can create a sign chart by dividing the number line into intervals using the critical values of x. We then choose a test point from each interval and evaluate the expression (x - 3)(x - 4) using that test point. If the result is positive, the expression is positive in that interval. If the result is negative, the expression is negative in that interval. Intervals: (-∞, 3), (3, 4), (4, ∞) Test point: -1, 3.5, 5 Expression: (-)(-)=+ (+)(-)= - (+)(+)=+ From the sign chart, we can see that the expression is less than or equal to 0 when x is between 3 and 4, inclusive. Therefore, the solution is: 3 ≤ x ≤ 4 So the correct option is: - 3 ≤ x ≤ 4

Read lesson note on Linear Inequalities (WAEC)

Read lesson note on Inequalities (JAMB)

Linear Inequalities

Inequalities

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