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JAMB UTME - Mathematics - 2014

Question 1 Report

If y = cos 3x, find \( \frac{\delta y}{\delta x} \)

\frac{1}{3} \sin 3 x

- \frac{1}{3} \sin 3 x

3 sin 3x

-3 sin 3x

Answer Details

To find the derivative of y = cos 3x, we need to use the chain rule of differentiation. The chain rule states that if y = f(g(x)), then the derivative of y with respect to x is given by the product of the derivative of f with respect to g multiplied by the derivative of g with respect to x. In other words, δy/δx = δf/δg * δg/δx. Using the chain rule, we have: δy/δx = δ(cos 3x)/δ(3x) * δ(3x)/δx The derivative of cos 3x with respect to 3x can be found using the chain rule again: δ(cos 3x)/δ(3x) = -sin(3x) The derivative of 3x with respect to x is simply 3. Substituting these values in the original equation, we get: δy/δx = -sin(3x) * 3 Simplifying, we have: δy/δx = -3 sin(3x) Therefore, the correct option is -3 sin 3x. In summary, the derivative of y = cos 3x is -3 sin 3x, which is obtained using the chain rule of differentiation.

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Sine, Cosine And Tangent Of An Angle

Trigonometry

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Question 2 Report

If P = {1,2,3,4,5} and \( \cup \) Q = {1,2,3,4,5,6,7}, list the elements in Q

{6}

{7}

{6,7}

{5,6}

Answer Details

Q = (P

\cup

Q) - P
{6,7}

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Question 5 Report

Find y, if

\[ \begin{pmatrix} 5 & -6 \\ 2 & -7 \end{pmatrix} \begin{pmatrix} 5 \\ 2 \end{pmatrix} = \begin{pmatrix} 7 \\ -11 \end{pmatrix} \]

8

5

3

2

Answer Details

(\begin{matrix} 5 & - 6 \\ 2 & - 7 \end{matrix}) (\begin{matrix} 5 \\ 2 \end{matrix}) = (\begin{matrix} 7 \\ - 11 \end{matrix})

By matrices multiplication;
5x - 6y = 7 ........(1)
2x - 7y = -11 ......(2)
2 x (1): 10x - 12y = 14 .......(3)
5 x (2): 10x - 35y = -55 ......(4)
(3) - (4): 23y = 69
y = 69/23 = 3

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Solution Of Linear Equations

Modular Arithmetic

Application Of Differentiation

Quadratic Equations

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Question 7 Report

P varies directly as Q and inversely as R. When Q = 36 and R = 16, P = 27. Find the relation between P, Q and R.

P = \frac{Q}{12 R}

P = \frac{12 Q}{R}

P = 12 Q R

P = \frac{12}{Q R}

Answer Details

P \propto \frac{Q}{R}

P = K \frac{Q}{R}

When Q = 36, R = 16, P = 27
Then substitute into the equation

27 = K \frac{36}{16}

$K = \frac{27 \times 16}{36}$
$K = 12$
So the equation connecting P, Q and R is
$P = \frac{12 Q}{R}$

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Simple Operations On Algebraic Expressions

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Question 8 Report

Evaluate the inequality \( \frac{x}{2} + \frac{3}{4} \leq \frac{5x}{6} - \frac{7}{12} \)

x \geq 4

x≤3

x≥−3

x≤−4

Answer Details

\frac{x}{2} + \frac{3}{4} \leq \frac{5 x}{6} - \frac{7}{12}

12 \frac{x}{2} + 12 \frac{3}{4} \leq 12 \frac{5 x}{6} - 12 \frac{7}{12}

6x + 9

\leq

10x - 7
6x - 10x

\leq

- 7 - 9
-4x

\leq

-16
-4x/-4

\geq

-16/-4
x

\geq

4

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Inequalities

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Question 9 Report

In how many ways can a team of 3 girls be selected from 7 girls?

\frac{7!}{3!}

\frac{7!}{4!}

\frac{7!}{3! 4!}

\frac{7!}{2! 5!}

Answer Details

A team of 2 girls can be selected from 7 girls in

^{7} C_{3}

= \frac{7!}{(7 - 3)! 3!}

= \frac{7!}{4! 3!} w a y s

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Permutation And Combination

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Question 10 Report

Evaluate \( \int \sin 2x \, dx \)

cos 2x + k

\frac{1}{2}

cos 2x + k

- \frac{1}{2}

cos 2x + k

-cos 2x + k

Answer Details

The value of the integral of sin(2x)dx is -(1/2)cos(2x) + k, where k is an arbitrary constant of integration. The integral of sin(2x) can be found using substitution or by recognizing that sin(2x) is the derivative of -(1/2)cos(2x). The constant of integration k is added to account for the fact that there are infinitely many functions that have the same derivative as sin(2x). The constant can take any value and is introduced to reflect the inherent uncertainty in finding an antiderivative.

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Integration

Trigonometry

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Question 11 Report

In the figure above, KL//NM, LN bisects < KNM. If angles KLN is \(54^\circ\) and angle MKN is \(35^\circ\), calculate the size of angle KMN.

91

^{\circ}

89

^{\circ}

37

^{\circ}

19

^{\circ}

Answer Details

In the diagram above,

α

= 54

^{\circ}

(alternate angles; KL||MN) < KNM = 2

α

(LN is bisector of < KNM) = 108

^{\circ}

35 $^{\circ}$ + < KMN + 108 $^{\circ}$ = 180 $^{\circ}$ (sum of angles of $△$ )

< KMN + 143 $^{\circ}$ = 180 $^{\circ}$

< KMN = 180 $^{\circ}$ - 143 $^{\circ}$

= 37 $^{\circ}$

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Angles And Intercepts On Parallel Lines

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Question 12 Report

Simplify \( \frac{2\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}} \)

3

\sqrt{6} - 7

3

\sqrt{6} - 7

3

\sqrt{6} - 1

3

\sqrt{6} + 1

Answer Details

= \frac{2 \sqrt{2} - \sqrt{3}}{\sqrt{2} + \sqrt{3}} \times \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} + \sqrt{3}}

= \frac{2 \sqrt{2} (\sqrt{2}) + (2 \sqrt{2}) (- \sqrt{3}) - \sqrt{3} (\sqrt{2}) - \sqrt{3} (- \sqrt{3})}{(\sqrt{2})^{2} - (\sqrt{3})^{2}}

= \frac{2 \times 2 - 2 \sqrt{6} - \sqrt{6} + 3}{2 - 3}

= \frac{4 - 3 \sqrt{6} + 3}{- 1}

= \frac{7 - 3 \sqrt{6}}{- 1}

= \frac{7}{- 1} - \frac{3 \sqrt{6}}{- 1}

= - 7 + 3 \sqrt{6}

= 3 \sqrt{6} - 7

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Functions And Relations

Surds (radicals)

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Question 15 Report

Evaluate Log28 + Log216 - Log24

3

4

5

6

Answer Details

= l o g_{2} \frac{8 \times 16}{4}

= l o g_{2} 32

= l o g_{2} 2^{5}

= 5 l o g_{2} 2

= 5 \times 1

= 5

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Question 18 Report

Find the value of

\[ \begin{vmatrix} 0 & 3 & 2 \\ 1 & 7 & 8 \\ 0 & 5 & 4 \end{vmatrix} \]

12

10

-1

-2

Answer Details

0 | \begin{matrix} 7 & 8 \\ 5 & 4 \end{matrix} | - 3 | \begin{matrix} 1 & 8 \\ 0 & 4 \end{matrix} | + 2 | \begin{matrix} 1 & 7 \\ 0 & 5 \end{matrix} |

= 0(28 - 40) - 3(4 - 0) + 2(5 - 0)
= 0(-12) - 3(4) + 2(5)
= 0 - 12 + 10
= -2

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Number Bases

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Question 19 Report

If \( \sin \theta = \frac{12}{13} \), find the value of \( 1 + \cos \theta \)

\frac{25}{13}

\frac{18}{13}

\frac{8}{13}

\frac{5}{13}

Sine, Cosine And Tangent Of An Angle

Number Bases

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Question 21 Report

A number is chosen at random from 10 to 30 both inclusive. What is the probability that the number is divisible by 3?

\frac{2}{15}

\frac{1}{10}

\frac{1}{3}

\frac{2}{5}

Answer Details

Sample space S = {10, 11, 12, ... 30}
Let E denote the event of choosing a number divisible by 3
Then E = {12, 15, 18, 21, 24, 27, 30} and n(E) = 7
Prob (E) =

\frac{n (E)}{n (E)}

Prob (E) =

\frac{7}{21}

Prob (E) =

\frac{1}{3}

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Probability

Surds (radicals)

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Question 27 Report

If \(y = 4x^3 - 2x^2 + x\), find \(\frac{\delta y}{\delta x}\)

8x2 - 2x + 1

8x2 - 4x + 1

12x2 - 2x + 1

12x2 - 4x + 1

Answer Details

If y = 4x3 - 2x2 + x, then;

\frac{δ y}{δ x}

= 3(4x2) - 2(2x) + 1
= 12x2 - 4x + 1

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Algebraic Expressions

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Question 28 Report

A cylindrical tank has a capacity of 6160m3. What is the depth of the tank if the radius of its base is 28cm?

8.0m

7.5m

5.0m

2.5m

Answer Details

Using

V = π r^{2} h

6160 = 22/7 x 28 x 28 x h

h = \frac{6160}{22 \times 4 \times 28}

h = 2.5 m

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Question 31 Report

From the Venn diagram above, the shaded parts represent

(P

\cap

Q)

\cup

(P

\cap

R)

(P

\cup

Q)

\cap

(P

\cap

R)

(P

\cup

Q)

\cup

(P

\cup

R)

(P

\cap

Q)

\cup

(P

\cup

R)

Answer Details

The Venn diagram above represents three sets: P, Q, and R. The shaded parts represent the elements that are in both P and Q (the intersection of P and Q), and the elements that are in both P and R (the intersection of P and R). Therefore, the shaded parts represent the set (P∩Q) and the set (P∩R). Option (A) (P∩Q)∪(P∩R) is the correct answer.

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Probability

Linear Inequalities

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Question 34 Report

If ∣∣∣−x12−14∣∣∣=−12, find x

-6

-2

3

6

Answer Details

Explanation:

| \begin{matrix} - x & 12 \\ - 1 & 4 \end{matrix} | = - 12

-4x - (-1)12 = -12
-4x + 12 = -12
-4x = -12 - 12
-4x = - 24
x = 6

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Question 35 Report

The gradient of a line joining \((x,4)\) and \((1,2)\) is \(\frac{1}{2}\). Find the value of \(x\)

5

3

-3

-5

Answer Details

Gradient m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

\frac{1}{2} = \frac{2 - 4}{1 - x}

1 - x = 2(2 - 4)
1 - x = 4 - 8
1 - x = -4
-x = -4 - 1
x = 5

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Question 36 Report

The mean of 2 - 4, 4 + t, 3 - 2t and t - 1 is

t

-t

2

-2

Answer Details

Mean x =

\frac{\sum x}{n}

= [(2 - t) + (4 + t) + (3 - 2t) + (2 + t) + (t - 1)

\div

] 5
= [11 - 1 + 3t - 3t]

\div

5
= 10

\div

5
= 2

View Answer

Question 39 Report

Calculate the mid point of the line segment y - 4x + 3 = 0, which lies between the x-axis and y-axis.

(\begin{matrix} 3 & - 3 \\ 8 & 2 \end{matrix})

(\begin{matrix} 3 & - 3 \\ 8 & 2 \end{matrix})

(\begin{matrix} 3 & 3 \\ 8 & 2 \end{matrix})

(\begin{matrix} - 2 & 2 \\ 2 & 2 \end{matrix})

(\begin{matrix} - 2 & 3 \\ 3 & 2 \end{matrix})

Answer Details

y - 4x + 3 = 0
When y = 0, 0 - 4x + 3 = 0
Then -4x = -3
x = 3/4
So the line cuts the x-axis at point (3/4, 0).
When x = 0, y - 4(0) + 3 = 0
Then y + 3 = 0
y = -3
So the line cuts the y-axis at the point (0, 3)
Hence the midpoint of the line y - 4x + 3 = 0, which lies between the x-axis and the y-axis is;

[\frac{1}{2} (x_{1} + x_{2}), \frac{1}{2} (y_{1} + y_{2})]

[\frac{1}{2} (\frac{3}{4} + 0), \frac{1}{2} (0 + - 3)]

[\frac{1}{2} (\frac{3}{4}), \frac{1}{2} (- 3)]

[\frac{3}{8}, \frac{- 3}{2}]

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Question 40 Report

What is the common ratio of the G.P. \( (\sqrt{10}+\sqrt{5})+(\sqrt{10}+2\sqrt{5})+\ldots \)?

\sqrt{2}

√5

3

5

Answer Details

Common ratio r of the G.P is

r = \frac{T_{n} + 1}{T_{n}} = \frac{T_{2}}{T_{1}}

r = \frac{\sqrt{10} + 2 \sqrt{5}}{\sqrt{10} + \sqrt{5}}

r = \frac{\sqrt{10} + 2 \sqrt{5}}{\sqrt{10} + \sqrt{5}} \times \frac{\sqrt{10} - \sqrt{5}}{\sqrt{10} - \sqrt{5}}

= \frac{(\sqrt{10}) (\sqrt{10}) + (\sqrt{10}) (- \sqrt{5}) + (2 \sqrt{5}) (\sqrt{10}) + (2 \sqrt{5}) (- \sqrt{5})}{(\sqrt{10})^{2} - (\sqrt{5})^{2}}

\frac{10 - \sqrt{50} + 2 \sqrt{50} - 10}{10 - 5}

\frac{\sqrt{50}}{5}

\frac{\sqrt{25 \times 2}}{5}

\frac{5 \sqrt{2}}{5}

\sqrt{2}

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Surds (radicals)

Sequence And Series

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Question 43 Report

If gt2 - k - w = 0, make g the subject of the formula

\frac{k + w}{t^{2}}

\frac{k - w}{t^{2}}

\frac{k + w}{t}

\frac{k - w}{t}

Answer Details

We can solve for g by manipulating the given equation, gt² - k - w = 0, to isolate g on one side of the equation. First, we can add k and w to both sides of the equation to obtain: gt² = k + w Next, we can divide both sides of the equation by t² to solve for g: g = (k + w)/t² Therefore, the solution is: g = (k + w)/t² Hence, the answer is: (k + w)/t²

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Change Of Subject Of A Formula/Relation

Solution Of Linear Equations

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Question 46 Report

Find the value of x in the figure above.

20

\sqrt{3}

cm

10

\sqrt{3}

cm

5

\sqrt{3}

cm

4

\sqrt{3}

cm

Answer Details

The figure above is a right triangle with sides x, 3 cm, and x√3 cm. Since it is a right triangle, we can use the Pythagorean theorem to find the value of x. The theorem states that the sum of the squares of the two shorter sides of a right triangle is equal to the square of the longest side, which is the hypotenuse. So, we have: x^2 + (3)^2 = (x√3)^2 Expanding the square on the right side: x^2 + 9 = x^2 * 3 Simplifying the equation: x^2 + 9 = 3x^2 Solving for x: 2x^2 = 9 x^2 = 4.5 Taking the square root of both sides: x = ±√(4.5) Since x has to be positive, we choose the positive square root: x = √(4.5) = 2√3 Finally, multiplying by 10: x = 10√3 So, the value of x in the figure is 10√3 cm.

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Surds (radicals)

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Question 47 Report

Evaluate \( \int (2x+3)^{\frac{1}{2}}\delta x \)

\frac{1}{12} (2 x + 3)^{6} + k

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

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

\frac{1}{12} (2 x + 3)^{\frac{3}{4}} + k

Answer Details

\int (2 x + 3)^{\frac{1}{2}} δ x

let u = 2x + 3,

\frac{δ y}{δ x} = 2

δ x = \frac{δ u}{2}

Now

\int (2 x + 3)^{\frac{1}{2}} δ x = \int u^{\frac{1}{2}} . \frac{δ x}{2}

= \frac{1}{2} \int u^{\frac{1}{2}} δ u

= \frac{1}{2} u^{\frac{3}{2}} \times \frac{2}{3} + k

= \frac{1}{3} u^{\frac{3}{2}} + k

= \frac{1}{3} (2 x + 3)^{\frac{3}{2}} + k

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Integration

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Question 48 Report

Numbers	1	2	3	4	5	6
Frequency	18	22	20	16	10	14

The table above represents the outcome of throwing a die 100 times. What is the probability of obtaining at least a 4?

\frac{1}{5}

\frac{1}{2}

\frac{2}{5}

\frac{3}{4}

Answer Details

Let E demote the event of obtaining at least a 4
Then n(E) = 16 + 10 + 14 = 40
Hence, prob (E) =

\frac{n (E)}{n (S)}

= \frac{40}{100}

= \frac{2}{5}

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Representation Of Data

Probability

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