Year : 
2019
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

71 - 80 of 80 Questions

# Question Ans
71.

If y = 8x\(^3\) - 3x\(^2\) + 7x - 1, find \(\frac{\mathrm d^2 y}{\mathrm d x^2}\).

A. 48x - 6

B. 11x\(^2\) + 6x - 7

C. 32x + 7

D. 24x\(^2\) - 6x + 7

Detailed Solution

y = 8x\(^3\) - 3x\(^2\) + 7x - 1
\(\frac{\mathrm d^2 y}{\mathrm d x^2} = \frac{\mathrm d}{\mathrm d x} (\frac{\mathrm d y}{\mathrm d x})\)
= \(\frac{\mathrm d}{\mathrm d x} (24x^2 - 6x + 7)\)
\(\frac{\mathrm d^2 y}{\mathrm d x^2} = 48x - 6\)
There is an explanation video available below.
72.

Differentiate \(\frac{2x}{\sin x}\) with respect to x.

A. \(2 \cot x \sec x (1 + \tan x)\)

B. \(2 \csc x - x \cot x\)

C. \(2x \csc x + \tan x\)

D. \(2\csc x(1 - x\cot x)\)

Detailed Solution

There is an explanation video available below.
73.

Find the distance between the points C(2, 2) and D(5, 6).

A. 13 units

B. 7 units

C. 12 units

D. 5 units

Detailed Solution

= \(\sqrt{(5 - 2)^2 + (6 - 2)^2}\)
= \(\sqrt{3^2 + 4^2}\)
= \(\sqrt{9 + 16}\)
= \(\sqrt{25}\)
= 5 units
There is an explanation video available below.
74.

Find the equation of a line perpendicular to the line 4y = 7x + 3 which passes through (-3, 1)

A. 7y + 4x + 5 = 0

B. 7y - 4x - 5 = 0

C. 3y - 5x + 2 = 0

D. 3y + 5x - 2 = 0

Detailed Solution

Equation: 4y = 7x + 3
\(\implies y = \frac{7}{4} x + \frac{3}{4}\)
Slope = coefficient of x = \(\frac{7}{4}\)
Slope of perpendicular line = \(\frac{-1}{\frac{7}{4}}\)
= \(\frac{-4}{7}\)
The perpendicular line passes (-3, 1)
\(\therefore\) Using the equation of line \(y = mx + b\)
m = slope and b = intercept.
\(y = \frac{-4}{7} x + b\)
To find the intercept, substitute y = 1 and x = -3 in the equation.
\(1 = \frac{-4}{7} (-3) + b\)
\(1 = \frac{12}{7} + b\)
\(b = \frac{-5}{7}\)
\(\therefore y = \frac{-4}{7} x - \frac{5}{7}\)
\(7y + 4x + 5 = 0\)
There is an explanation video available
75.

Table:
The table above is the distribution of data with mean equals to 3. Find the value of y.

A. 5

B. 2

C. 3

D. 6

Detailed Solution

Mean = \(\frac{\sum fx}{\sum f}\)
\(3 = \frac{26 - y}{3y + 2}\)
\(3(3y + 2) = 26 - y\)
\(9y + 6 = 26 - y\)
\(9y + y = 26 - 6\)
\(10y = 20 \implies y = 2\)
There is an explanation video available below.
76.

The histogram above represents the number of candidates who did Further Mathematics examination in a school. How many candidates scored more than 40?

A. 100

B. 79

C. 150

D. 90

Detailed Solution

Number of students that scored above 40 = 55 + 45 + 30 + 15 + 5
= 150 students.
There is an explanation video available below.
77.

In the circle above, with centre O and radius 7 cm. Find the length of the arc AB, when < AOB = 57°.

A. 5.32 cm

B. 4.39 cm

C. 7.33 cm

D. 6.97 cm

Detailed Solution

Length of arc = \(\frac{\theta}{360°} \times 2 \pi r\)
= \(\frac{57}{360} \times 2 \times \frac{22}{7} \times 7\)
= 6.97 cm
There is an explanation video available below.
78.

Calculate the volume of the regular three dimensional figure drawn above, where < ABC = 90° (a right- angled triangle).

A. 394 cm\(^3\)

B. 425 cm\(^3\)

C. 268 cm\(^3\)

D. 540 cm\(^3\)

Detailed Solution

|AC| = |DF| = 13 cm
Using Pythagoras theorem,
|AC|\(^2\) = |AB|\(^2\) + |BC|\(^2\)
13\(^2\) = 12\(^2\) + |BC|\(^2\)
|BC|\(^2\) = 169 - 144 = 25
|BC| = \(\sqrt{25}\)
= 5 cm
Volume of triangular prism = \(\frac{1}{2} \times base \times length \times height\)
= \(\frac{1}{2} \times 5 \times 12 \times 18\)
= 540 cm\(^3\)
There is an explanation video available below.
79.

If \(\sin x = \frac{4}{5}\), find \(\frac{1 + \cot^2 x}{\csc^2 x - 1}\).

A. \(\frac{13}{2}\)

B. \(\frac{25}{9}\)

C. \(\frac{3}{13}\)

D. \(\frac{4}{11}\)

Detailed Solution

\(\sin x = \frac{opp}{Hyp} = \frac{4}{5}\)
5\(^2\) = 4\(^2\) + adj\(^2\)
adj\(^2\) = 25 - 16 = 9
adj = \(\sqrt{9}\) = 3
\(\tan x = \frac{4}{3}\)
\(\cot x = \frac{1}{\frac{4}{3}} = \frac{3}{4}\)
\(\cot^2 x = (\frac{3}{4})^2 = \frac{9}{16}\)
\(\csc x = \frac{1}{\sin x}\)
= \(\frac{1}{\frac{4}{5}} = \frac{5}{4}\)
\(\csc^2 x = (\frac{5}{4})^2 = \frac{25}{16}\)
\(\therefore \frac{1 + \cot^2 x}{\csc^2 x - 1} = \frac{1 + \frac{9}{16}}{\frac{25}{16} - 1}\)
= \(\frac{25}{16} \div \frac{9}{16}\)
= \(\frac{25}{9}\)
There is an explanation video available below.
80.

Evaluate \(\frac{2\sin 30 + 5\tan 60}{\sin 60}\), leaving your answer in surd form.

A. \(\frac{2\sqrt{3}}{3} + 10\)

B. \(\frac{3\sqrt{2} - 1}{5}\)

C. \(\frac{3\sqrt{2} + 1}{5}\)

D. \(\frac{2\sqrt{3}}{3} - 10\)

Detailed Solution

\(\frac{2\sin 30 + 5\tan 60}{\sin 60}\)
\(\sin 30 = \frac{1}{2}; \tan 60 = \sqrt{3}; \sin 60 = \frac{\sqrt{3}}{2}\)
\(\therefore \frac{2\sin 30 + 5\tan 60}{\sin 60} = \frac{2(\frac{1}{2}) + 5(\sqrt{3})}{\frac{\sqrt{3}}{2}}\)\)
= \(\frac{1 + 5\sqrt{3}}{\frac{\sqrt{3}}{2}}\)
= \(\frac{2(1 + 5\sqrt{3})}{\sqrt{3}}\)
= \(\frac{2 + 10\sqrt{3}}{\sqrt{3}}\)
Rationalizing, we get
= \(\frac{2\sqrt{3} + 30}{3}\)
= \(\frac{2}{3} \sqrt{3} + 10\)
There is an explanation video available below.
71.

If y = 8x\(^3\) - 3x\(^2\) + 7x - 1, find \(\frac{\mathrm d^2 y}{\mathrm d x^2}\).

A. 48x - 6

B. 11x\(^2\) + 6x - 7

C. 32x + 7

D. 24x\(^2\) - 6x + 7

Detailed Solution

y = 8x\(^3\) - 3x\(^2\) + 7x - 1
\(\frac{\mathrm d^2 y}{\mathrm d x^2} = \frac{\mathrm d}{\mathrm d x} (\frac{\mathrm d y}{\mathrm d x})\)
= \(\frac{\mathrm d}{\mathrm d x} (24x^2 - 6x + 7)\)
\(\frac{\mathrm d^2 y}{\mathrm d x^2} = 48x - 6\)
There is an explanation video available below.
72.

Differentiate \(\frac{2x}{\sin x}\) with respect to x.

A. \(2 \cot x \sec x (1 + \tan x)\)

B. \(2 \csc x - x \cot x\)

C. \(2x \csc x + \tan x\)

D. \(2\csc x(1 - x\cot x)\)

Detailed Solution

There is an explanation video available below.
73.

Find the distance between the points C(2, 2) and D(5, 6).

A. 13 units

B. 7 units

C. 12 units

D. 5 units

Detailed Solution

= \(\sqrt{(5 - 2)^2 + (6 - 2)^2}\)
= \(\sqrt{3^2 + 4^2}\)
= \(\sqrt{9 + 16}\)
= \(\sqrt{25}\)
= 5 units
There is an explanation video available below.
74.

Find the equation of a line perpendicular to the line 4y = 7x + 3 which passes through (-3, 1)

A. 7y + 4x + 5 = 0

B. 7y - 4x - 5 = 0

C. 3y - 5x + 2 = 0

D. 3y + 5x - 2 = 0

Detailed Solution

Equation: 4y = 7x + 3
\(\implies y = \frac{7}{4} x + \frac{3}{4}\)
Slope = coefficient of x = \(\frac{7}{4}\)
Slope of perpendicular line = \(\frac{-1}{\frac{7}{4}}\)
= \(\frac{-4}{7}\)
The perpendicular line passes (-3, 1)
\(\therefore\) Using the equation of line \(y = mx + b\)
m = slope and b = intercept.
\(y = \frac{-4}{7} x + b\)
To find the intercept, substitute y = 1 and x = -3 in the equation.
\(1 = \frac{-4}{7} (-3) + b\)
\(1 = \frac{12}{7} + b\)
\(b = \frac{-5}{7}\)
\(\therefore y = \frac{-4}{7} x - \frac{5}{7}\)
\(7y + 4x + 5 = 0\)
There is an explanation video available
75.

Table:
The table above is the distribution of data with mean equals to 3. Find the value of y.

A. 5

B. 2

C. 3

D. 6

Detailed Solution

Mean = \(\frac{\sum fx}{\sum f}\)
\(3 = \frac{26 - y}{3y + 2}\)
\(3(3y + 2) = 26 - y\)
\(9y + 6 = 26 - y\)
\(9y + y = 26 - 6\)
\(10y = 20 \implies y = 2\)
There is an explanation video available below.
76.

The histogram above represents the number of candidates who did Further Mathematics examination in a school. How many candidates scored more than 40?

A. 100

B. 79

C. 150

D. 90

Detailed Solution

Number of students that scored above 40 = 55 + 45 + 30 + 15 + 5
= 150 students.
There is an explanation video available below.
77.

In the circle above, with centre O and radius 7 cm. Find the length of the arc AB, when < AOB = 57°.

A. 5.32 cm

B. 4.39 cm

C. 7.33 cm

D. 6.97 cm

Detailed Solution

Length of arc = \(\frac{\theta}{360°} \times 2 \pi r\)
= \(\frac{57}{360} \times 2 \times \frac{22}{7} \times 7\)
= 6.97 cm
There is an explanation video available below.
78.

Calculate the volume of the regular three dimensional figure drawn above, where < ABC = 90° (a right- angled triangle).

A. 394 cm\(^3\)

B. 425 cm\(^3\)

C. 268 cm\(^3\)

D. 540 cm\(^3\)

Detailed Solution

|AC| = |DF| = 13 cm
Using Pythagoras theorem,
|AC|\(^2\) = |AB|\(^2\) + |BC|\(^2\)
13\(^2\) = 12\(^2\) + |BC|\(^2\)
|BC|\(^2\) = 169 - 144 = 25
|BC| = \(\sqrt{25}\)
= 5 cm
Volume of triangular prism = \(\frac{1}{2} \times base \times length \times height\)
= \(\frac{1}{2} \times 5 \times 12 \times 18\)
= 540 cm\(^3\)
There is an explanation video available below.
79.

If \(\sin x = \frac{4}{5}\), find \(\frac{1 + \cot^2 x}{\csc^2 x - 1}\).

A. \(\frac{13}{2}\)

B. \(\frac{25}{9}\)

C. \(\frac{3}{13}\)

D. \(\frac{4}{11}\)

Detailed Solution

\(\sin x = \frac{opp}{Hyp} = \frac{4}{5}\)
5\(^2\) = 4\(^2\) + adj\(^2\)
adj\(^2\) = 25 - 16 = 9
adj = \(\sqrt{9}\) = 3
\(\tan x = \frac{4}{3}\)
\(\cot x = \frac{1}{\frac{4}{3}} = \frac{3}{4}\)
\(\cot^2 x = (\frac{3}{4})^2 = \frac{9}{16}\)
\(\csc x = \frac{1}{\sin x}\)
= \(\frac{1}{\frac{4}{5}} = \frac{5}{4}\)
\(\csc^2 x = (\frac{5}{4})^2 = \frac{25}{16}\)
\(\therefore \frac{1 + \cot^2 x}{\csc^2 x - 1} = \frac{1 + \frac{9}{16}}{\frac{25}{16} - 1}\)
= \(\frac{25}{16} \div \frac{9}{16}\)
= \(\frac{25}{9}\)
There is an explanation video available below.
80.

Evaluate \(\frac{2\sin 30 + 5\tan 60}{\sin 60}\), leaving your answer in surd form.

A. \(\frac{2\sqrt{3}}{3} + 10\)

B. \(\frac{3\sqrt{2} - 1}{5}\)

C. \(\frac{3\sqrt{2} + 1}{5}\)

D. \(\frac{2\sqrt{3}}{3} - 10\)

Detailed Solution

\(\frac{2\sin 30 + 5\tan 60}{\sin 60}\)
\(\sin 30 = \frac{1}{2}; \tan 60 = \sqrt{3}; \sin 60 = \frac{\sqrt{3}}{2}\)
\(\therefore \frac{2\sin 30 + 5\tan 60}{\sin 60} = \frac{2(\frac{1}{2}) + 5(\sqrt{3})}{\frac{\sqrt{3}}{2}}\)\)
= \(\frac{1 + 5\sqrt{3}}{\frac{\sqrt{3}}{2}}\)
= \(\frac{2(1 + 5\sqrt{3})}{\sqrt{3}}\)
= \(\frac{2 + 10\sqrt{3}}{\sqrt{3}}\)
Rationalizing, we get
= \(\frac{2\sqrt{3} + 30}{3}\)
= \(\frac{2}{3} \sqrt{3} + 10\)
There is an explanation video available below.