**Exam Body:**

If the binary operation \(\ast\) is defined by m \(\ast\) n = mn + m + n for any real number m and n, find the identity of the elements under this operation### Options

A) e = 1

B) e = -1

C) e = -2

D) e = 0

Exam Body: JAMB

m \(\ast\) n = mn + m + n

m \(\ast\) e = me + m + e, e \(\ast\) m = e

∴ me + m + e, m(e + 1)e - e = 0

e + 1 = 0

∴ e = -1

p = \(\begin{vmatrix} x & 3 & 0 \\ 2 & y & 3\\ 4 & 2 & 4 \end{vmatrix}\)### Options

Q = \(\begin{vmatrix} x & 2 & z \\ 3 & y & 2\\ 0 & 3 & z \end{vmatrix}\) Where p^{T} is the transpose P calculate /p^{T}/ when x = 0, y = 1 and z = 2

A) 48

B) 24

C) -24

D) -48

Exam Body: JAMB

p = \(\begin{vmatrix} 0 & 3 & 0 \\ 2 & 1 & 3\\ 4 & 2 & 2 \end{vmatrix}\)

P^{T} = \(\begin{vmatrix}0 & 2 & 4 \\ 2 & 1 & 3\\ 0 & 3 & 2 \end{vmatrix}\)

/p^{T}/ = \(\begin{vmatrix}0 & 2 & 4 \\ 3 & 1 & 3\\ 0 & 3 & 2 \end{vmatrix}\)

= 0[2 - 6] - 2[6 - 0] + 4[9 - 0]

= 0 - 12 + 36 = 24

p = \(\begin{vmatrix} x & 3 & 0 \\ 2 & y & 3\\ 4 & 2 & 4 \end{vmatrix}\)### Options

Q = \(\begin{vmatrix} x & 2 & z \\ 3 & y & 2\\ 0 & 3 & z \end{vmatrix}\)

PQ is equivalent to

A) PP^{T}

B) pp^{-1}

C) qp

D) pp

Exam Body: JAMB

p = \(\begin{vmatrix} 0 & 3 & 0 \\ 2 & 1 & 3\\ 4 & 2 & 2 \end{vmatrix}\)

Q = \(\begin{vmatrix} 0 & 2 & 4 \\ 3 & 1 & 2\\ 0 & 3 & 2 \end{vmatrix}\) = p^{T}

pq = pp^{T}

If the angles of quadrilateral are (P + 10)^{o}(2P - 30)^{o}(3P + 20)^{o} and 4p^{o}, find p### Options

A) 63

B) 40

C) 36

D) 28

Exam Body: JAMB

Determine the distance on the earth's surface between two town P (lat^{o}N, Long 20^{o}N) and Q(Lat 60^{o}N, Long 25^{o}W) (Radius of the earth = 6400km)### Options

A) \(\frac{800\pi}{9}\)km

B) \(\frac{800\sqrt{3\pi}}{9}\)km

C) 800\(\pi\) km

D) 800\(\sqrt{3\pi}\) km

Exam Body: JAMB

If in the diagram, FG is parallel to KM, find the value of x### Options

A) 75^{o}

B) 95^{o}

C) 105^{o}

D) 125^{o}

Exam Body: JAMB

x is a point due east of point Y on a coast Z is another point on the coast but 63m due south of y. If the distance ZX is 12Km. Calculate the bearing of Z from X### Options

A) 240^{o}

B) 210^{o}

C) 150^{o}

D) 60^{o}

Exam Body: JAMB

The locus of a point which is equidistant from two given fixed points is the### Options

A) perpendicular bisector of the straight line joining them

B) parallel line to the straight line joining them

C) transverse to the straight line joining them

D) angle bisector of 90^{o} which the straight line joining them makes with the horizontal

Exam Body: JAMB

What is the perpendicular distance of a point (2, 3) from the line 2x - 4y + 3 = 0?### Options

A) \(\frac{\sqrt{5}}{2}\)

B) \(\frac{\sqrt{5}}{20}\)

C) \(\frac{5}{\sqrt{13}}\)

D) 6

Exam Body: JAMB

2x - 4y + 3 = 0

Required distance = \(\frac{(2 \times 2) + 3(-4) + 3}{\sqrt{2^2} + (-4)^2}\)

= \(\frac{4 - 12 + 3}{\sqrt{20}}\)

= \(\frac{-5}{-2\sqrt{5}}\}

= -\{\frac{\sqrt{5}{2}\)

find then equation line through (5, 7) parallel to the line 7x + 5y = 12### Options

A) 5x + 7y = 120

B) 7x + 5y = 70

C) x + y = 7

D) 15x + 17y = 90

Exam Body: JAMB

Equation (5, 7) parallel to the line 7x + 5y = 12

5Y = -7x + 12

y = \(\frac{-7x}{5}\) + \(\frac{12}{5}\)

Gradient = \(\frac{-7}{5}\)

∴ Required equation = \(\frac{y - 7}{x - 5}\) = \(\frac{-7}{5}\) i.e. 5y - 35 = -7x + 35

5y + 7x = 70