Mathematics Past Questions - Page 104


Exam Body:

Correct answers are highlighted in green.

Question 1031

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

Explanation:

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

Question 1032

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

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

Options

A) 48

B) 24

C) -24

D) -48

Exam Body: JAMB

Explanation:

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

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

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

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

= 0 - 12 + 36 = 24

Question 1033

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

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

Options

A) PPT

B) pp-1

C) qp

D) pp

Exam Body: JAMB

Explanation:

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

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

pq = ppT

Question 1034

If the angles of quadrilateral are (P + 10)o(2P - 30)o(3P + 20)o and 4po, find p

Options

A) 63

B) 40

C) 36

D) 28

Exam Body: JAMB

Question 1035

Determine the distance on the earth's surface between two town P (latoN, Long 20oN) and Q(Lat 60oN, Long 25oW) (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


Question 1036

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

Options

A) 75o

B) 95o

C) 105o

D) 125o

Exam Body: JAMB

Question 1037

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) 240o

B) 210o

C) 150o

D) 60o

Exam Body: JAMB

Question 1038

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 90o which the straight line joining them makes with the horizontal

Exam Body: JAMB

Question 1039

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

Explanation:

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}\)

Question 1040

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

Explanation:

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