JAMB Mathematics Past Questions & Answers - Page 264

1,316.

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

A.

63

B.

40

C.

36

D.

28

Correct answer is C

The sum of angles in a quadrilateral = 360°

\(\therefore (p + 10) + (2p - 30) + (3p + 20) + 4p = 360\)

\(10p = 360° \implies p = \frac{360}{10} = 36°\)

1,317.

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

A.

PPT

B.

pp-1

C.

qp

D.

pp

Correct answer is A

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

1,318.

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

A.

48

B.

24

C.

-24

D.

-48

Correct answer is B

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

1,319.

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

A.

e = 1

B.

e = -1

C.

e = -2

D.

e = 0

Correct answer is D

Identity(e) : a \(\ast\) e = a

m \(\ast\) e = m...(i)

m \(\ast\) e = me + m + e

Because m \(\ast\) e = m 

: m = me + m + e

m - m = e(m + 1)

e = \(\frac{0}{m + 1}\)

e = 0

1,320.

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

A.

e = 1

B.

e = -1

C.

e = -2

D.

e = 0

Correct answer is B

Identity(e) : a \(\ast\) e = a

m \(\ast\) e = m...(i)

m \(\ast\) e = me + m + e

Because m \(\ast\) e = m 

: m = me + m + e

m - m = e(m + 1)

e = \(\frac{0}{m + 1}\)

e = 0