JAMB Mathematics Past Questions & Answers - Page 304

1,516.

The determinant of matrix \(\begin{pmatrix} x & 1 & 0 \\ 1-x & 2 & 3 \\ 1 & 1+x & 4\end{pmatrix}\) in terms of x is

A.

-3x2 - 17

B.

-3x2 + 9x - 1

C.

3x2 + 17

D.

3x2 - 9x + 5

Correct answer is B

\(\begin{vmatrix} x & 1 & 0 \\ 1-x & 2 & 3 \\ 1 & 1+x & 4\end{vmatrix}\) = x\(\begin{vmatrix}2 & 3 \\ 1+x & 4\end{vmatrix}\) - \(\begin{vmatrix}1-x & 3 \\ 1 & 4\end{vmatrix}\) = 0

= x[8 - 3(1 + x)] - [4(1 - x)-3] - 0 = x[5 - 3x] - [1 - 4x]

= 5x - 3x2 -1 + 4x

= -3x2 + 9X - 1

1,517.

The binary operation \(\oplus\) is defined by x \(\ast\) y = xy - y - x for all real values x and y. If x \(\ast\) 3 = 2\(\ast\), find x

A.

-1

B.

4

C.

1

D.

5

Correct answer is C

x \(\ast\) y = xy - y - x, x \(\ast\) 3 = 3x - 3 - x = 2x - 3

2 \(\ast\) x = 2x - x - 2 = x - 2

∴ 2x - 3 = x - 2

x = -2 + 3

= 1

1,518.

The identity element with respect to the multiplication shown in the diagram below is \(\begin{array}{c|c} \otimes & p & p & r & s \\ \hline p & r & p & r & p
\\ q & p & q & r & s\\ r & r & r & r & r\\ s & q & s & r & q\end{array}\)

A.

p

B.

q

C.

r

D.

s

Correct answer is B

The identity element (e) under an operation, say \(\otimes\), is the element such that for any given element under the operation, say a,

\(a \otimes e = e \otimes a = a\)

From the table, q is the identity element.

\(p \otimes q = q \otimes p = p\)

Same as all through.

1,519.

The sum of the first three terms of a geometric progression is half its sum to infinity. Find the positive common ratio of the progression.

A.

\(\frac{1}{4}\)

B.

\(\sqrt{\frac{3}{2}}\)

C.

\(\frac{1}{\sqrt{3}}\)

D.

\(\frac{1}{\sqrt{2}}\)

Correct answer is B

Let the G.p be a, ar, ar2, S3 = \(\frac{1}{2}\)S

a + ar + ar2 = \(\frac{1}{2}\)(\(\frac{a}{1 - r}\))

2(1 + r + r)(r - 1) = 1

= 2r3 = 3

= r3 = \(\frac{3}{2}\)

r(\(\frac{3}{2}\))\(\frac{1}{3}\) = \(\sqrt{\frac{3}{2}}\)

1,520.

If p + 1, 2P - 10, 1 - 4p2are three consecutive terms of an arithmetic progression, find the possible values of p

A.

-4, 2

B.

-3, \(\frac{4}{11}\)

C.

-\(\frac{4}{11}\), 2

D.

5, -3

Correct answer is C

2p - 10 = \(\frac{p + 1 + 1 - 4P^2}{2}\) (Arithmetic mean)

= 2(2p - 100 = p + 2 - 4P2)

= 4p - 20 = p + 2 - 4p2

= 4p2 + 3p - 22 = 0

= (p - 2)(4p + 11) = 0

∴ p = 2 or -\(\frac{4}{11}\)