JAMB Mathematics Past Questions & Answers - Page 185

921.

Find the values of x for which the expression \(\frac{(x - 3)(x - 2)}{x^2 + x - 2}\)

A.

1, -2

B.

-1, 2

C.

2, 3

D.

-1. -2

E.

-2, -3

Correct answer is A

to find the values of x for which the expression is underlined, let x2 + x - 2 = 0

By factorizing, we have (x + 2)(x - 1) = 0

when x + 2 = 0, when x - 1 = 0, x = -2 or x = 1

The two values are -2 and 1

922.

Express 37.05 x 0.0042 in standard form

A.

15.561 x 102

B.

1.5561 x 10-4

C.

1.556 x 10-1

D.

1.5561 x 101

E.

1.55 x 101

Correct answer is C

37.05 x 0.0042 in standard form

\(\begin{array}{c|c}No. & log \\\hline 37.05 & 1.5688\\ 0.0042 & 3.6232 \\ \hline & 1.1920\end{array}\)

= 0.1556

= 1.556 x 10-1

923.

Solve for x, If \(\frac{\frac{2}{x}}{\frac{1}{p^2} + \frac{1}{p^2}}\) = m

A.

\(\frac{4pq}{m(p + q)}\)

B.

\(\frac{2p^2q^2}{m(q^2 + p^2)}\)

C.

\(\frac{2pq}{m(q^2 + p^2)}\)

D.

\(\frac{2p^2q^2}{m(p^2)}\)

Correct answer is B

\(\frac{1}{p^2}\) + \(\frac{1}{q^2}\) = \(\frac{q^2 + p^2}{p^2 + q^2}\)

\(\frac{\frac{2}{x}}{\frac{p^2 + q^2}{p^2 q^2}}\)

m = \(\frac{2p^2q^2}{x(p^2 + q^2)}\)

= m2p2q2 = m x (p2 + q2)

x = \(\frac{2p^2q^2}{m(q^2 + p^2)}\)

924.

The currency used in a country is called 'Matimalik'(M) and is of base seven. A lady in that country bought 4 bags of rice at M56 per bag and and 3 tins of milk at M4 per tin. What is the total cost of the item she bought?

A.

M2467

B.

M2427

C.

M2367

D.

M3417

E.

M3387

Correct answer is D

4 bags of rice - M 56 each

3 tins of milk - M 4 each

\(M 56 \times 4 = M 323\)

\(M 4 \times 3 = M 15\)

\(M (323 + 15) = M 341\)

 

925.

If sin \(\theta\) = \(\frac{m - n}{m + n}\); Find the value of 1 + tan2\(\theta\)

A.

\(\frac{(m^2 + n^2)}{m + n}\)

B.

\(\frac{(m^2 + n^2 + 2mn)}{4mn}\)

C.

\(\frac{2(m^2 + n^2 + mn)}{m + n}\)

D.

\(\frac{(m^2 + n^2 + mn)}{m + n}\)

Correct answer is B

\((m + n)^{2} = (m - n)^{2} + x^{2}\)

\(m^{2} + 2mn + n^{2} = m^{2} - 2mn + n^{2} + x^{2}\)

\(x^{2} = 4mn\)

\(x = \sqrt{4mn} = 2\sqrt{mn}\)

1 + tan2\(\theta\) = sec2\(\theta\)

= \(\frac{1}{cos^2\theta}\)

\(\cos \theta = \frac{2\sqrt{mn}}{(m + n)}\)

\(\frac{1}{\cos \theta} = \frac{(m + n)}{2\sqrt{mn}}\)

\(\sec^{2} \theta = \frac{(m + n)^{2}}{4mn}\)

= \(\frac{(m^2 + n^2 + 2mn)}{4mn}\)