JAMB Mathematics Past Questions & Answers - Page 290

1,446.

If x - 1 and x + 1 are both factors of the equation x\(^3\) + px\(^2\) + qx + 6 = 0, evaluate p and q

A.

-6, -1

B.

6, 1

C.

1, -1

D.

6, -6

Correct answer is A

\(x^{3} + px^{2} + qx + 6 = 0\)

f(x - 1) = 0; f(1) = 0.

\(1^{3} + p(1^{2}) + q(1) + 6 = 0 \implies p + q = -7 ... (1)\)

f(x + 1) = 0; f(-1) = 0.

\((-1)^{3} + p(-1^{2}) + q(-1) + 6 = 0 \implies p - q = -5 ... (2)\)

Subtract (2) from (1).

\(2q = -2 \implies q = -1\)

\(p - (-1) = -5 \implies p = -5 - 1 = -6\)

\((p, q) = (-6, -1)\)

1,447.

If S = (x : x\(^2\) = 9, x > 4), then S is equal to

A.

4

B.

{0}

C.

\(\emptyset\)

D.

{\(\emptyset\)}

Correct answer is C

No explanation has been provided for this answer.

1,448.

Four members of a school first eleven cricket team are also members of the first fourteen rugby team. How many boys play for at least one of the two teams?

A.

25

B.

21

C.

16

D.

3

Correct answer is B

Number of people playing both rugby and cricket = 4 Number that play cricket only = 11 - 4 = 7 Number that play rugby only = 14 - 4 = 10. Number that play for at least one of the teams = 4 + 7 + 10 = 21.

1,449.

Simplify \(\frac{\sqrt{12} - \sqrt{3}}{\sqrt{12} + \sqrt{3}}\)

A.

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

B.

9

C.

16cm

D.

3

Correct answer is A

\(\frac{\sqrt{12} - \sqrt{3}}{\sqrt{12} + \sqrt{3}}\)

\(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\)

\(\therefore \frac{\sqrt{12} - \sqrt{3}}{\sqrt{12} + \sqrt{3}} = \frac{2\sqrt{3} - \sqrt{3}}{2\sqrt{3} + \sqrt{3}}\)

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

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

1,450.

Evaluate \(\frac{(81)^{\frac{3}{4}} - (27)^{\frac{1}{3}}}{3 \times 2^3}\)

A.

27

B.

1

C.

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

D.

\(\frac{1}{8}\)

Correct answer is B

\(\frac{(81)^{\frac{3}{4}} - (27)^{\frac{1}{3}}}{3 \times 2^3}\)

= \(\frac{(3^{4})^{\frac{3}{4}} - (3^{3})^{\frac{1}{3}}}{3 \times 2^{3}}\)

= \(\frac{3^{3} - 3^{1}}{24}\)

= \(\frac{27 - 3}{24} = 1\)