JAMB Mathematics Past Questions & Answers - Page 288

1,436.

Find the sum to infinity of the following sequence \(1, \frac{9}{10}, (\frac{9}{10})^{2}, (\frac{9}{10})^{3}\)

A.

\(\frac{1}{10}\)

B.

\(\frac{9}{10}\)

C.

\(\frac{10}{9}\)

D.

10

Correct answer is D

\(S_{\infty} = \frac{a}{1 - r}\) (Sum to infinity of a G.P)

\(a = 1; r = \frac{9}{10}\)

\(S_{\infty} = \frac{1}{1 - \frac{9}{10}}\)

= \(\frac{1}{\frac{1}{10}} = 10\)

1,437.

If a \(\ast\) b = + \(\sqrt{ab}\), evaluate 2 \(\ast\)(12 \(\ast\) 27)

A.

12

B.

9

C.

6

D.

2

Correct answer is C

\(2 \ast (12 \ast 27)\)

\(12 \ast 27 = + \sqrt{12 \times 27}\)

= \(+ \sqrt{324} = 18\)

\(2 \ast 18 = + \sqrt{2 \times 18}\)

= \(+ \sqrt{36} = 6\)

1,438.

Which of the following binary operations is cumulative on the set of integers?

A.

a \(\ast\) b = a + 2b

B.

a \(\ast\) b = a + b - ab

C.

a \(\ast\) b = a2 + b

D.

a \(\ast\) b = \(\frac{a(b + 1)}{2}\)

Correct answer is B

\(a \ast b = a + b - ab\)

\(b \ast a = b + a - ba\)

On the set of integers, the two above are cumulative as multiplication and addition are cumulative on the set of integers.

1,439.

Express \(\frac{5x - 12}{(x - 2)(x - 3)}\) in partial fractions

A.

\(\frac{2}{x + 2} - \frac{3}{x - 3}\)

B.

\(\frac{2}{x - 2} + \frac{3}{x - 3}\)

C.

\(\frac{2}{x - 3} - \frac{3}{x - 2}\)

D.

\(\frac{5}{x - 3} - \frac{4}{x - 2}\)

Correct answer is B

\(\frac{5x - 12}{(x - 2)(x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3}\)

= \(\frac{A(x - 3) + B(x - 2)}{(x - 2)(x - 3)}\)

\(\implies 5x - 12 = Ax - 3A + Bx - 2B\)

\(A + B = 5 ... (i)\)

\(-(3A + 2B) = -12 \implies 3A + 2B = 12 ... (ii)\)

From (i), \(A = 5 - B\)

\(3(5 - B) + 2B = 12\)

\(15 - 3B + 2B = 12 \implies B = 3\)

\(A + 3 = 5 \implies A = 2\)

\(\frac{5x - 12}{(x - 2)(x - 3)} = \frac{2}{x - 2} + \frac{3}{x - 3}\)

1,440.

Simplify \(\frac{x^2 - 1}{x^3 + 2x^2 - x - 2}\)

A.

\(\frac{1}{x + 2}\)

B.

\(\frac{x - 1}{x + 1}\)

C.

\(\frac{x - 1}{x + 2}\)

D.

\(\frac{1}{x - 2}\)

Correct answer is A

\(\frac{x^2 - 1}{(x - 1)(x + 2)(x + 1)}\) = \(\frac{(x -
1)(x + 1)}{(x - 1)(x + 2)(x + 1)}\)

= \(\frac{1}{x + 2}\)