Which of the following binary operations is cumulative on the set of integers?
a \(\ast\) b = a + 2b
a \(\ast\) b = a + b - ab
a \(\ast\) b = a2 + b
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.
Express \(\frac{5x - 12}{(x - 2)(x - 3)}\) in partial fractions
\(\frac{2}{x + 2} - \frac{3}{x - 3}\)
\(\frac{2}{x - 2} + \frac{3}{x - 3}\)
\(\frac{2}{x - 3} - \frac{3}{x - 2}\)
\(\frac{5}{x - 3} - \frac{4}{x - 2}\)
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), \(A = 5 - B\)
\(3(5 - B) + 2B = 12\)
\(15 - 3B + 2B = 12 \implies B = 3\)
\(A + 3 = 5 \implies A = 2\)
\(\frac{5x - 12}{(x - 2)(x - 3)} = \frac{2}{x - 2} + \frac{3}{x - 3}\)
Simplify \(\frac{x^2 - 1}{x^3 + 2x^2 - x - 2}\)
\(\frac{1}{x + 2}\)
\(\frac{x - 1}{x + 1}\)
\(\frac{x - 1}{x + 2}\)
\(\frac{1}{x - 2}\)
Correct answer is A
\(\frac{x^2 - 1}{(x - 1)(x + 2)(x + 1)}\) = \(\frac{(x -
1)(x + 1)}{(x - 1)(x + 2)(x + 1)}\)
= \(\frac{1}{x + 2}\)
The 4th term of an A.P. is 13 while the 10th term is 31. Find the 21st term.
175
85
64
45
Correct answer is C
a + 3d = 13
a + 9d = 31
6d = 18
= d = 3
a = 13 - 9
= 4
a + 20d = 4 + (20 x 3)
= 64
Solve the inequality (x - 3)(x - 4) \(\leq\) 0
3 \(\leq\) x \(\leq\) 4
3 < x < 4
3 \(\leq\) x < 4
3 < x \(\leq\) 4
Correct answer is A
(x - 3)(x - 4) \(\leq\) 0
Case 1 (+, -) = x - 3 \(\geq\) 0, X - 4 \(\geq\) 0
= X \(\leq\) 3, x \(\geq\) 4
= 3 < x \(\geq\) 4 (solution)
Case 2 = (-, +) = x - 3 \(\leq\) 0, x - 4 \(\geq\) 0
= x \(\leq\) 3, x \(\geq\) 4
therefore = 3 \(\leq\) x \(\leq\) 4