0, 1
1, 2
2, 2
0, 2
Correct answer is A
x \(\ast\) y = xy
x \(\ast\) 2 = x2
x \(\ast\) 2 = x
∴ x2 - x = 0
x(x - 1) = 0
x = 0 or 1
3 < x < 4
-4 < x < -3
-2 < x < -1
-3 < x < 0
Correct answer is B
\(\frac{x + 1}{3}\) > \(\frac{1}{X + 3}\) = \(\frac{x + 1}{3}\) > \(\frac{x + 3}{X + 3}\)
= (x + 1)(x + 3)2 > 3(x + 3) = (x + 1)[x2 + 6x + 9] > 3(x + 3)
x3 + 7x2 + 15x + 9 > 3x + 9 = x3 + 7x2 + 12x > 0
= x(x + 3)9x + 4) > 0
Case 1 (+, +, +) = x > 0 , x + 3 > 0, x + 4 > 0
= x > -4 (solution only)
Case 2 (+, -, -) = x > 0, x + 4 < 0
= x > 0, x < -3, x < -4 = x < -3(solution only)
Case 3 (-, +, -) = x < 0, x > -3, x < -4 = x < -0, -4 < x < 3(solutions)
Case 4 (-, -, +) = x < 0, x + 3 < 0, x + 4 > 0
= x < 0, x < -5, x > -4 = x < -0, -4 < x < -3(solution)
combining the solutions -4 < x < -3
Solve the inequality y2 - 3y > 18
-3 < y < 6
y < -3 or y > 6
y > -3 or y > 6
y < 3 or y < 6
Correct answer is A
y2 - 3y > 18 = 3y - 18 > 0
y2 - 6y + 3y - 18 > 0 = y(y - 6) + 3 (y - 6) > 0
= (y + 3) (y - 6) > 0
Case 1 (+, +) \(\to\) (y + 3) > 0, (y - 6) > 0
= y > -3 y > 6
Case 2 (-, -) \(\to\) (y + 3) < 0, (y - 6) < 0
= y < -3, y < 6
Combining solution in case 1 and Case 2
= x < -3y < 6
= -3 < y < 6
Simplify \(\frac{1}{p}\) - \(\frac{1}{q}\) \(\div\) \(\frac{p}{q}\) - \(\frac{q}{p}\)
\(\frac{1}{p - q}\)
\(\frac{-1}{p + q}\)
\(\frac{1}{pq}\)
\(\frac{1}{pq(p - q)}\)
Correct answer is B
\(\frac{1}{p}\) - \(\frac{1}{q}\) \(\div\) \(\frac{p}{q}\) - \(\frac{q}{p}\) = \(\frac{q - p}{pq}\) ÷ \(\frac{p^2 - q^2}{pq}\)
\(\frac{q - p}{pq}\) x \(\frac{pq}{p^2q^2}\) = \(\frac{q - p}{p^2 - q^2}\)
\(\frac{-(p - q)}{(p + q)(p - q)}\)
= \(\frac{-1}{p + q}\)
Divide the expression x3 + 7x2 - x - 7 by -1 + x2
-x3 + 7x2 - x - 7
-x3 = 7x + 7
x - 7
x + 7
Correct answer is D
No explanation has been provided for this answer.