Find the equation of the line passing through (0, -1) and parallel to the y- axis.
y = -1
y = 0
x = 0
x = -1
Correct answer is C
No explanation has been provided for this answer.
Find the coefficient of \(x^{4}\) in the binomial expansion of \((1 - 2x)^{6}\).
320
240
-320
-240
Correct answer is B
\((1 - 2x)^{6} = ^{6}C_{0}(1^{6}) + ^{6}C_{1}(1^{5})(-2x^{1}) + ^{6}C_{2}(1^{4})(-2x^{2}) + ^{6}C_{3}(1^{3})(-2x^{3}) + ^{6}C_{4}(1^{2})(-2x^{4}) + ...\)
\(\text{The coefficient of} x^{4} = 15 \times 16 = 240\)
If \(f(x) = \frac{1}{2 - x}, x \neq 2\), find \(f^{-1}(-\frac{1}{2})\).
4
0
-2
-4
Correct answer is A
\(f(x) = \frac{1}{2 - x}, x \neq 2\)
\(f(y) = \frac{1}{2 - y}\)
\(x = \frac{1}{2 - y}\) (Let x = f(y))
\(2x - xy = 1 \implies y = \frac{2x - 1}{x}\)
\(\therefore f^{-1}(x) = \frac{2x - 1}{x}\)
\(f^{-1}(-\frac{1}{2}) = \frac{2(-\frac{1}{2}) - 1}{-\frac{1}{2}}\)
= \(-2 \times -2 = 4\)
Given that \((\sqrt{3} - 5\sqrt{2})(\sqrt{3} + \sqrt{2}) = p + q\sqrt{6}\), find q.
4
-4
-5
-7
Correct answer is B
\((\sqrt{3} - 5\sqrt{2})(\sqrt{3} + \sqrt{2}) = 3 + \sqrt{6} - 5\sqrt{6} - 10\)
= \(-7 - 4\sqrt{6} = p + q\sqrt{6}\)
\(\therefore q = -4\)
Given that \(\frac{1}{8^{2y - 3y}} = 2^{y + 2}\).
\(\frac{1}{5}\)
\(\frac{7}{8}\)
1
\(1\frac{1}{5}\)
Correct answer is C
\(\frac{1}{8^{2y - 3y}} = \frac{1}{8^{-y}} = 8^{y}\)
\(8^{y} = 2^{y + 2} \implies (2^{3})^{y} = 2^{y + 2}\)
\(3y = y + 2 \implies 2y = 2\)
\(y = 1\)