If \(f(x) = \frac{4}{x} - 1, x \neq 0\), find \(f^{-1}(7)\).
\(\frac{-3}{7}\)
\(0\)
\(\frac{1}{2}\)
\(4\)
Correct answer is C
\(f(x) = \frac{4}{x} - 1\). Let y = f(x)
\(y = \frac{4 - x}{x} \implies xy + x = 4\)
\(x(y + 1) = 4 \therefore x = \frac{4}{y + 1}\)
\(f^{-1}(7) = \frac{4}{7 + 1} = \frac{1}{2}\)
\(p \vee q\)
\(p \vee \sim q\)
\(p \wedge \sim q\)
\(p \wedge q\)
Correct answer is C
No explanation has been provided for this answer.
\(\frac{-1}{2}\)
\(0\)
\(\frac{2}{3}\)
\(2\)
Correct answer is A
Given the formula for p * q as: \(p + q + 2pq\) and its identity element is 0, such that if, say, t is the inverse of p, then
\(p * t = 0\), then \(p + t + 2pt = 0 \therefore p + (1 + 2p)t = 0\)
\(t = \frac{-1}{1 + 2p}\) is the formula for the inverse of p and is undefined on R when
\(1 + 2p) = 0\) i.e when \(2p = -1; p = \frac{-1}{2}\).
Express \(\frac{8 - 3\sqrt{6}}{2\sqrt{3} + 3\sqrt{2}}\) in the form \(p\sqrt{3} + q\sqrt{2}\)
\(7\sqrt{3} - \frac{17\sqrt{2}}{3}\)
\(7\sqrt{2} - \frac{17\sqrt{3}}{3}\)
\(-7\sqrt{2} + \frac{17\sqrt{3}}{3}\)
\(-7\sqrt{3} - \frac{17\sqrt{2}}{3}\)
Correct answer is B
Given \(\frac{8 - 3\sqrt{6}}{2\sqrt{3} + 3\sqrt{2}}\),
first, we rationalise by multiplying through with \(2\sqrt{3} - 3\sqrt{2}\) (the inverse of the denominator).
\((\frac{8 - 3\sqrt{6}}{2\sqrt{3} + 3\sqrt{2}})(\frac{2\sqrt{3} - 3\sqrt{2}}{2\sqrt{3} - 3\sqrt{2}})\)
= \(\frac{16\sqrt{3} - 24\sqrt{2} - 18\sqrt{2} + 18\sqrt{3}}{4(3) - 6\sqrt{6} + 6\sqrt{6} - 9(2)}\)
= \(\frac{34\sqrt{3} - 42\sqrt{2}}{-6} = 7\sqrt{2} - \frac{17\sqrt{3}}{3}\)
\({x : -5 < x < 5}\)
\({x : -5 \leq x \leq 5}\)
\({x : -5 \leq x < 5}\)
\({x : -5 < x \leq 5}\)
Correct answer is A
\(P = {x : -2 < x < 5}\) and \(Q = {x: -5 < x < 2}\)
\((P \cup Q) = {x : -5 < x < 5}\)