Correct Answer: C. log \(\frac{2}{9}\)

Explanation

\(\frac{1}{3}\) log8 + \(\frac{1}{3}\) log 64 - 2 log6

= log 2 + log 4 - log 36

= log \(\frac{8}{36}\)

= log \(\frac{2}{9}\)

Correct Answer: B. 3

Correct Answer: A. -30

Explanation

\(\int (3x^{2} - 2x - 12) \mathrm {d} x = \frac{3x^{2 + 1}}{2 + 1} - \frac{2x^{1 + 1}}{2} - 12x\)

= \(x^{3} - x^{2} - 12x\)

\((x^{3} - x^{2} - 12x)|_{-2}^{3} = ((3^{3}) - (3^{2}) - 12(3)) - ((-2^{3}) - (-2^{2}) - 12(-2))\)

= \((27 - 9 - 36) - (-8 - 4 + 24) = -18 - 12 = -30\)

Correct Answer: D. 672

Explanation

Let the power of \(2x^{2}\) be t and the power of \(\frac{1}{x} \equiv x^{-1}\) = 9 - t.

\((2x^{2})^{t}(x^{-1})^{9 - t} = x^{0}\)

Dealing with x alone, we have

\((x^{2t})(x^{-9 + t}) = x^{0} \implies 2t - 9 + t = 0\)

\(3t - 9 = 0 \therefore t = 3\)

The binomial expansion is then,

\(^{9}C_{3} (2x^{2})^{3}(x^{-1})^{6} = \frac{9!}{(9-3)! 3!} \times 2^{3}\)

= 84 x 8

= 672

Correct Answer: B. \(16 \sqrt{3}\) N

Correct Answer: C. \(\frac{5?6}{12}\)

Explanation

(\(\frac{3√6}{√5} + \frac{√54}{3√5}\))\(^{-1}\)

= \(\frac{√5(3√5)}{3√6 + 3√6}\)

= \(\frac{3*5}{6√6} = \frac{5}{2√6}\)

= \(\frac{5*2√6}{2√6+2√6} = \frac{10√6}{4*6}\)

= \(\frac{5√6}{12}\)

Explanation

(a) If we let P(x, y) be the coordinates of the point, then we would have;

P(x, y) = (\(\frac{(3)(-2) - (2)(7)}{3 -2}\), \(\frac{(3)(7) - (2)(5)}{3- 2}\)) = (\(\frac{-6 - 14}{1}\)), \(\frac{21 - 18}{1}\)) = (-20, 31)

(b), They took the L.C.M. and arrived at \(\frac{2}{1 - \sqrt{2}}\) - \(\frac{2}{2 - \sqrt{2}}\) = \(\frac{2(2 + \sqrt{2}) - 2 (1 - \sqrt{2})}{(1 - \sqrt{2})(2 + \sqrt{2})}\)

Simplify and rationalize to get -4 - \(\sqrt{2}\)

Correct Answer: D. \(y \geq \frac{1}{x^{2}}\)

Explanation

\(\log_{10}y + 3\log_{10}x \geq \log_{10}x\)

\(\implies \log_{10}y \geq \log_{10}x - 3 \log_{10}x \)

\(\log_{10}y \geq -2\log_{10}x = \log_{10}y \geq \log_{10}x^{-2}\)

\(\log_{10}y \geq \log_{10}(\frac{1}{x^{2}}) \implies y \geq \frac{1}{x^{2}}\)

Correct Answer: C. 120

Explanation

In a circular seating arrangement, we fix the position of one person and then rotate the others, so we have

\((6-1)! = 5! = 5\times4\times3\times2 = 120\)

Explanation

(a) \(f : x \to x^{2} + 1\)

\(g : x \to 5 - 3x\)

(i) \(f(x) = x^{2} + 1\)

Let y = f(x).

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

\(x = \sqrt{y - 1}\)

\(\implies f^{-1} (x) = \sqrt{x - 1}\)

\(Dom(f^{-1} (x)) = x \geq 1\)

(ii) \(g(x) = 5 - 3x\)

Let y = g(x)

\(y = 5 - 3x \implies 3x = 5 - y\)

\(x = \frac{5 - y}{3}\)

\(g^{-1} (x) = \frac{5 - x}{3}\)

\(g^{-1} (2) = \frac{5 - 2}{3} = 1\)

(b) \(\int \frac{x + 3}{x^{2} + 6x + 9} \mathrm {d} x\)

\(\frac{x + 3}{x^{2} + 6x + 9} = \frac{x + 3}{(x + 3)^{2}}\)

= \(\frac{1}{x + 3}\)

\(\therefore \int \frac{x + 3}{x^{2} + 6x + 9} \mathrm {d} x = \int \frac{1}{x + 3} \mathrm {d} x\)

= \(\ln |x + 3| + c\)