Correct Answer: D. 64

Explanation

\(\log_{2} y^{\frac{1}{2}} = \log_{5} 125\)

\(\log_{2} y^{\frac{1}{2}} = \log_{5} 5^{3} = 3\log_{5} 5 = 3\)

\(\log_{2} y^{\frac{1}{2}} = 3 \implies y^{\frac{1}{2}} = 2^{3} = 8\)

\(y = 8^{2} = 64\)

Correct Answer: C. \(k < \frac{25}{8}\)

Correct Answer: B. 79.7°

Explanation

\(m . n = |m||n|\cos \theta\)

\((3i + 4j) . (2i - j) = 6 - 4 = 2\)

\(2 = |(3i + 4j)||(2i - j)| \cos \theta\)

\(|3i + 4j| = \sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5\)

\(|2i - j| = \sqrt{2^{2} + (-1)^{2}} = \sqrt{5}\)

\(2 = 5(\sqrt{5})(\cos \theta)\)

\(\cos \theta = \frac{2}{5\sqrt{5}} = 0.08\sqrt{5}\)

\(\theta = \cos^{-1} 0.1788 = 79.7°\)

Correct Answer: A. 9

Explanation

\(f(x) = mx^{2} - 6x - 3\)

\(f '(x) = \frac{\mathrm d y}{\mathrm d x} = 2mx - 6\)

\(f'(1) = 2m - 6 = 12\)

\(2m = 18 \implies m = 9\)

Explanation

P(at most two winning prizes) = 1 - [P(all winning) + P(three winning)]

P(three winning) = \(P(ABCD') + P(ABC'D) + P(AB'CD) + P(A'BCD)

P(A) = \(\frac{1}{5}\), P(A') = \(\frac{4}{5}\)

P(B) = \(\frac{1}{4}\), P(B') = \(\frac{3}{4}\)

P(C) = \(\frac{1}{3}\), P(C') = \(\frac{2}{3}\)

P(D) = \(\frac{1}{2}\), P(D') = \(\frac{1}{2}\)

P(three winning) = \((\frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2}) + (\frac{1}{5} \times \frac{1}{4} \times \frac{2}{3} \times \frac{1}{2}) + (\frac{1}{5} \times \frac{3}{4} \times \frac{1}{3} \times \frac{1}{2}) + (\frac{4}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2})\)

= \(\frac{1}{120} + \frac{2}{120} + \frac{3}{120} + \frac{4}{120}\)

= \(\frac{10}{120}\)

P(all winning) = \(\frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2}\)

= \(\frac{1}{120}\)

P(at most 2 winning) = \(1 - [\frac{1}{120} + \frac{10}{120}]\)

= \(\frac{109}{120}\)

Correct Answer: C. \(9\pi\)

Explanation

Equation of a circle: \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Given that \(x^{2} + y^{2} - 4x + 8y + 11 = 0\)

Expanding the equation of a circle, we have: \(x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2}\)

Comparing this expansion with the given equation, we have

\(2a = 4 \implies a = 2\)

\(-2b = 8 \implies b = -4\)

\(r^{2} - a^{2} - b^{2} = -11 \implies r^{2} = -11 + 2^{2} + 4^{2} =9\)

\(r = 3\)

\(Area = \pi r^{2} = \pi \times 3^{2}\)

= \(9\pi\)

Explanation

Let \(x^{\frac{1}{3}} = b\) so that the equation is

\((x^{\frac{1}{3}})^{2} - 5x^{\frac{1}{3}} + 6 = 0\)

= \(b^{2} - 5b + 6 = 0\)

\(b^{2} - 3b - 2b + 6 = 0\)

\(b(b - 3) - 2(b - 3) = 0 \implies b = \text{2 or 3}\)

\(x^{\frac{1}{3}} = 2 \implies x = 2^{3} = 8\)

\(x^{\frac{1}{3}} = 3 \implies x = 3^{3} = 27\)

Correct Answer: D. \(2 - \sqrt{3}\)

Explanation

\(\tan 15 = \tan (60 - 45)\)

\(\tan (x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}\)

\(\tan (60 - 45) = \frac{\tan 60 - \tan 45}{1 + \tan 60 \tan 45}\)

= \(\frac{\sqrt{3} - 1}{1 + (\sqrt{3} \times 1)}\)

= \(\frac{\sqrt{3} - 1}{1 + \sqrt{3}}\)

Rationalizing by multiplying denominator and numerator by \(1 - \sqrt{3}\),

\(\tan 15 = 2 - \sqrt{3}\)

Correct Answer: B. -6

Correct Answer: C. \(\sqrt{3}\)

Explanation

\(\tan (x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}\)

\(\implies \frac{\tan 80 - \tan 20}{1 + \tan 80 \tan 20} = \tan (80 - 20) = \tan 60°\)

\(\tan 60 = \frac{\sin 60}{\cos 60} = \frac{\sqrt{3}}{2} ÷ \frac{1}{2}\)

= \(\sqrt{3}\)