Correct Answer: A. \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)

Explanation

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Correct Answer: C. \(p \wedge \sim q\)

Explanation

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Correct Answer: B. \(\frac{8}{5}\)

Explanation

\(2x^{2} - 6x + 5 = 0 \implies a = 2, b = -6, c = 5\)

\(\alpha + \beta = \frac{-b}{a} = \frac{-(-6)}{2} = 3\)

\(\alpha\beta = \frac{c}{a} = \frac{5}{2} \)

\(\frac{\beta}{\alpha} + \frac{\alpha}{\beta} = \frac{\beta^{2} + \alpha^{2}}{\alpha\beta}\)

\(\frac{(\alpha + \beta)^{2} - 2\alpha\beta}{\alpha\beta} = \frac{3^{2} - 2(\frac{5}{2})}{\frac{5}{2}}\)

= \(\frac{4}{\frac{5}{2}} = \frac{8}{5}\)

Correct Answer: B. Some strong wrestlers are not weightlifters

Explanation

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Correct Answer: C. \(\frac{1}{13}(-5i + 12j)\)

Explanation

The unit vector \(\hat{n} = \frac{\overrightarrow{r}}{|r|}\)

\(\hat{n} = \frac{-5i + 12j}{\sqrt{(-5)^{2} + (12)^{2}} \)

= \(\frac{-5i + 12j}{13} \)

Correct Answer: C. \((R \cap P) \subset (R \cap U)\)

Explanation

All the statements are false except option C.

\(R \cap P = {3, 5, 7} and R \cap U = {2, 3, 5, 7, 11}\)

\(\therefore (R \cap P) \subset (R \cap U)\)

Explanation

(a) 3 log\(_2\) x = y \(\to\) 2\(^y\) = x\(^3\).............(1)

Similarly, log\(_2\) 4x = y + 4 can be written as 2\(^{y + 4}\) = 4x............(2)

Substituting for 2y in equation (2) to have 16x\(^3\) = 4x and to form a cubic equation 16x\(^3\) - 4x = 0.

Then, using factorizing to have 4x(4x\(^2\) - 1) = 0 and solving for x to get x = 0 or x = \(\frac{1}{2}\) or \(\frac{1}{2}\)

Next, substituting for x in equation (1), when x = 0, y has no solution.

Also, when x = -\(\frac{1}{2}\) y has no solution and when x = \(\frac{1}{2}\), 2\(^y\) = (\(\frac{1}{2}\))\(^3\) = 2\(^{-3}\)

Therefore, y = -3

(b). Substitute to have -3*5 = (-3)\(^2\) - 2(-3) (5) + 5\(^2\) = 2\(^n\) and when simplified to get 64 = 2\(^n\). However, 2\(^6\) = 2\(^n\) so that n = 6.

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

Explanation

\(\sin \theta = \frac{\sqrt{3}}{2} \implies opp = \sqrt{3}; hyp = 2\)

\(adj^{2} = 2^{2} - (\sqrt{3})^{2} = 1 \implies adj = 1\)

\(\cos \theta = \frac{1}{2}\)

\(\sin 2\theta = \sin (180 - \theta) = \sin \theta = \frac{\sqrt{3}}{2}\)

\(\cos 2\theta = \cos (180 - \theta) = -\cos \theta = -\frac{1}{2}\)

\(\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\)

= \(- \sqrt{3}\)

Correct Answer: A. -67

Explanation

\(f(x) = 3x^{3} + 8x^{2} + 6x + k\)

\(f(2) = 3(2^{3}) + 8(2^{2}) + 6(2) + k = 1\)

\(\implies 24 + 32 + 12 + k = 1\)

\(68 + k = 1 \therefore k = 1 - 68 = -67\)

Correct Answer: B. -3

Explanation

\(P = begin{pmatrix} y - 2 & y - 1 \\ y - 4 & y + 2 \end{pmatrix}\)

\(|P| = (y - 2)(y + 2) - (y - 1)(y - 4) = (y^{2} - 4) - (y^{2} - 5y + 4) = -23\)

\(5y - 8 = -23 \implies 5y = -23 + 8 = -15\)

\(y = \frac{-15}{5} = -3\)