Correct Answer: C. \(1, -\frac{1}{3}\)

Explanation

At turning point, \(\frac{\mathrm d y}{\mathrm d x} = 0\).

Given \(x^{3} - x^{2} - x + 6 \)

\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 2x - 1 = 0 \)

\(3x^{2} - 3x + x - 1 = 0 \implies (3x + 1)(x - 1) = 0\)

\(x = \frac{-1}{3}, 1\)

Explanation

p = \(1 - \frac{6Σ(d_i)^2}{n(n^2-1}\)

p = Spearman's rank correlation coefficient

di = difference between the two ranks of each observation

n = number of observation

P = \(1 - \frac{6(4+1+1+0+16+0+0+0)}{8(8^2-1)}\)

p = \(1 - \frac{6(22)}{8(63)}\)

p = \(1 - \frac{132}{504}\)

p = \(1 - \frac{11}{42}\)

p = \(\frac{11}{42}\) or 0.7381

Correct Answer: A. \(\begin{pmatrix} 16 & 36 \end{pmatrix}\)

Explanation

\(|Q| = \begin{vmatrix} -1 & -2 \\ 3 & 2 \end{vmatrix}\)

= \(-2 - (-6) = 4\)

\(4P = 4\begin{pmatrix} 4 & 9 \end{pmatrix}\)

= \(\begin{pmatrix} 16 & 36 \end{pmatrix}\)

Correct Answer: D. 24

Correct Answer: A. x > -1/6

Correct Answer: C. \(\frac{2x^3}{3}\) + 2x + k

Explanation

(a) Sample - space

(i) Same digit = {(1,1), (2,2),(3,3), (4,4),(5,5), (6,6)}

P(same digit) = \(\frac{6}{36}\)

(ii) P(total score greater than 5) = \(\frac{26}{36}\)

= \(\frac{13}{18}\)

(b) \(\frac{x^{2} + y^{2}}{y^{2} - x^{2}} = \frac{(\cos 30)^{2} + (\sin 30)^{2}}{(\sin 30)^{2} - (\cos 30)^{2}}\)

= \(\frac{(\frac{\sqrt{3}}{2})^{2} + (\frac{1}{2})^{2}}{(\frac{1}{2})^{2} - (\frac{\sqrt{3}}{2})^{2}}\)

= \(\frac{\frac{3}{4} + \frac{1}{4}}{\frac{1}{4} - \frac{3}{4}}\)

= \(\frac{1}{-\frac{1}{2}}\)

= \(-2\).

Correct Answer: B. \(\frac{p - q}{a(p + q)}\)

Correct Answer: B. (-2,-3)

Explanation

y= 3x\(^2\) + 11x +7;

dy/dx = 6x + 11= -1;

6x = -12;

x=-2

y= 3x\(^2\) + 11x + 7

At x-2,

y= 3(-2)\(^2\) +11(-2) + 7

= 12 - 22 + 7= -3

p(x, y) = (-2,-3)

Correct Answer: C. 2.4

Explanation

\(\begin{array}{c|c} x & f & fx\\ \hline 1 & 6 & 6 \\

2 & 8 & 16\\ 3 & 8 & 18\\ 4 & 5 & 20\end{array}\)

mean x = \(\frac{\sum fx}{\sum f}\)

= \(\frac{60}{25}\)

x = 2.4