Correct Answer: A. \(\frac{-(2x + y)}{x}\)

Explanation

\(\frac{\mathrm d}{\mathrm d x}(x^2 + xy - 5) = \frac{\mathrm d (x^{2})}{\mathrm d x} + \frac{\mathrm d (xy)}{\mathrm d x} - \frac{\mathrm d (5)}{\mathrm d x} = 0\)

= \(2x + x\frac{\mathrm d y}{\mathrm d x} + y = 0\)

\(\implies x\frac{\mathrm d y}{\mathrm d x} = -(2x + y)\)

\(\frac{\mathrm d y}{\mathrm d x} = \frac{-(2x + y)}{x}\)

Correct Answer: A. 28

Explanation

\(f(4) = 3(4) - 2 = 12 - 2 = 10\)

\(f(-3) = 5(-3) - 3 = -15 - 3 = -18\)

\(f(4) - f(-3) = 10 - (-18) = 28\)

Correct Answer: B. \(\frac{1}{2}\)

Explanation

\(T_{n} = ar^{n - 1}\)

\(T_{4} = ar^{4 - 1} = ar^{3} = 192\)

\(T_{9} = ar^{9 - 1} = ar^{8} = 6\)

Dividing \(T_{9}\) by \(T_{4}\),

\(r^{8 - 3} = \frac{6}{192}\)

\(r^{5} = \frac{1}{32} = (\frac{1}{2})^{5}\)

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

Correct Answer: C. \(\frac{16}{3}\)

Explanation

\(\int_{0}^{2} (8x - 4x^{2}) \mathrm {d} x = (\frac{8x^{1 + 1}}{2} - \frac{4x^{2+1}}{3})|_{0}^{2}\)

= \((4x^{2} - \frac{4x^{3}}{3}) |_{0}^{2}\)

= \((4(2^2) - \frac{4(2^3)}{3})\)

= \(16 - \frac{32}{3} = \frac{16}{3}\)

Explanation

Assumed mean, A = 19.

(a) \(\bar{x} = A + \frac{\sum fd}{\sum f}\)

= \(19 + \frac{-54}{22}\)

= \(19 - 2.455\)

= \(16.545 \approxeq 16.5\) years.

(b) Standard deviation , \(\sigma = \sqrt{\frac{\sum fd^{2}}{\sum f} - (\frac{\sum fd}{\sum f})}\)

= \(\sqrt{\frac{360}{22} - (\frac{-54}{22})}\)

= \(\sqrt{16.364 - (2.45)^{2}}\)

= \(\sqrt{16.364 - 6.025}\)

= \(\sqrt{10.339}\)

= 3.215 \(\approxeq\) 3.22 years.

Correct Answer: B. \(x< \frac{-1}{2}, x > 3\)

Explanation

\(x^{3} + y^{3} - 2xy = 11\)

Differentiating implicitly,

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

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

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

At (2, -1) , Gradient = \(\frac{2(-1) - 3(2^{2})}{3(-1)^{2} - 2(2)}\)

= \(\frac{-2 - 12}{3 - 4}\)

= \(\frac{-14}{-1} = 14\)

Correct Answer: B. -3

Explanation

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

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

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

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

Correct Answer: C. \(\begin{pmatrix} 0.8 \\ 0.6 \end{pmatrix}\)

Explanation

\(PQ = \begin{pmatrix} 7 - 3 \\ 4 - 1 \end{pmatrix}\)

\(= \begin{pmatrix} 4 \\ 3 \end{pmatrix}\)

\(\hat{n} = \frac{\overrightarrow{PQ}}{|PQ|} \)

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

\(\hat{n} = \frac{1}{5}\begin{pmatrix} 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 0.8 \\ 0.6 \end{pmatrix}\)

Correct Answer: B. \(15a^{4}b^{2}\)

Explanation

\((a - b)^{6} = ^{6}C_{0}(a)^{6}(-b)^{0} + ^{6}C_{1}(a)^{5}(-b)^{1} + ^{6}C_{2}(a)^{4}(-b)^{2} + ...\)

Third term = \(^{6}C_{2}(a)^{4}(-b)^{2} = \frac{6!}{(6-2)! 2!}(a^4)(b^2)\)

= \(15a^{4}b^{2}\)