Correct Answer: C. \(6x - \frac{2}{x^{3}}\)

Explanation

\(y = 3x^{2} + \frac{1}{x^{2}} \equiv y = 3x^{2} + x^{-2}\)

\(\frac{\mathrm d y}{\mathrm d x} = 6x - 2x^{-3}\)

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

Correct Answer: C. 3.5

Explanation

where Log\(_2\) 8√2 → Log\(_2\) √128

→ Log\(_2\) 128\(^\frac{1}{2}\)

= \(\frac{1}{2}\) * (Log\(_2\) 128) → \(\frac{1}{2}\) * (Log\(_2\) 2\(^7\))

= 7 * \(\frac{1}{2}\) * (Log\(_2\) 2)

where (Log\(_2\) 2) = 1

→ 7 * \(\frac{1}{2}\) * 1

= \(\frac{7}{2}\) or 3.5

Explanation

(a)

(b)

(i) Median, \(Q_{2} = \frac{1}{2} \text{Nth score}\)

= \(\frac{1}{2} \text{60th score} = 30th\) score.

= \(48.4\)

(ii) \(Q_{1} = 35.5 ; Q_{3} = 57.5\)

Quartile deviation = \(\frac{1}{2} (Q_{3} - Q_{1})\)

= \(\frac{1}{2} (57.5 - 35.5) = \frac{1}{2} (22)\)

= \(11.0\)

Correct Answer: A.\(\frac{-x}{1 - x}, x \neq 1\)

Explanation

\(x * y = x + y - xy\)

Let \(x^{-1}\) be the inverse of x, so that

\(x * x^{-1} = x + x^{-1} - x(x^{-1}) = 0\)

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

\(x^{-1}(x - 1) = x \implies x^{-1} = \frac{x}{x - 1}\)

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

Correct Answer: D. \(h=\frac{mr}{n}\)

Explanation

\(h(m+n) = m(h+r)\\

hm+hn=hm+mr\\

hn=mr\\

h=\frac{mr}{n}\)

Correct Answer: D. ±2

Explanation

(a) Area of the major sector = \(\frac{\theta}{360°} \times \pi r^{2}\)

= \(\frac{360 - 120}{360} \times \frac{22}{7} \times 7 \times 7\)

= \((\frac{2}{3} \times 154) cm^{2}\)

Volume = \(\text{Area of the major sector} \times \text{height (cross section)}\)

= \(\frac{2}{3} \times 154 \times 15 = 1540 cm^{3}\)

(b) Curved surface area = \(\frac{240}{360} \times 2\pi rh\)

= \(\frac{240}{360} \times 2 \times \frac{22}{7} \times 7 \times 15\)

= \(440 cm^{2}\)

Area of rectangular surface = \(2 \times l \times b\)

= \(2 \times 7 \times 15 = 210 cm^{2}\)

Area of top and bottom of cylinder (major sector)

= \(2 \times \frac{240}{360} \times \frac{22}{7} \times 7 \times 7\)

= \(205.33 cm^{2}\)

\(\therefore \text{Total surface area} = 440 + 205.33 + 210 = 855.33 cm^{2}\)

Correct Answer: B. N29,040

Explanation

(a) \(\log_{10} (\frac{30}{16}) - 2 \log_{10} (\frac{5}{9}) + \log_{10} (\frac{400}{243})\)

= \(\log_{10} (\frac{30}{16}) - \log_{10} (\frac{5}{9})^{2} + \log_{10} (\frac{400}{243})\)

= \(\log_{10} (\frac{30}{16} \times \frac{400}{243} \div \frac{25}{81})\)

= \(\log_{10} (\frac{30}{16} \times \frac{400}{243} \times \frac{81}{25})\)

= \(\log_{10} (10) = 1\)

(b) \(\sqrt{\frac{P}{Q}} = \sqrt{\frac{3.6 \times 10^{-3}}{2.25 \times 10^{6}}\)

= \(\sqrt{\frac{36 \times 10^{-4}}{2.25 \times 10^{6}}\)

= \(\frac{6 \times 10^{-2}}{1.5 \times 10^{3}}\)

= \(4 \times 10^{-2 - 3} = 4 \times 10^{-5}\)

Explanation

(a) \(2x^{2} + (p + 1)x + q = 0\)

The roots of the equation are 1 and 3, therefore (x - 1)(x - 3) = 0.

Equation : \(x^{2} - 3x - x + 3 = 0\)

\(x^{2} - 4x + 3 = 0\)

\(\therefore 2x^{2} + (p + 1)x + q \equiv x^{2} - 4x + 3\)

\(\frac{2x^{2} + (p + 1)x + q}{2} = x^{2} + \frac{(p + 1)}{2} x + \frac{q}{2}\)

\(\implies \frac{q}{2} = 3\)

\(q = 6\)

\(\frac{p + 1}{2} = -4 \implies p + 1 = -8\)

\(p = -9\)

(p, q) = (-9, 6).

(b) \(W \propto \frac{1}{d^{2}} \)

\(W = \frac{k}{d^{2}}\)

\(k = Wd^{2}\)

\(W = \frac{80 \times (6400)^{2}}{d^{2}}\) (equation of variation).

\(W = \frac{80 \times (6400)^{2}}{(6400 + 800)^{2}}\)

\(W = \frac{80 \times (6400)^{2}}{(7200)^{2}}\)

= 63.21 kg