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\(\int \sin 3x \mathrm d x = - \frac{\cos 3x}{3} + c\) = \(-\frac{1}{3} (\cos 3x) + c\)

Area of the circle (A) = \(\pi r^{2}\) \(\frac{\mathrm d A}{\mathrm d r} = 2\pi r\) \(r = 5cm\); \(\frac{\mathrm d r}{\mathrm d t} = 0.2\) \(\frac{\mathrm d A}{\mathrm d t} = \frac{\mathrm d A}{\mathrm d r} . \frac{\mathrm d r}{\mathrm d t}\) \(\frac{\mathrm d A}{\mathrm d t} = 2\pi \times 5 \times 0.2 = 2\pi\)

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

x\(^3\) - 5x\(^2\) - x + 5 = 0. \(x^{2}(x - 5) - 1(x - 5) = 0\) \((x^2 - 1)(x - 5) = 0 \implies (x - 1)(x + 1)(x - 5) = 0\) \(\therefore x = 1, -1, 5\)

The time (t) to do the work is inversely proportional to the number of workers (n). \(\implies t \propto \frac{1}{n}\) \(t = \frac{k}{n}\) \(5 = \frac{k}{45} \implies k = 45 \times 5 = 225\) \(\therefore t = \frac{225}{n}\) For 25 men, \(t = \frac{225}{25} = 9\) \(\therefore\) 25 men will do the work in 9 days.

\(\frac{x + 2}{4} - \frac{2x - 3}{3}\) < 4 \(\frac{3(x + 2) - 4(2x - 3)}{12}\) < 4 3x + 6 - 8x + 12 < 48 18 - 5x < 48 \(\implies\) - 5x < 48 - 18 -5x < 30 \(\implies \)x > - 6

\(y = 1 - 2x - 3x^2\) At maximum point, \(\frac{\mathrm d y}{\mathrm d x} = 0\) \(\frac{\mathrm d y}{\mathrm d x} = -2 - 6x\) \(-2 - 6x = 0 \implies -2 = 6x\) \(x = -\frac{1}{3}\) At x = \(-\frac{1}{3}\), y = \(1 - 2(-\frac{1}{3}) - 3(-\frac{1}{3})^{2}\) = \(1 + \frac{2}{3} - \frac{1}{3}\) = \(\frac{4}{3}\)

\(T_{n} = a + (n - 1) d\) (nth term of an AP). \(T_{9} = 5T_{5}\) \(a + 8d = 5(a + 4d) \implies a + 8d = 5a + 20d\) \(5a - a + 20d - 8d = 0 \implies 4a + 12d = 0\) \(a + 3d = 0\)

\(\frac{x}{r + a} = \frac{a}{r}\) \(\implies rx = ar + a^{2}\) \(rx - ra = a^{2} \implies r = \frac{a^2}{x - a}\)

Let y = f(x). \(y = 3x + 4 \implies x = \frac{y - 4}{3}\) Replace y with x. \(y = \frac{x - 4}{3} \)

-2 is the solution implies x = -2. 2x + 1 - 3c = 2c + 3x - 7 2(-2) + 1 - 3c = 2c + 3(-2) - 7 -4 + 1 - 3c = 2c - 6 - 7 -3 + 13 = 2c + 3c \(\implies\) c = 2.

\(p \ast q = pq + p + q\) \(2 \ast (3 \ast 4) \) \(3 \ast 4 = 12 + 3 + 4 = 19\) \(2 \ast 19 = 38 + 2 + 19 = 59\)

The series is geometric with common ratio \(\frac{1}{3}\). \(S_{\infty} = \frac{a}{1 - r}\) = \(\frac{1}{1 - \frac{1}{3}} \) = \(\frac{1}{\frac{2}{3}}\) = \(\frac{3}{2}\)

\(x \propto \sqrt{n}\) \(x = k \sqrt{n}\) \(9 = k \sqrt{9} \implies 9 = 3k\) \(k = 3\) \(x = 3 \sqrt{n}\) When n = 17/9, \(x = 3 \times \sqrt{\frac{17}{9}} = \sqrt{17}\)

Diameter = 4\(\sqrt{3}\) cm radius = 2\(\sqrt{3}\) cm Area of major sector = \(\frac{\theta}{360} \times \pi r^{2}\) \(\theta = 360 - 120 = 240^\circ\) = \(\frac{240}{360} \times \pi \times 12\) = \(8\pi cm^{2}\)