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\(y = 1 - 2x - 3x^2\) 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 = \(\frac{\theta}{360} \times \pi r^{2}\) \(\theta = 360 - 120 = 240^\circ\) = \(\frac{240}{360} \times \pi \times 12\) = \(8\pi cm^{2}\)

\(\tan \theta = \frac{opposite}{adjacent} = \frac{4}{3}\) Hyp\(^2\) = 4\(^2\) + 3\(^2\) Hyp = 5. \(\sin \theta = \frac{4}{5}; \cos \theta = \frac{3}{5}\) \(\sin^{2} \theta - \cos^{2} \theta = \frac{16}{25} - \frac{9}{25}\) = \(\frac{7}{25}\)

The line equidistant from x = 1 and x = 7 is \(x = \frac{1 + 7}{2} \implies x = 4\)

Volume of a frustrum with top of radius R and bottom r and height h = \(\frac{1}{3} \pi (R^{2} + Rr + r^{2})\) V = \(\frac{1}{3} \pi \times 4 \times (6^2 + 6 \times 4 + 4^2)\) = \(\frac{304}{3} \pi cm^{3}\)

The formula for the sum of the interior angles of a regular polygon = (2n - 4) x 90^\circ Given: Sum = 20 right angles (2n - 4) x 90^\circ = 20 \times 90^\circ ⇒ 2n - 4 = 20 2n = 24; n = 12.

2y = 4x - 8 \(\implies\) y = 2x - 4. When x = 0, y = -4. When y = 0, x = 2. The midpoint between (0, -4) and (2, 0) = \((\frac{0 + 2}{2}, \frac{-4 + 0}{2})\) = \((1, -2)\)

\(\tan 45 = \frac{x}{10.4}\) = 10.4m \(height of the tree is equal to 1.6m + 10.4m=12m\)