Question Library

\(\frac{0.21 \times 0.072 \times 0.00054}{0.006 \times 1.68 \times 0.063}\) = \(\frac{9 \times 10^{-10}}{7 \times 10^{-8}}\) = \(1.2857 \times 10^{-2}\) = \(0.012857 \approxeq 0.01286\) (to 4 s.f)

Let the number of goats be X C.P of X goats = N4000X S.P of X goats = (75/100) x 4000X = N3000X 3000X = 180,000 => X = 180000/3000 = 60. X = 60 goats

dy/dx = 2x - 3 y = ∫2x - 3 dx => y = x2 - 3x + c 3 = (0)2 - 3(0) + c => c = 3 Thus y = x2 - 3x + 3

\(y = \sin^{2} (5x)\) Let u = sin 5x \(\frac{\mathrm d u}{\mathrm d x} = 5 \cos 5x\) \(\therefore y = u^{2}\) \(\frac{\mathrm d y}{\mathrm d u} = 2u\) \(\frac{\mathrm d y}{\mathrm d x} = 2u . 5 \cos 5x\) = \(10u \cos 5x\) = \(10 \sin 5x \cos 5x\)

y = 3x\(^2\) - 2x + 5 Slope = \(\frac{\mathrm d y}{\mathrm d x} = 6x - 2\) At x = 1, Slope : 6(1) - 2 = 4.

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

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. \(y = \frac{x - 4}{3} \)