Question Library

2x + x = 180^\circ, => 3x = 180^\circ, and thus x = 60^\circ Each exterior angle = 60^\circ but size of ext. angle = 360^\circ/n Therefore 60^\circ = 360^\circ/n n = 360^\circ/60^\circ = 6 sides

x2 + y2 = (\sqrt2)2 x2 + y2 = 2 but y = x Thus; x2 + x2 = 2 2x2 = 2 x2 = + or - 1 But x2 + y2 = 2 12 + y2 = 2 1 + y2 = 2 y2 = 2 - 1 y2 = 1 y = + or - 1 Thus point (x,y) = (1,1) only.

\(\int^{\pi}_{0}\frac{cos^{2}\theta-1}{sin^{2}\theta}d\theta\) = \(\int^{\pi}_{0}\frac{-sin^{2}\theta}{sin^{2}\theta}\) = \(\int^{\pi}_{0}d\theta = -\pi\)

y = 2x cos2x - sin2x \(\frac{dy}{dx}\) = 2 cos2x +(-2x. 2sin2x) - 2cos2x = 2 cos2x - 2x sin2x - cos2x = -2x 2sin2x = -4 x (\(\frac{\pi}{4}\)) sin2 x (\(\frac{\pi}{4}\)) = -(\pi) x 1 = - \pi

∫30 \pi(y+1) dy = \pi[y\(^2 + y]^3_0\) = \pi(9/2 + 3) = 15\pi/2

\(V = \frac{2}{3} \pi r^{3}\) Given: \(\frac{\mathrm d V}{\mathrm d t} = 18\pi m^{3} s^{-1}\) \(\frac{\mathrm d V}{\mathrm d t} = \frac{\mathrm d V}{\mathrm d r} \times \frac{\mathrm d r}{\mathrm d t}\) \(\frac{\mathrm d V}{\mathrm d r} = 2\pi r^{2}\) \(18\pi = 2\pi r^{2} \times \frac{\mathrm d r}{\mathrm d t}\) \(\frac{\mathrm d r}{\mathrm d t} = \frac{18\pi}{2\pi r^{2}} = \frac{9}{r^{2}}\) The rate of change of the radius when r = 6m, \(\frac{\mathrm d r}{\mathrm d t} = \frac{9}{6^{2}} = \frac{1}{4}\) = \(0.25 ms^{-1}\)

P (X or Y) = P(X) + P(Y), when they are independent as given. 0.7 = 0.4 + P(Y) P(Y) = 0.7 - 0.4 = 0.30

Mean = \(\frac{{0 + (x+2) + (3x+6) + (4x+8)}}{4}\) = 4 => 0 + (x+2) + (3x+6) + (4x+8) = 16 8x + 16 = 16 x = 0 Now prepare a table showing the deviation of each of 0, (x+2), (3x+6), and (4x+8). Adding the deviations will give 12. Thus M.D = \(\frac{12}{4}\) = 3

Number of letter in MATHEMATICS = 11 Number of letter M = 2 Number of letter T = 2 Number of letter A = 2 Arrangement = 11!/(2! 2! 2!) ways

Angle corresponding to 4 is 40/240 x 360/1 = 60^\circ Note that total angle in a circle is 360^\circ. Note also that sum of the frequencies given is 240.

U = {1, 2, 3, 4, 5,..., 20} E1 = {3, 6, 9, 12, 15, 18} E2 = {4, 8, 12, 16, 20} P(E\(_2\)) = \(\frac{5}{20}\) P(not E\(_2\)) = 1 - (\(\frac{5}{20}\)) = \(\frac{15}{20} = \frac{3}{4}\)

Hint: prepare a three-columned table, one for (x), another for the (deviation), and the last for the (squared-deviation). Sum of (x) = 15x, algebraic sum of (deviation) = zero (0) Sum of (squared-deviation) = 10x\(^2\) Variance = \(\frac{∑(squared-deviation)}{\text{n}}\) = \(\frac{10x^2}{5}\) = 2x\(^2\)

Range = Highest - lowest number => 10-4 = 6 Mode is the number with highest occurrence => Mode = 10 Sum = 6 + 10 = 16

No of ways of choosing 1 man, 2 women = 5C1 x 3C2 No of ways of choosing 2 men, 1 woman = 5C2 x 3C1 Summing, => (5C1 x 3C2) + (5C2 x 3C1) = 15 + 30 = 45

Hints: 1. Integrate the given first derivative of f(x) at the boundaries, (0,0) Then solve accordingly to get f(x) = y = \(\frac{3x^2}{2}\) + 2x