Question Library

x\(^2\) + 3\(^3\) = 5\(^2\) x\(^2\) + 9 = 25 x\(^2\) = 25 – 9 x\(^2\) = 16 x = \sqrt 16 = 4 cm ∴ The locus is a circle of radius 4 cm with the center O

Gradient = \(\frac{y_2 - y_1}{x_2 - x_1}\) \(\frac{2}{3}\) = \(\frac{4 - p}{p - (-1)}\) \(\frac{2}{3}\) = \(\frac{4 - p}{p + 1}\) 3(4-p) = 2(p+1) 12 - 3p = 2p + 2 12 – 2 = 2p + 3p 10 = 5p p= \(\frac{10}{5}\) = 2

A(4,2) and B(1,r), AB = 3 units 3 = \sqrt (x2- x1)2 + (y2-y1)2 3 = \sqrt (4-1)2 + (2-r)2 3 = \sqrt 32 + (2-r)2 3 = \sqrt 9 + 4 – 4r + r2 3 = \sqrt r2 - 4r + 13 9 = r2 - 4r + 13 By squaring both sides r2 - 4r + 4 = 0 (r-2)(r-2) = 0 r = 2

y = 3 cos 4x Applying product rule: \(\frac{dy}{dx}\) = u\(\frac{dv}{dx}\) + v\(\frac{du}{dx}\) Where u = 3 and v = cos 4x \(\frac{du}{dx}\) of 3 = 0 and \(\frac{dv}{dx}\) of cos 4x = -4sin 4x \(\frac{dy}{dx}\) = 3 \(\times\) -4 sin 4x + cos 4x \(\times\) 0 \(\frac{dy}{dx}\) = -12 sin 4x

\(Sin 45 - cos 30\\ \frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\\ =\frac{\sqrt{2}-\sqrt{3}}{2}\)

Tan 60^\circ = \(\frac{300}{\text{x}}\) x = \(\frac{300}{tan 60}\) x = \(\frac{300}{\sqrt{3}}\) x = \(\frac{300}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\) x = \(\frac{300\sqrt{3}}{3}\) x = 100\(\sqrt{3}\)m

x = (2+3t)(5t-4) Let u = 2+3t ∴ \(\frac{\text{du}}{\text{dt}}\) = 3 and v = 5t-4 ∴ \(\frac{\text{dv}}{\text{dt}}\) = 5 dx/dt = \(\frac{\text{Vdu}}{\text{dt}}\) + \(\frac{\text{Udv}}{\text{dt}}\) = (5t-4)3 + (2+3t)5 = 15t - 12 + 10 + 15t = 30t - 2 = 30 x \(\frac{4}{5}\) - 2 = 24 - 2 = 22

S = t\(^3\) - t\(^2\) - t + 5 ds/dt = 3t\(^2\) - 2t - 1 As ds = 0 3t\(^2\) - 2t - 1 = 0 (3t+1)(t-1) = 0 ∴ t = 1 or -1/3 At min pt t = 1 S = t\(^3\) - t\(^2\) - t + 5 put t = 1 = 1\(^3\) - \1(^2\) - 1 + 5 = 1 - 1 - 1 + 5 = 4

∫sec2\theta d\theta = ∫ 1/cos2 d\theta ∫(cos)-2 d\theta, let u = cos \theta ∴∫u-2 = 1/u + c ∫cos \theta = sin \theta + c ∫sec-2\theta = 1/u sin \theta + c = (sin\theta / cos\theta) + c = Tan \theta + c

20 + 2x + 60 + 40 + x + 50 = 260 170 + 3x = 260 3x = 260 - 170 3x = 90 x = 90/3 = 30 At least 4 days 4 days = 40 5 days = 30 6 days = 50 Total = 120

Sectorial angle of apple 100 + 100 + 115 + x = 30o 315 + x = 360 x = 360 - 315 x = 45o 360o represents 400 1o represents 400 / 360 45o represents (400/360) x (45/1) = 50 apples

P(pass) = 2/3 P(not pass) = 1 - 2/3 = 1/3 P(not passing any of the 3 Exams) = 1/3 x 1/3 x 1/3 = 1/27

x f fx 5 3 15 8 2 16 6 4 24 k i k ∑f=10 ∑fx=55+k Mean = \(\frac{∑fx}{∑f}\) 5.7 = \(\frac{55 + k}{10}\) 57 = 55 + k k = 57 - 55 k = 2

x ∴x x-\(\overline{X}\) /x-\(\overline{X}\)/ x 1 -1 1 2x 2 0 0 x+1 2 0 0 3x 3 1 1 ∑/x-\(\overline{X}\)/ = 2 Mean(\(\overline{X}\)) = \(\frac{ x + 2x + x + 1 + 3x}{4}\) = 2 2 = \(\frac{7x + 1}{4}\) 8 = 7x + 1 7x = 8 - 1 = 7 x = \(\frac{7}{7}\) = 1 Mean deviation = \(\frac{∑(| x - \overline{X}|)}{n}\) M.D = \(\frac{2}{4}\) = \(\frac{1}{2}\) = 0.5

= \((^{5}C_{2} \times ^{3}C_{1}) + (^{5}C_{1} \times ^{3}C_{2})\) = \((10 \times 3) + (5 \times 3)\) = 45.