Question Library

(2\(\sqrt{y} - 3y\)) and (3y + 2\(\sqrt{y}\)) = 6y\(\sqrt{y}\) + 4y - 9y\(^2\) - 6y\(\sqrt{y}\) = 4y - 9y\(^2\)

kx2 - 4x + 1 = 0, comparing with ax2 + bx + c = 0 a = k, b = -4, c = 1 for equal root b2 = 4ac (-4)2 = 4k k = \(\frac{16}{4}\) = 4

x + 2\(\sqrt{x} - 8 = 0, Let \sqrt{x} = y\) x = \(y^2\) \(y^2 + 2y\) - 8 = 0 (y + 4)(x - 2) = 0 y = -4 or 2 x = 16 or 4

\(5^{x + 1} + 5^x = 150\) according to the laws of indices \(5^1(5^x) + 5^x = 150\) let \(5^x\) = y 5(y) + y = 150 6y = 150 y = \(\frac{150}{6}\) = 25 but y = \(5^x\) = 25 \(5^x\) = 25 = \(5^2\) equating powers since bases are the same x = 2

\(2^{x + y} = 32, 3^{3y - x}\) = 27 \(2^{x + y} = 2^5\) \(3^{3y - x} = 3^3\) equating powers, we have x + y = 5 ------------1 3y - x = 3 ------------2 solving simultaneously(any method) x = 3 and y = 2

\(\sqrt{\frac{1 - cos x}{1 + cos x}}\) = a a2 = \(\frac{1 - cosx}{1 + cosx}\) \(\frac{1 - cosx}{1 + cosx}\) = \(\frac{1 - cosx}{1 - cosx}\) = \(\frac{(1 - cosx)^2}{cos^2 x}\) a2 = \(\frac{(1 - cos x)^2}{sin^2 x}\) a = \(\frac{1 - cos x}{sin x}\)

S = curved surface area = \(\pi\)rL = \(\frac{22}{7}\) x 6 x 10 = 188.57cm2

3 tan \(\theta\) + 2 = -1 3 tan \(\theta\) \(\frac{-3}{3}\) = -1 \(\theta\) = tan -1(-1) \(\theta\) = 360o - 45o = 315o \(\theta\) = 180 - 45o = 135o

log 10.5 = log \(\frac{21}{2}\) = log 21 - log 2 = log(3 x 7) - log 2 = log 3 + log 7 - log 2 = 1.10 + 1.90 - 0.69 = 3 - 0.69 = 2.31

102 + log105 = log10 10100 + log105 = log105 x 10100

\(\frac{5}{x + 1} - \frac{3}{1 - x} - \frac{7x - 1}{x^{2} - 1} = \frac{?}{x + 1}\) \(\frac{5(x - 1) - 3(- (x + 1)) - (7x - 1)}{x^{2} - 1}\) = \(\frac{5x - 5 + 3x + 3 - 7x + 1}{x^{2} - 1}\) = \(\frac{x - 1}{x^{2} - 1}\) = \(\frac{x - 1}{(x - 1)(x + 1)}\) = \(\frac{1}{x + 1}\). The numerator = 1.

F(X) = x\(^3\) - 4x\(^2\) + cx + d = (X + 1) Q(X) + R x = -1, R = 0,f(-1) = -1\(^3\) - 4(-1)\(^2\) + c(-1) + d = 0 -1 - 4 - c + d = 0 d - c = 5................(i) f(-2) = -2\(^3\) - 4(-2)\(^2\) + c(-2) + d = 1 = -8 - 16 - 2c + d = 1 -8 - 16 - 2c + d = 1 -24 - 2c + d = 1 d - 2c = 1 + 24 d - 2c = 25.................(ii) \(\frac{d - c = 5}{-c = 20}\) d - c = 5 c = -20 d - (-20) = 5 d + 20 = 5 d = 5 - 20 = -15 c = -20, d = -15

f(x + 1) = 3x2 - x + 4 f(0) = f(x + 1) x + 1 = 0 ===> x = -1 f(0) = 3(-1)2 - (-1) + 4 f(0) = 3 + 1 + 4 = 8

V = \(\pi\)r2h = \(\frac{22}{7}\) x 52 x 12 - \(\frac{22}{7}\) x 32 x 12 = \(\frac{22}{7}\) x 12(52 - 32) = \(\frac{4224}{7}\)

No expanded working has been attached for this question yet. The verified answer is Option B.