Question Library

x\(^2\) - (sum of roots) x + (product of roots) = 0 x\(^2\) - \(\frac{5x}{2}\) + 4 = 0 2x\(^2\) - 5x + 8 = 0

\((\log_{3} x)^{2} - 6(\log_{3} x) + 9 = 0\) Let \(\log_{3} x = a\). \(a^{2} - 6a + 9 = 0\) \(a^{2} - 3a - 3a + 9 = 0\) \(a(a - 3) - 3(a - 3) = 0\) \((a - 3)(a - 3) = 0\) \(\implies a = 3 (twice)\) \(\log_{3} x = 3 \implies x = 3^{3} = 27\)

\(-1 < 2x - 5 \leq 5\) \(\implies -1 + 5 < 2x - 5 + 5 \leq 5 + 5\) \(4 < 2x \leq 10\) \(\implies 2 < x \leq 5 \) = 3, 4, 5.

10x2 - 13x - 3 = 0 = 10x2 - 15x + 2x - 3 = 0 5x(2x - 3) + 2x - 3 = 0 = (5x + 1)(2x - 3) = 0 5x + 1 = 0 or 2x - 3 = 0 x = -\(\frac{1}{5}\) or \(\frac{3}{2}\)

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

\(2^{x + y} = 32, 3^{3y - x}\) = 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

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

x2 - 2x = 8 = 8 x2 - 2x - 8 = 0 (x2 - 4x) + (2x - 8) = 0 When x + 2 = 0 x = -2, when x - 4 = 0 x = 4 -2 or 4

a(b + c) + \(\frac{5}{d}\) - 2 = 0 ab + ac + \(\frac{5}{d}\) - 2 = 0 abd + \(\frac{acd}{d}\) + 5 - 2 = ab + ac + 5 - 2d acd = 2d - 5 - abd c = \(\frac{2d - 5 - abd}{ad}\)

0 < \(\frac{x + 3}{x - 1}\) < 2 Put x = 0, -3 and 9 0 < \(\frac{9 + 3}{9 - 1}\) \(\leq\) 2 i.e. 0 < 1.5 \(\leq\) 2 (true) but 0 < \(\frac{0 + 3}{0 - 1}\) \(\leq\) 2 i.e. 0 < -3 \(\leq\) 2 (not true) -3 \(\leq\) 2 -3 is not greater than 0

2x + 5y = 11.......(i) 7x + 4y = 2.......(ii) (i) x 7 \(\to\) 14x + 35y = 77.........(iii) (ii) x 2 \(\to\) 14x + 8y = 4........(iv) (iii) - (iv) 27y = 73 y = \(\frac{73}{27}\)

Given the equation \[x^3 - 12x - 16 = 0\] with \( x = -2 \) as a solution, we can factor it as follows: Divide by \( x + 2 \) gives: \[x^3 - 12x - 16 = (x + 2)(x^2 - 2x - 8)\] Finding Roots of \( x^2 - 2x - 8 = 0 \) Using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{36}}{2}\] This gives:\[x = 4 \quad \text{and} \quad x = -2 \text{ (with multiplicity 2)}\]

\(N \propto P_{A}\); \(N \propto P_{B}\); \(N \propto \frac{1}{D^{2}}\) \(\therefore N \propto \frac{P_{A} P_{B}}{D^{2}}\) \(N = \frac{k P_{A} P_{B}}{D^{2}}\)

\(T_1\) = 153 \(T_6 = \frac{17}{27}\) The equation for the nth term is ar\(^{n-1}\) \(T_6: 153r^5\) = \(\frac{17}{27}\) divide through by 153 r\(^5\) = \(\frac{17}{27}\) x \(\frac{1}{153}\) take 5th root of both sides r = \(\frac{1}{3}\) The sum of n terms is \(S_n = \frac{a(1 - r^n)}{1 - r }\) So, \(S_4 = \frac{153(1 - (\frac{1}{3})^4)}{1 - \frac{1}{3}}\) = \(\frac{680}{3}\)