Question Library

\(\frac{1}{p} = \frac{a^{2} + 2ab + b^{2}}{a - b}\) \(\frac{1}{q} = \frac{a + b}{a^{2} - 2ab + b^{2}}\) \(\frac{1}{p} = \frac{(a + b)^{2}}{a - b}\) \(\frac{1}{q} = \frac{a + b}{(a - b)^{2}}\) \(\therefore p = \frac{a - b}{(a + b)^{2}}\) \(\frac{p}{q} = p \times \frac{1}{q} = \frac{a - b}{(a + b)^{2}} \times \frac{a + b}{(a - b)^{2}}\) = \(\frac{1}{(a + b)(a - b)}\) = \(\frac{1}{a^{2} - b^{2}}\)

\(x \propto \frac{1}{\sqrt[3]{y}} \implies x = \frac{k}{\sqrt[3]{y}}\) When y = 8, x = 1 \(1 = \frac{k}{\sqrt[3]{8}} \implies 1 = \frac{k}{2}\) \(k = 2\) \(\therefore x = \frac{2}{\sqrt[3]{y}}\) When x = 3, \(3 = \frac{2}{\sqrt[3]{y}} \implies \sqrt[3]{y} = \frac{2}{3}\) \(y = (\frac{2}{3})^{3} = \frac{8}{27}\)

\(\frac{a^{3} - b^{3} - c^{\frac{1}{2}}}{b - a - c} = \frac{(-3)^{3} - (2)^{3} - 4^{\frac{1}{2}}}{2 - (-3) - 4}\) = \(\frac{-37}{1} = -37\)

\(g(y) = \frac{y - 3}{11} + \frac{11}{y^{2} - 9}\) \(\therefore g(y + 3) = \frac{(y + 3) - 3}{11} + \frac{11}{(y + 3)^{2} - 9}\) \(g(y + 3) = \frac{y}{11} + \frac{11}{y^{2} + 6y + 9 - 9}\) \(g(y + 3) = \frac{y}{11} + \frac{11}{y^{2} + 6y}\) = \(\frac{y}{11} + \frac{11}{y(y + 6)}\)

\((x^{2} + x)^{2} - (2x + 2)^{2}\) = \((x^{2} + x + 2x + 2)(x^{2} + x - (2x + 2))\) = \((x^{2} + 3x + 2)(x^{2} - x - 2)\) = \(((x + 1)(x + 2))((x + 1)(x - 2))\) = \((x + 1)^{2} (x + 2)(x - 2)\)

\(\frac{x - y}{x^{\frac{1}{3}} - y^{\frac{1}{3}}}\) ( x - y ) = (\(x^{\frac{1}{3}})^3 - (y^{\frac{1}{3}})^3\) = (\(x^{\frac{1}{3}}\) - \(y^{\frac{1}{3}}\))(\(x^{\frac{2}{3}}\) + \(x^{\frac{1}{3}}\)\(y^{\frac{1}{3}}\) + \(y^{\frac{2}{3}}\)) SO, \(\frac{x - y}{x^{\frac{1}{3}} - y^{\frac{1}{3}}}\) = \(\frac{(x^{\frac{1}{3}} - y^{\frac{1}{3}})(x^{\frac{2}{3}} + x^{\frac{1}{3}}y^{\frac{1}{3}} + y^{\frac{2}{3}})}{x^{\frac{1}{3}} - y^{\frac{1}{3}}}\) Common factor in numerator and denominator cancels out. Final answer = (\(x^{\frac{2}{3}}\) + \(x^{\frac{1}{3}}\)\(y^{\frac{1}{3}}\) + \(y^{\frac{2}{3}}\))

\(x^{2} - x - 1 = 0\) Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) a = 1, b = -1, c = -1. \(x = \frac{-(-1) \pm \sqrt{(-1)^{2} - 4(1)(-1)}}{2(1)}\) \(x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}\) \(x = \frac{1 + 2.24}{2} ; x = \frac{1 - 2.24}{2}\) \(x = \frac{3.24}{2}; x = \frac{-1.24}{2}\) \(x = 1.62 ; x = -0.61 \) \(x \approxeq 1.6; -0.6\)

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

The second graph is \((x^{2} - 2x - 1) + (2 + x - x^{2})\) = \(1 - x\)

Let the first term and common difference = a & d respectively. \(T_{n} = \text{nth term} = a + (n - 1) d\) (A.P) Given: \(T_{4} = -6 \implies a + 3d = -6 ... (i)\) \(T_{8} + T_{9} = 72\) \(\implies a + 7d + a + 8d = 72 \implies 2a + 15d = 72 ... (ii)\) From (i), \(a = -6 - 3d\) \(\therefore\) (ii) becomes \(2(-6 - 3d) + 15d = 72\) \(-12 - 6d + 15d = 72 \implies 9d = 72 + 12 = 84\) \(d = \frac{84}{9} = 9\frac{1}{3}\)

\(T_{n} = ar^{n - 1}\) (nth term of a G.P) \(T_{4} = ar^{3} = 189\) \(7 \times r^{3} = 189 \implies r^{3} = 27\) \(r = \sqrt[3]{27} = 3\) \(S_{n} = \frac{a(r^{n} - 1)}{r - 1}\) \(S_{3} = \frac{7(3^{3} - 1)}{3 - 1} \) = \(\frac{7 \times 26}{2} = 91\)

\(\frac{0.24633}{0.03060}\) multiplying throughout by 100,000 = \(\frac{24633}{3060}\) = 8.05 = 8.1

No of students offering Physics are \(\frac{12}{360}\) x 150 = 5

From the distribution, Mode = 1 and Median = \(\frac{2 + 2}{2}\) = 2 = 1, 2

Prime numbers between 1 and 6 are 2, 3, and 5 x = 2 + 3 + 5 = 10 H.C.F. of 6, 9, 15 = 3 ∴ y = 3 X x Y = 10 x 3 = 30