Question Library

The equation of a straight line is given as \(y = mx + b\) where m = the slope of the line b = intercept Given points A(3, 12) and B(5, 22), the slope = \(\frac{22 - 12}{5 - 3}\) = \(\frac{10}{2}\) = 5 Hence, the equation of the line is \(y = 5x + 2\).

Q(m, n) and R(n, -4) Midpoint : P(2, m) \(\implies (\frac{m + n}{2}, \frac{n - 4}{2}) = (2, m)\) \(m + n = 2 \times 2 \implies m + n = 4 ... (i)\) \(n - 4 = 2 \times m \implies n - 4 = 2m ... (ii)\) Solving (i) and (ii) simultaneously, m = 0 and n = 4.

\(\begin{vmatrix} 2 & -4 \\ x & 9 \end{vmatrix} = 58\) \(\implies (2 \times 9) - (-4 \times x) = 58\) \(18 + 4x = 58 \implies 4x = 58 - 18 = 40\) \(x = 10\)

\(y = 6x^3 + 2x^{-2} - x^{-3}\) \(\frac{\mathrm d y}{\mathrm d x} = 18x^2 - 4x^{-3} + 3x^{-4}\)

\(\frac{d}{dx} [\log (4x^3 - 2x)]\) ... (1) Let u = 4x\(^3\) - 2x. \(\frac{\mathrm d}{\mathrm d x} (\log (4x^3 - 2x)) = (\frac{\mathrm d}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\) \(\frac{\mathrm d}{\mathrm d u} (\log u)\) = \(\frac{1}{u}\) \(\frac{\mathrm d u}{\mathrm d x} = 12x^2 - 2\) \(\therefore \frac{d}{dx} [\log (4x^3 - 2x)] = \frac{12x^2 - 2}{u}\) = \(\frac{12x^2 - 2}{4x^3 - 2x}\)

\(f(x) = 3x^3 + 4x^2 + x - 8\) \(f(-2) = 3(-2)^3 + 4(-2)^2 + (-2) - 8\) = \(-24 + 16 - 2 - 8\) = -18

\(\frac{4x - 6}{3} \leq \frac{3 + 2x}{2}\) 2(4x - 6) \(\leq\) 3(3 + 2x) 8x - 12 \(\leq\) 9 + 6x 8x - 6x \(\leq\) 9 + 12 2x \(\leq\) 21 \(x \leq \frac{21}{2}\)

-7 \(\leq\) 9 - 8x < 16 - x -7 \(\leq\) 9 - 8x and 9 - 8x < 16 - x -7 - 9 \(\leq\) -8x and -8x + x < 16 - 9 -16 \(\leq\) -8x and -7x < 7 \(\therefore\) x \(\leq\) 2 and -1 < x -1 < x \(\leq\) 2.

\(T_n = 2^{2n - 1}\) \(T_1 = 2^{2(1) - 1} \) = 2 \(T_2 = 2^{2(2) - 1}\) = 8 \(T_3 = 2^{2(3) - 1}\) = 32 \(T_4 = 2^{2(4) - 1}\) = 128 \(T_1 + T_2 + T_3 + T_4 = 2 + 8 + 32 + 128\) = 170

\(\int_{-1} ^{2} (2x^2 + x) \mathrm {d} x\) = \([\frac{2x^{2 + 1}}{3} + \frac{x^{1 + 1}}{2}]_{-1} ^{2}\) = \([\frac{2x^{3}}{3} + \frac{x^{2}}{2}]_{-1} ^{2}\) = \((\frac{2(2)^{3}}{3} + \frac{2^2}{2}) - (\frac{2(-1)^{3}}{3} + \frac{(-1)^{2}}{2})\) = \((\frac{16}{3} + 2) - (\frac{-2}{3} + \frac{1}{2})\) = \(\frac{22}{3} - (-\frac{1}{6})\) = \(\frac{22}{3} + \frac{1}{6}\) = \(\frac{15}{2}\) = \(7\frac{1}{2}\)

\(p \propto \frac{1}{\sqrt{q}}\) \(\implies p = \frac{k}{\sqrt{q}}\) when p = 3, q = 16. \(3 = \frac{k}{\sqrt{16}}\) \(k = 3 \times 4 = 12\) \(\therefore p = \frac{12}{\sqrt{q}}\) when p = 4, \(4 = \frac{12}{\sqrt{q}} \implies \sqrt{q} = \frac{12}{4}\) \(\sqrt{q} = 3 \implies q = 3^2 \) \(q = 9\)

200 mangoes at 4 for N2.50 \(\implies\) Total cost price = \(\frac{200}{4} \times N 2.50\) = N 125.00 Since 30 mangoes got spoilt \(\implies\) Left over = 200 - 30 = 170 mangoes 170 mangoes at 2 for N 2.40 \(\implies\) Total selling point = \(\frac{170}{2} \times N 2.40\) = N 204.00 Profit : N (204.00 - 125.00) = N 79.00 % profit = \(\frac{79}{125} \times 100%\) = 63.2% profit.

S.I = \(\frac{PRT}{100}\) \(\implies\) N 4890 = \(\frac{8550 \times 3 \times x}{100}\) \(x = \frac{4890 \times 100}{8550 \times 3}\) \(x = 19.06%\) \(x \approxeq 19%\)

- Friction depends on the nature of the surfaces in contact

Circular motion has uniform