Question Library

Let f(x) = x2 - x - k Then by the factor theorem, (x - 4): f(4) = (4)2 - (4) - k = 0 16 - 4 - k = 0 12 - k = 0 k = 12

Let f(p) = 6p3 - p2 - 47p + 30 Then by the remainder theorem, (p - 3): f(3) = remainder R, i.e. f(3) = 6(3)3 - (3)2 - 47(3) + 30 = R 162 - 9 - 141 + 30 = R 192 - 150 = R R = 42

P \(\propto\) mu, p \(\propto \frac{1}{q}\) p = muk ................ (1) p = \(\frac{1}{q}k\).... (2) Combining (1) and (2), we get P = \(\frac{mu}{q}k\) 4 = \(\frac{m \times u}{1}k\) giving k = \(\frac{4}{6} = \frac{2}{3}\) So, P = \(\frac{mu}{q} \times \frac{2}{3} = \frac{2mu}{3q}\) Hence, P = \(\frac{2 \times 6 \times 4}{3 \times \frac{8}{5}}\) P = \(\frac{2 \times 6 \times 4 \times 5}{3 \times 8}\) p = 10

\(r \propto \frac{1}{\sqrt{s}}, r \propto \frac{1}{\sqrt{t}}\) \(r \propto \frac{1}{\sqrt{s}}\) ..... (1) \(r \propto \frac{1}{\sqrt{t}}\) ..... (2) Combining (1) and (2), we get \(r = \frac{k}{\sqrt{s} \times \sqrt{t}} = \frac{k}{\sqrt{st}}\) This gives \(\sqrt{st} = \frac{k}{r}\) By taking the square of both sides, we get st = \(\frac{k^2}{r^2}\) s = \(\frac{k^2}{r^{2}t}\)

3(x + 2) > 6(x + 3) 3x + 6 > 6x + 18 3x - 6x > 18 - 6 -3x > 12 x < -4

|x - 2| < 3 implies -(x - 2) < 3 .... or .... +(x - 2) < 3 -x + 2 < 3 .... or .... x - 2 < 3 -x < 3 - 2 .... or .... x < 3 + 2 x > -1 .... or .... x < 5 combining the two inequalities results, we get; -1 < x < 5

Using Tn = \(\frac{3n + 1}{n + 1}\), T1 = \(\frac{3(1) + 1}{(1) + 1} = \frac{4}{2}\) T2 = \(\frac{3(2) + 1}{(2) + 1} = \frac{7}{3}\) T3 = \(\frac{3(3) + 1}{(3) + 1} = \frac{10}{4}\)

x * y = x + 2y (given) 3 * 4 = 3 + 2(4) = 11 Hence, 2 * (3 * 4) = 2 * 11 = 2 + 2(11) = 2 + 22 = 24

2P + Q = 2\(\begin{pmatrix} 5 & 3 \\ 2 & 1 \end{pmatrix}\) + \(\begin{pmatrix} 4 & 2 \\ 3 & 5 \end{pmatrix}\) = \(\begin{pmatrix} 10 & 6 \\ 4 & 2 \end{pmatrix}\) + \(\begin{pmatrix} 4 & 2 \\ 3 & 5 \end{pmatrix}\) = \(\begin{pmatrix} 14 & 8 \\ 7 & 7 \end{pmatrix}\)

Let A = \(\begin{pmatrix} 5 & 3 \\ 6 & 4 \end{pmatrix}\) Then |A| = \(\begin{pmatrix} 5 & 3 \\ 6 & 4 \end{pmatrix}\) = 20 - 18 = 2 Hence A-1 = \(\frac{1}{|A|}\begin{pmatrix} 4 & -3 \\ -6 & 5 \end{pmatrix}\) = \(\frac{1}{2}\begin{pmatrix} 4 & -3 \\ -6 & 5 \end{pmatrix}\) = \(\begin{pmatrix} 4 \times 1/2 & -3 \times 1/2 \\ -6 \times 1/2 & 5 \times 1/2 \end{pmatrix}\) = \(\begin{pmatrix} 2 & -\frac{3}{2} \\ -3 & \frac{5}{2} \end{pmatrix}\)

y = x sin x Where u = x and v = sin x Then \(\frac{\delta u}{\delta x}\) = 1 and \(\frac{\delta v}{\delta x}\) = cos x By the chain rule, \(\frac{\delta y}{\delta x} = v\frac{\delta u}{\delta x} + u\frac{\delta v}{\delta x}\) = (sin x)1 + x cos x = sin x + x cos x

\(y = (2x + 2)^{3}\) \(\frac{\mathrm d y}{\mathrm d x} = 3(2x + 2)^{3 - 1} . 2\) = \(6(2x + 2)^{2}\)

A = \(\pi\)r2, \(\frac{\delta A}{\delta r}\) = 2πr So, using \(\frac{\delta A}{\delta t}\) = \(\frac {\delta A}{\delta r}\) x \(\frac {\delta A}{\delta t}\) = 2\(\pi\)r x 0.02 = 2\(\pi\) x 7 x 0.02 = 2 x \(\frac{22}{7}\) x 0.02 = 0.88cm2s-1

\(\sum x\) = (t + 2) + (2t + 4) + (3t + 2) + 2t = 8t N = 4_ ∴ Mean, x = \(\frac{\sum x}{N} = \frac{8t}{4} = 2t\) = 2t

Using x = \(\frac{\sum x}{N}\) in each case, we get; \(\sum_{6}^{i=1} x_i\) = 10 x 7 = 70 \(\sum_{7}^{i=1} x_i\) = 2 + 4 + 8 + 14 + 16 + 18 = 62 Hence the missing number can be obtained from \(\sum_{6}^{i=1} x_i - \sum_{7}^{i=1} x_i\) = 70 - 62 = 8 So, all the seven numbers are 2, 4, 8, 8, 14, 16, 18 Mode = 8