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Since (x - 1), is a factor, when the polynomial is divided by (x - 1), the remainder = zero \(\therefore (x - 1) = 0\) x = 1 Substitute in the polynomial the value x = 1 = \(p(1)^3 + q(1)^2 + 11(1) - 6 = 0\) p + q + 5 = 0 .....(i) Also since x - 3 is a factor, \(\therefore\) x - 3 = 0 x = 3 Substitute \(p(3)^3 + q(3)^2 + 11(3) - 6 = 0\) 27p + 9q = -27 ......(2) Combine eqns. (i) and (ii) Multiply equation (i) by 9 to eliminate q 9p + 9q = -45 Subtract (ii) from (i), \(18p = 18\) \(\therefore\) p = 1 Put p = 1 in (i), \(1 + q = -5 \implies q = -6\) \((p, q) = (1, -6)\)

\(\frac{a}{a - x}\) = \(\frac{b}{x - b}\) \(\frac{1}{1 - x}\) = \(\frac{3}{x - 3}\) ∴ 3(1 - x) = x - 3 3 - 3x = x - 3 Rearrange 6 = 4x; x = \(\frac {6}{4}\) = \(\frac{3}{2}\)

\(\frac{1}{r - 1}\) + \(\frac{2}{r + 1}\) = \(\frac{3}{r}\) Multiply through by r(r -1) which is the LCM = (r)(r + 1) + 2(r)(r - 1) = 3(r - 1)(r + 1) = r2 + r + 2r2 - 2r 3r2 - 3 = 3r2 r = 3r2 - 3 -r = -3 ∴ r = 3

\(\frac{x - 3}{(1 - x)(x + 2)}\) = \(\frac{p}{1 - x}\) + \(\frac{Q}{x + 2}\) Multiply both sides by LCM i.e. (1 - x(x + 2)) ∴ x - 3 = p(x + 2) + Q(1 - x) When x = +1 (+1) - 3 = p(+1 + 2) + Q(1 - 1) -2 = 3p + 0(Q) 3p = -2 ∴ p = \(\frac{-2}{3}\)

\(\frac{1}{x}\) > 2 = \(\frac{x}{x^2}\) > 2 x > 2x2 = 2x2 2 - x 0, 2x - 1 0, x 0 x < 0, x , \(\frac{1}{2}\) = 0

In an AP, Tn = a + (n - 1)d T6 = a + 5d = 11 The first term = a = 1 ∴ T6 = 1 + 5d = 11 5d = 11 - 1 5d = 10 ∴ d = 2

log10 r + log10 r2 + log10 r4 + log10 r8 + log10 r16 + log10 r32 = 63 log10r63 = 63 63 = 1063 ∴ r = 10

\(\frac{(n + 1)(n + 2)}{2}\) If n = 1, the expression becomes 3 n = 2, the expression becomes 6 n = 4, the expression becomes 15 n = 5, the expression becomes 21

a \(\oplus\) b = ab The set of all national rules Q, is closed under the operations, additions, subtraction, multiplication and division. Since a \(\oplus\) b = ab; b \(\oplus\) a = ba = ab The number 1 is the identity element under multiplication

The inverse of 2 is 6 since 2 x 6 = 12; under mod 10 12 = 2 which is also the value required

\(\begin{pmatrix} 1 & 1 \\ 3 & y \end{pmatrix}\)\(\begin{pmatrix} x \\ 1 \end{pmatrix}\) = \(\begin{pmatrix} 4 \\ 1\end{pmatrix}\) = x + 1 = 4 x = 4 - 1 = 3 3x + y =1 3(3) = y = 1 = 9 + y = 1 y = 1 - 9 = -8

Total number of pupils = 7 + 27 + 6 = 40 n(at least 11 years old) = 27 + 7 = 34. P(picking a pupil who is at least 11 years old) = \(\frac{34}{40} = \frac{17}{20}\).

The sum of angles of a quadrilateral is 360o ∴ (3y - x - z)o + 3xo + (2z - 2y - x)o + po = 360o Where P is the fourth angle 3y - x - z + 3x + 2z - 2y - x + p = 360o p = 360 - (x + y + z) ∴ p = (360 - x - y - z)o

Internal dimension are 50cm, 36cm and 20cm internal volume = 50 x 36 x 20cm3 1000 x 36cm3 = 36000cm3 External dimension are 54cm x 40cm x 22cm = 2160cm2 x 22cm = 47520cm3 Volume of wood = Ext. volume - Int. volume = 47,520cm3 - 36,000cm3 = 11,520cm3

Perimeter = OP + OQ + PQ = 8 + 8 + PQ length PQ = \(\frac{\theta}{360 \times 2\pi r}\) = \(\frac{45}{360}\) x 2 x \(\pi\) x 8 = 2\(\pi\) Perimeter of sector 2r + L Where l = length of arc and r = radius ∴ P = 2(8) + 2\(\pi\) = 16 + 2\(\pi\)