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y = 3x2 - x3 dy/dx = 6x - 3x2 as dy/dx = 0 6x - 3x2 = 0 3x (2 - x) = 0 this implies that 2 -x = 0 and 3x = 0 x = 2 (or) 0 But = dy/dx = 6x - 3x2 d2y/dx2 = 6 - 6x at x = 2 = 6 - 6(2) = -6 y = 3x2 - x3 = 3(2)2 - 23 = 12 - 8 = 4

\(Mean = \frac{30+56+31+55+43+44}{6}\) = \(\frac{259}{6}\) = 43.167 Median = 30,31,43,44,55,56 = \(\frac{43+44}{2}\) =\( \frac{87}{2}\) = 43.5 Median-mean = 43.5 - 43.17 = 0.33

Range = (Highest - Lowest) items = 27 - 3 = 24

Mean = \(\frac{(0 \times 1) + (1 \times 2) + (2 \times 2) + (3 \times 1) + (4 \times 9)}{1 + 2 + 2 + 1 + 9}\) = \(\frac{(0 + 2 + 4 + 3 + 36)}{15}\) = \(\frac{45}{15}\) =3

Let the 4th angle be = x ∴ x + 45 + 90 + 135 = 360 x + 270 = 360 x = 360 - 270 x = 90 ∴ smallest angle = 45o 45o = N28.00 1o = 28.00/45 135o = (28.00/45) * (135/1) = N84.00

To solve the equation \(^{n}P_{3} - 6(^{n}C_{4}) = 0\), we start by using the formulas for permutations and combinations: \(^{n}P_{r} = \frac{n!}{(n-r)!}\) \(^{n}C_{r} = \frac{n!}{r!(n-r)!}\) \(^{n}P_{3} = \frac{n!}{(n-3)!}\) \(^{n}C_{4} = \frac{n!}{4!(n-4)!}\) \(\frac{n!}{(n-3)!} - 6 \left(\frac{n!}{4!(n-4)!}\right) = 0\) Factor out \(n!\) Factoring \(n!\) from both terms gives: \(n! \left(\frac{1}{(n-3)!} - \frac{6}{4!(n-4)!}\right) = 0\) Since \(n! \neq 0\), we can simplify to: \(\frac{1}{(n-3)!} - \frac{6}{4!(n-4)!} = 0\) Recall that \(4! = 24\), so: \(\frac{6}{4!(n-4)!} = \frac{6}{24(n-4)!} = \frac{1}{4(n-4)!}\) The equation now is: \(\frac{1}{(n-3)!} - \frac{1}{4(n-4)!} = 0\) Multiply through by \(4(n-3)!(n-4)!\) gives: \(4(n-4)! - (n-3)! = 0\) Recall that \((n-3)! = (n-3)(n-4)!\), so: \(4(n-4)! - (n-3)(n-4)! = 0\) \((n-4)!\) Factoring out \((n-4)!\) gives: \((n-4)! \left(4 - (n-3)\right) = 0\) Since \((n-4)! \neq 0\), we have: \(4 - (n-3) = 0\) Solving for \(n\): \(n - 3 = 4 \implies n = 7\) The value of n is: 7

\(^{4}C_2 \times ^{3}C_1\) = \(\left(\frac{4!}{(4-2)!2!}\right)\times\left(\frac{3!}{(3-1)!1!}\right)\) = \(\frac{4!}{2!2!}\times \frac{3!}{2!}\) = \(\frac{4 \times 3 \times 2!}{2 \times 1 \times 2!} \times \frac{3 \times 2!}{2!} = 18\)

Black balls = 5 Red balls = 3 P(B) = \(\frac{5}{8}\), P(R) = \(\frac{3}{8}\) P( a Black and a Red) = black ball and red ball or red ball and black ball i.e BR + RB = (\(\frac{5}{8}\)) * (\(\frac{3}{7}\)) + (\(\frac{3}{8}\)) * (\(\frac{5}{7}\)) = (\(\frac{15}{56}\)) + (\(\frac{15}{56}\)) = \(\frac{30}{56}\) = \(\frac{15}{28}\)

12 + 20 + x + 21 + x-1 +28 = 120 2x + 80 = 120 2x = 120 - 80 2x = 40 x = 20

\(\frac{4}{a-3}-\frac{1}{a+3}\) = \(\frac{4(a+2)-1(a-3)}{(a-3)(a+2)}\) \(\frac{4a+8 -a +3}{(a-3)(a+2)}\) = \(\frac{3a+11}{(a-3)(a+2)}\)

Q' ∩ R Q' = U - Q Q' = {a, b, c, d, g, h, i} R = {c, e, h, g} Q' ∩ R = {c, h, g}

Since PQ = QR, \Delta PQR is a right angle isosceles triangle ∴ ∠PQR = 45o => The bearing of P from R = 180 + 45 = 225o

From the given table, the most common mark is 3 with a frequency of 18

\theta = 30 x 2(∠ at center twice ∠ at circumference) ∴ \theta = 60^\circ Length of arc = \(\frac{\theta}{360}\times 2\pi r\) = \(\frac{60}{360}\times 2 \times \frac{22}{7}\times \frac{21}{1}\) = 22cm

Locus of a point P(x,y) which is equidistant from Q(0,0) and R(2,1) is the perpendicular bisector of the straight line joining Q and R Mid point QR = \(\frac{x_2 + x_1}{2}, \frac{y_2 + y_1}{2}\) = \(\frac{2 + 0}{2}, \frac{1 + 0}{2}\) = \((1, \frac{1}{2})\) Gradient of QR = \(\frac{y_{2} - y_{1}}{x_{2} - x_{1}}\) = \(\frac{1 - 0}{2 - 0}\) = \(\frac{1}{2}\) Gradient of PM = \(\frac{-1}{\frac{1}{2}}\) = -2 Equation of PM = y - y\(_1\) = m(x-x_\(_1\)) i.e y - 1/2 = -2(x-1) 2y - 1 = -4(x-1) 2y - 1 = -4x + 4 2y + 4x = 5