Let   f ( x ) =  2 x³ + 3 x² - k x + 5

Using remainder theorem,

F ( 2 ) = 7

$$\begin{gathered} \therefore 2 ( 2 {)}^3 + 3 ( 2 {)}^2 - \mathrm{k} ( 2 ) + 5 = 7 \\ \therefore 2 ( 8 ) + 3 ( 4 ) - \mathrm{k} ( 2 ) + 5 = 7 \\ \therefore 16 + 12 - 2 \mathrm{k} + 5 = 7 \\ \therefore 2 \mathrm{k} = 16 + 12 + 5 - 7 \\ \therefore 2 \mathrm{k} = 26 \\ \therefore \mathrm{k} = 13 \end{gathered}$$

A² = 9 A + mI

$$\begin{gathered} \Rightarrow {\mathrm{A}}^2 - 9 \mathrm{A} = \mathrm{mI} ............( \mathrm{i} ) \\ {\mathrm{A}}^2 = \mathrm{A} \mathrm{A} \\ = \left[\begin{array}{cc} 2 & 0\\ - 1 & 7\end{array} \right] \left[\begin{array}{cc} 2 & 0\\ - 1 & 7\end{array} \right] \\ = \left[ \begin{array}{cc} 4 & 0\\ - 9& 49\end{array} \right] \\ \mathrm{Substitute} {\mathrm{A}}^2 \mathrm{in} ( \mathrm{i} ) \\ {\mathrm{A}}^2 - 9 \mathrm{A} = \mathrm{mI} \end{gathered}$$

$$\begin{gathered} \Rightarrow \left[ \begin{array}{cc} 4 & 0\\ - 9& 49\end{array} \right] - 9 \left[\begin{array}{cc} 2& 0\\ - 1& 7\end{array} \right] = \mathrm{m} \left[ \begin{array}{cc}1& 0\\ 0& 1\end{array} \right] \\ \Rightarrow \left[ \begin{array}{cc} 4 & 0\\ - 9& 49\end{array} \right] - \left[\begin{array}{cc} 18& 0\\ - 9& 63\end{array} \right] = \left[ \begin{array}{cc}\mathrm{m}& 0\\ 0& \mathrm{m}\end{array} \right] \\ \Rightarrow \left[ \begin{array}{cc}- 14& 0 \\ 0 & - 14\end{array} \right] = \left[ \begin{array}{cc}\mathrm{m}& 0\\ 0& \mathrm{m}\end{array} \right] \\ \Rightarrow \mathrm{m} = - 14 \end{gathered}$$

$$\begin{gathered} \mathrm{Mean} = \frac{\mathrm{Sum} \mathrm{of} \mathrm{all} \mathrm{observations}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{observations}} \\ \therefore 68 = \frac{45 + 52 + 60 + \mathrm{x} + 69 + 70 + 26 + 81 + 94}{9} \\ \Rightarrow 68 = \frac{497 + \mathrm{x}}{9} \\ \Rightarrow 612 = 497 + \mathrm{x} \\ \Rightarrow \mathrm{x} = 612 - 497 \\ \Rightarrow \mathrm{x} = 115 \end{gathered}$$

Data in ascending order

26,  45,  52,  60,  69,  70,  81,  94,  115

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{observations} \mathrm{is} \mathrm{odd}, \mathrm{the} \mathrm{median} \mathrm{is} \mathrm{the} {\left( \frac{\mathrm{n} + 1}{2} \right)}^{\mathrm{th}} \mathrm{observation} \\ \Rightarrow \mathrm{Median} = {\left( \frac{9 + 1}{2}\right)}^{\mathrm{th}} \mathrm{observation} = 5^{\mathrm{th}} \mathrm{observation} \end{gathered}$$

Hence, the median is  69.

$$\begin{gathered} ( \mathrm{i} ) \mathrm{Slope} \mathrm{of} \mathrm{PQ} = \frac{3 - \mathrm{k}}{1 - 3 \mathrm{k} - 6} \\ \Rightarrow \frac{1}{2} = \frac{3 - \mathrm{k}}{- 3 \mathrm{k} - 5} \\ \Rightarrow - 3 \mathrm{k} - 5 = 2 ( 3 - \mathrm{k} ) \\ \Rightarrow - 3 \mathrm{k} - 5 = 6 - 2 \mathrm{k} \\ \Rightarrow \mathrm{k} = - 11 \end{gathered}$$

( ii ) Substituting  k  in  p  and  Q,  we get

P ( 6, K ) = P ( 6, - 11 )   and 

Q ( 1 - 3 k, 3 )  =  Q ( 1 - 3 ( - 11 ), 3 )  =  Q ( 1 +  33, 3 )  =  Q ( 34, 3 )

$$\begin{gathered} \therefore \mathrm{Midpoint} \mathrm{of} \mathrm{PQ} = \left( \frac{6 + 34}{2}, \frac{- 11 + 3}{2} \right) \\ = \left( \frac{40}{2}, \frac{- 8}{2} \right) \\ = ( 20, - 4 ) \end{gathered}$$

$$\begin{gathered} {\mathrm{cosec}}^2 57° - {\tan }^2 33° + \cos 44° \mathrm{cosec} 46° - \sqrt{2} \cos 45° - {\tan }^2 60° \\ = {\mathrm{cosec}}^2 ( 90° - 33° )° - {\tan }^2 33° + \cos 44° \mathrm{cosec} ( 90° -44 )° - \sqrt{2} \cos 45° - {\tan }^2 60° \\ = {\sec }^{2 }33° - {\tan }^2 33° + \cos 44° \sec 44° - \sqrt{2} \cos 45° - {\tan }^2 60° \\ = 1 + 1 - \sqrt{2} \cos 45° - {\tan }^2 60° \\ = 1 + 1 - \sqrt{2} \left( \frac{1}{\sqrt{2}} \right) - {\left( \sqrt{3} \right)}^2 \\ = 2 - 1 - 3 \\ = - 2 \end{gathered}$$