Let the required equation of the straight line trend is
y = a + bX
Since Σx = 0. \(a = \frac{\Sigma y}{x} = \frac{16.9}{5}\)
\(b = \frac{\Sigma xy}{\Sigma x^2} = \frac{13.3}{10} = 1.33\)
Hence, the straight line trend is
y = 3.38 + 1.33 (x - 2002)
∴ The trend for 2007 is
y_t = 3.38 + 1.33 (2007 - 2002)
⇒  y_t = 3.38 + 1.33 (5)
⇒  y_t = 3.38 + 6.65
⇒  y_t = 10.03

\(E(X)=0\times \frac { 1 }{ 5 } +1(\frac { 2 }{ 5 } )+2\left( \frac { 1 }{ 5 } \right) +3\times \frac { 1 }{ 5 } \)
\(=\frac { 2 }{ 5 } +\frac { 2 }{ 5 } +\frac { 3 }{ 25 } =\frac { 7 }{ 5 } \)
\(\\ \therefore 2E(X)=\frac { 14 }{ 5 } ...(1)\)
\(E({ Y }^{ 2 })=0\times \frac { 1 }{ 5 } +{ 1 }^{ 2 }(\frac { 3 }{ 10 } )+{ 2 }^{ 2 }(\frac { 2 }{ 5 } )+{ 3 }^{ 2 }(\frac { 1 }{ 10 } )\)
\(=\frac { 3 }{ 10 } +\frac { 8 }{ 5 } +\frac { 9 }{ 10 } =\frac { 3+16+9 }{ 10 } \)
\(=\frac { 28 }{ 10 } =\frac { 14 }{ 5 } ..(2)\)
From (1) and (2), E(Y²) = 2 E(X).

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 4 } }{ sin } 3x\ sin\ 2x\ dx\)
= \(\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 4 } }{ 2sin } 3xsin2x\quad dx\)
= \(\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left[ cos(3x-2x)-cos(3x+2x) \right] } dx\)
\(\left[ \because 2sin(sinD=cos(C-D))-cos(C+D) \right] \)
= \(\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 4 } }{ (cosx-cos5x)dx } \)
= \(\frac { 1 }{ 2 } { \left[ sinx-\frac { sin5x }{ 5 } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
= \(\frac { 1 }{ 2 } \left[ \left( sin\frac { \pi }{ 4 } -\frac { 1 }{ 5 } sin5\frac { \pi }{ 4 } \right) -\left( sin0-\frac { 1 }{ 5 } sin5(0) \right) \right] \)
= \(\frac { 1 }{ 2 } \left[ \frac { 1 }{ \sqrt { 2 } } -\frac { 1 }{ 5 } \left( \frac { -1 }{ \sqrt { 2 } } \right) \right] \quad \quad \left[ \because sin0=0 \right] \)

= \(\frac { 1 }{ 2 } \left[ \frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 5\sqrt { 2 } } \right] \)
= \(\frac { 1 }{ 2 } \left( \frac { 5+1 }{ 5\sqrt { 2 } } \right) =\frac { 1 }{ 2 } \times \frac { 6 }{ 5\sqrt { 2 } } =\frac { 3 }{ 5\sqrt { 2 } } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } } \)
= \(\frac { 3\sqrt { 2 } }{ 10 } \)

\(\therefore I=\int { \frac { { x }^{ 7 } }{ { x }^{ 5 }+1 } } dx\)
= \(\int { \left( { x }^{ 6 }-{ x }^{ 5 }+{ x }^{ 4 }-{ x }^{ 3 }+{ x }^{ 2 }-x+1-\frac { 1 }{ x+1 } \right) } dx\)
\(I=\frac { { x }^{ 7 } }{ 7 } -\frac { { x }^{ 6 } }{ 6 } +\frac { { x }^{ 5 } }{ 5 } -\frac { { x }^{ 4 } }{ 4 } +\frac { { x }^{ 3 } }{ 3 } -\frac { { x }^{ 2 } }{ 2 } +x-log\left| x+1 \right| +c\)

The auxilary equation is m² + 14m + 49 = 0
⇒ (m + 7)² = 0
⇒ m = -7, -7
The roots are real and equal
∴ CF is (Ax+ B) e^-7x
[∴ (D-7)² = 0, when D = 7]
PI₁ = \(\frac { e^{ -7x } }{ (D-7)^{ 2 } } =\frac { { x }^{ 2 } }{ 2 } \).e^-7x
PI₂ = \(\frac { 4.{ e }^{ 0x } }{ (D-7)(D-7) } =\frac { 4.e^{ 0x } }{ (0-7)(0-7) } =\frac { 4 }{ 49 } \)
∴ The general solution is y = CF + PI₁ + PI₂
⇒ y = (Ax+B)e-7x + \(\frac { { x }^{ 2 } }{ 2 } e^{ -7x }+\frac { 4 }{ 49 } \).

2x-y = 4

y² = 4x and 2x = 4 + y ⇒ 4x = 8 + 2y
y² = 8+2y ⇒ y²-2y-8 = 0
(y-4)(y+2) = 0
y = -2, 4.
Required area \(=\int _{ -2 }^{ 4 }{ ({ x }_{ 1 }-{ x }_{ 2 })dy } \)
\(=\int _{ -2 }^{ 4 }{ \left( \frac { y+4 }{ 2 } -\frac { { y }^{ 2 } }{ 4 } \right) dy } \)
Where x₁ is the line  x₂ is the parabola
\(=\int _{ -2 }^{ 4 }{ \frac { y+4 }{ 2 } dy } -\frac { 1 }{ 4 } \int _{ -2 }^{ 4 }{ { y }^{ 2 }dy } \)
\(=\frac { 1 }{ 2 } { \left[ \frac { { y }^{ 2 } }{ 2 } +4y \right] }_{ -2 }^{ 4 }-\frac { 1 }{ 4 } { \left[ \frac { { y }^{ 3 } }{ 3 } \right] }_{ -2 }^{ 4 }\\ \)
\(=\frac { 1 }{ 2 } \left[ \left( \frac { 16 }{ 2 } +16 \right) -\left( \frac { 4 }{ 2 } -8 \right) \right] -\frac { 1 }{ 4 } \left[ \frac { 64 }{ 3 } +\frac { 8 }{ 3 } \right] \)
\(=\frac { 1 }{ 2 } \left[ 24+6 \right] -\frac { 1 }{ 4 } \left[ \frac { 64+8 }{ 3 } \right] \)
\(=\frac { 1 }{ 2 } (30)-\frac { 1 }{ 4 } \left( \frac { 72 }{ 3 } \right) \)
\(=15-\frac { 1 }{ 4 } (24)=15-6\)
= 9 sq.units.

The given non-homogeneous equations can be written as
\(\left( \begin{matrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) \)

Case (i):
When  \(k\neq 1\) and \(k\neq 2\)
\(\rho (A)=\rho (A,B)=3=\) Number of unknowns
\(\therefore \) The system has unique solution 
Case (ii):
When k = 1
\(\left[ A,B \right] \sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right) \)
\(\rho (A)=\rho\) (A, B) = 1
\(\therefore \) The system is consistent and has infinitely many solutions.
Case (iii):
When k = - 2
\(\left[ A,B \right] \sim \left( \begin{matrix} 1 & 1 & -2 \\ 0 & -3 & 3 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 1 \\ 0 \\ -3 \end{matrix} \right) \)
\(\rho (A)=2\rho (A,B)\)= 3
\(\Rightarrow \rho (A)\neq 2\rho (A,B)\)
\(\therefore\) The system is inconsistent and has no solution.