we know that
\(\int _{ 0 }^{ \infty }{ { e }^{ -2x }{ x }^{ 5 }dx } \) = \(\frac { n! }{ { a }^{ n+1 } } \)
\(∴\int _{ 0 }^{ \infty }{ { e }^{ -2x }{ x }^{ 5 } } dx=\frac { 5! }{ { 2 }^{ 5+1 } } =\frac { 5! }{ { 2 }^{ 6 } } \)

\(\left( \frac { dy }{ dx } \right) ^{ 3 }+y=x-\frac { 1 }{ \left( \frac { dy }{ dx } \right) } \)
⇒ \(\left( \frac { dy }{ dx } \right) ^{ 3 }+y=\frac { x\left( \frac { dy }{ dx } \right) -1 }{ \left( \frac { dy }{ dx } \right) } \)
⇒ \(\left( \frac { dy }{ dx } \right) ^{ 4 }+y\left( \frac { dy }{ dx } \right) =x\left( \frac { dy }{ dx } \right) -1\)
The highest derivative is of first order and its power is 3.
∴ Order is 1 and degree is 4.

The value of test statistic which separates the critical (or rejection) region and the acceptance region is called the critical value or significant value.

Let \(A=\left( \begin{matrix} 5 & 6 \\ 7 & 8 \end{matrix} \right) \)
Order of A is 2 \(\times\) 2
\(\therefore \rho (A)\le 2\)  [Since minimum of (2, 2) is 2]
Consider the second order minor
\(\left| \begin{matrix} 5 & 6 \\ 7 & 8 \end{matrix} \right| =40-42\)
= \(-2\neq 0\)
There is a minor of order 2, which is not zero
\(\therefore \rho (A)=2\)

Let A =\(\begin{pmatrix} 1 & 5 \\ 3 & 9 \end{pmatrix}\)
Order of A is 2 \(\times\) 2
∴ \(\rho \) (A) \(\le \) 2
Consider the second order minor
\(\begin{vmatrix} 1 & 5 \\ 3 & 9 \end{vmatrix}=-6\neq 0\)
There is a minor of order 2, which is not zero.
∴ \(\rho \) (A) \(\le \) 2

Given
The difference table is

To findy when x = x ⇒ x₀+ nh = x ⇒ 1 + n (1) = x ⇒ n = x-1
Newton's forward interpolation formula is
y_(x=x) = \(\frac { n }{ 1! } { \triangle y }_{ 0 }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n+1)(n+2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }\)+ .....
y_(x=x) = 1 + (x-1)(-2) + \(\frac { (x-1)(x-2) }{ 2 } (4)+\frac { (x-1)(x-2)(x-3) }{ 6 } (-8)+\frac { (x-1)(x-2)(x-3)(x-4) }{ 24 } (16)\)
⇒ y = 1 - 2x + 2(x² - 3x + 2) - \(\frac{4}{3}\) (x - 1) (x - 2) (x - 3) + \(\frac{2}{3}\) (x - 1) (x - 2) (x - 3) (x -4)
⇒ y = 3 - 2x + 2x² - 6x + 4 - \(\frac{4}{3}\) [(x² - 3x + 2) (x- 3)] + \(\frac{2}{3}\) [(x² - 3x + 2)(x² - 7x + 12)]
⇒ y = 2x² - 8x+ 7- \(\frac{4}{3}\) [x³- 3x²+ 2x- 3x² + 9x- 6] + \(\frac{2}{3}\) [x⁴ - 3x³ + 2x² - 7x³ + 21x² - 14x + 12x² - 36x + 24]
⇒ y = 2x²-8x+7- \(\frac{4}{3}\) x³ + 4x² - \(\frac{8}{3}\) x + 4x² - 12x + 8 + \(\frac { { 12x }^{ 4 } }{ 3 } -\frac { 20 }{ 3 } { x }^{ 3 }+\frac { 70 }{ 3 } { x }^{ 2 }-\frac { 100x }{ 3 } +\frac { 48 }{ 3 } \)
⇒ y = \(\frac{2}{3}\) x⁴ + x³ \(\left( \frac { -4 }{ 3 } \frac { -20 }{ 3 } \right) \) + x²\(\left( 2+4+4+\frac { 70 }{ 3 } \right) \) + x \(\left( -8-\frac { 8 }{ 3 } -12-\frac { 100 }{ 3 } \right) \) + 31
⇒ y =  \(\frac{2}{3}\) x⁴ - 8x³ + \(\frac{100}{3}\) x² - 56x + 31 which is the required polynomial.

Slope of the normal at any point P(x, y) = -\(\frac { dx }{ dy } \)
Let Q be (1, 0)
Slope of the normal PQ is \(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
i.e, \(\frac { y-0 }{ x-1 } =\frac { y }{ x-1 } \)
∴ \(\frac { dx }{ dy } =\frac { y }{ x-1 } \) ⇒ \(\frac { dx }{ dy } =\frac { y }{ 1-x } \), which is the differential equation
i.e., (1− x)dx = ydy
ഽ(1−x)dx = ഽydy + c
\(x-\frac { { x }^{ 2 } }{ 2 } =\frac { { y }^{ 2 } }{ 2 } +c\)   ....(1)
Since it passes through (1,2)
1 - \(\frac { 1 }{ 2 } =\frac { 4 }{ 2 } +c\)
\(c=\frac { 1 }{ 2 } -2=\frac { 4 }{ 2 } +c\)
Put \(c = \frac { -3 }{ 2 } \) in (1)
\(x-\frac { { x }^{ 2 } }{ 2 } =\frac { { y }^{ 2 } }{ 2 } -\frac { 3 }{ 2 } \)
2x − x² = y² − 3
⇒ y² = 2x−x² + 3, which is the equation of the curve

Let I = \(\int _{ 0 }^{ 1 }{ \log { \left( \frac { 1 }{ x } -1 \right) } } dx\)
\(I=\int _{ 0 }^{ 1 }{ \log { \left( \frac { 1-x }{ x } \right) } } dx\) ---(1)
By the property, \(\int _{ 0 }^{ a }{ f\left( x \right) dx=\int _{ 0 }^{ a }{ f\left( a-x \right) } } dx\)
\(I=\int _{ 0 }^{ 1 }{ \log { \left( \frac { 1-(1-x) }{ 1-x } \right) } } dx\)
\(I=\int _{ 0 }^{ 1 }{ \log { \left( \frac { 1-1+x }{ 1-x } \right) } } dx\)
\(=\int _{ 0 }^{ 1 }{ \log { \left( \frac { x }{ 1-x } \right) } } dx\)  ---(2)
Adding (1) and (2) we get,
\(2I=\int _{ 0 }^{ 1 }{ \left[ \log { \left( \frac { 1-x }{ x } \right) } +\log { \left( \frac { x }{ 1-x } \right) } \right] } dx\)

[∵ log m + log n= log mn]
 \(=\int _{ 0 }^{ 1 }{ \log { 1 } } dx=0\)   [∵ log 1 = 0]
2I = 0
⇒ I = 0

Here \(\triangle =\left| \begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 & -1 \end{matrix} \right| =13\neq 0\)
\(\therefore \) We can apply Cramer’s Rule and the system is consistent and it has unique solution.
\({ \triangle }_{ x }=\left| \begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 & -1 \end{matrix} \right| =-13\)
\( { \triangle }_{ y }=\left| \begin{matrix} 1 & 4 & 1 \\ 2 & 1 & 3 \\ 3 & 1 & -1 \end{matrix} \right| =39\)
\( { \triangle }_{ z }=\left| \begin{matrix} 1 & 1 & 4 \\ 2 & -1 & 1 \\ 3 & 2 & 1 \end{matrix} \right| =26\)
\(\therefore \) By Cramer’s rule
\(x=\frac { { \triangle }x }{ { \triangle } } =\frac { -13 }{ 13 } =-1\)
\( y=\frac { { \triangle }y }{ { \triangle } } =\frac { 39 }{ 13 } =3\)
\( z=\frac { { \triangle }z }{ { \triangle } } =\frac { 26 }{ 13 } =2\)
\(\therefore \) The solution is (x, y, z) = (−1, 3, 2)

Since only four values of f(x) are given, the polynomial which fits the data is of degree three. Hence fourth differences are zeros.
(ie) Δ⁴y₀ = 0, ∴ (E−1)⁴y = 0
(E⁴ - 4E³ + 6E² - 4E +1)y₀ = 0
E⁴y₀ - 4E³y₀+ 6E²y₀ - 4Ey₀ +y₀  = 0
y₄ − 4y₃ + 6y₂ − 4y₁ + y₀ = 0
67.4 − 4y₃+ 6(54.1) − 4(4.2) + 45 = 0
240.2 = 4y₃ ∴ y₃ = 60.05

(D²−4D−1)y = e^−3x
The auxiliary equation is
m²−4m−1 = 0
(m−2)²−4−1 = 0
(m− 2)² = 5
\(m-2=±\sqrt { 5 } \)
\(m=2±\sqrt5\)
C.F = Ae^{\((2+\sqrt5)x\)} + Be^{\((2-\sqrt5)x\)}
\(PI=\frac { 1 }{ \phi (D) } f(x)\)
= \(\frac { 1 }{ { D }^{ 2 }-4D-1 } { e }^{ -3x }\)
\(=\frac { 1 }{ { (-3) }^{ 2 }-4(-3)-1 } { e }^{ -3x }\) (Replace D by −3)
\(=\frac { 1 }{ 9+12-1 } ​​{ e }^{ -3x }\)
\(=\frac { e^{ -3x } }{ 20 } \)
Hence the general solution is y = C.F+P.I
⇒ y = Ae^{\((2+\sqrt5)x\)} + Be^{\((2-\sqrt5)x\)}+\(\frac { e^{ -3x } }{ 20 } \)

Let the distribution of heights be normally distributed with mean mean 68.22 and standard deviation = 3.286
\(Z=\frac { X-\mu }{ \sigma } =\frac { X-69.25 }{ 3.286 } \)
When X = 74
\(Z=\frac { X-\mu }{ \sigma } =\frac { 74-69.25 }{ 3.286 } =1.4455\)
Now P(Z > 74) = P(Z > 1.44)
= 0.5 – 0.4251
= 0.0749
Expected number of children to be over 74 cm out of 1200 children
= 1200 × 0.0749 ≈ 90 children

ഽ\(\frac { dx }{ x^{ 2 }-3x+2 } \) = ഽ\(\frac { dx }{ { \left( x-\frac { 3 }{ 2 } \right) }^{ 2 }{ -\left( \frac { 1 }{ 2 } \right) }^{ 2 } } \)
= \(\frac { 1 }{ 2\left( \frac { 1 }{ 2 } \right) } \log\left| \frac { \left( x-\frac { 3 }{ 2 } \right) -\frac { 1 }{ 2 } }{ \left( x-\frac { 3 }{ 2 } \right) +\frac { 1 }{ 2 } } \right| +c\)
= \(\log\left| \frac { 2x-4 }{ 2x-2 } \right| +c\)
= \(\log\left| \frac { x-2 }{ x-1 } \right| +c\)

By completing the squares

\(x^{ 2 }-3x+2={ \left( x-\frac { 3 }{ 2 } \right) }^{ 2 }{ -\left( \frac { 9 }{ 4 } \right) }+2\) \(={ \left( x-\frac { 3 }{ 2 } \right) }^{ 2 }-\frac { 1 }{ 4 } \) \(={ \left( x-\frac { 3 }{ 2 } \right) }^{ 2 }{ -\left( \frac { 1 }{ 2 } \right) }^{ 2 }\)

Given supply function p = 7 + x and x = 5
When x₀ = 5, p₀ = 7 + 5 = 12
∴ p₀x₀ = x₁₂ = 60
Producer's Surplus
\(={ p }_{ 0 }{ x }_{ 0 }-\int _{ g }^{ x }{ (x)dx } \)
\(=60-\int _{ 0 }^{ 5 }{ (7+x)dx } \)
\(=60-{ \left[ 7x+\frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 5 }\)
\(=60-\left[ 7(5)+\frac { { 5 }^{ 2 } }{ 2 } \right] \)
\(=60-\left[ 35+\frac { 25 }{ 2 } \right] \)
= 60-(35+12.5)
= 60-47.5
P.S = 12.5 = \(\frac{25}{2}\) units.

Given f'(x) = x + b, f(1) = 5 and f(2) = 13
f'(x) = x + b
\(\Rightarrow \int { f'(x)dx=\int { \left( x+b \right) dx } } \)
[ ∴ Integration is the reverse process of differentiation]
\(\Rightarrow f(0)=\frac { { x }^{ 2 } }{ 2 } +bx+c\)...(1)
Given f(1) = 5
\(\Rightarrow 5=\frac { { 1 }^{ 2 } }{ 2 } +b(1)+c\)
\(\Rightarrow 5=\frac { 1 }{ 2 } +b(1)+c\Rightarrow 5-\frac { 1 }{ 2 } =b+c\)
\(\Rightarrow \frac { 10-2 }{ 2 } =b+c\Rightarrow b+c=\frac { 9 }{ 2 } \)
\(\Rightarrow 2b+2c=9\)..(2)
\(Also\quad f(2)=13\Rightarrow 13=\frac { { 2 }^{ 2 } }{ 2 } +b(2)+c\)
⇒ 13 = 2 + 2b + c
⇒ 13-2 = 2b + c
⇒ 2b + c = 11    ---(3)
(2) - (3) ⟶ 2b + 2c = 9
-2b + -c = -11
c = -2
Substituting c = -2 in (3) we get
2b - 2 = 11⇒ 2b = 11+2 ⇒ 2b = 13
\(\Rightarrow b=\frac { 13 }{ 2 } \)
Substituting \(b=\frac { 13 }{ 2 } \), c = -2 in(1) we get,
\(f(x)=\frac { { x }^{ 2 } }{ 2 } +\frac { 13 }{ 2 } x-2\)