Clearly 1 is the identity element of (G₁)
Inverse of 1 is 1 [∴ (1)(1) = 1]
Inverse of i is -i [∴ (i) (-i) = -i² = 1]
Inverse of -1 is -1 [∴ (-1)(-1) = 1]
Inverse of is i [∴ (-i)(i) = -i² = 1]

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { cot \ x } }{ \sqrt { cot \ x } +\sqrt { tan \ x } } dx } \) ....(1)
By the property, \(\int _{ 0 }^{ a }{ f(x)dx } =\int _{ 0 }^{ a }{ f(a-x) } dx\)
∴ I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { cot(\frac { \pi }{ 2 } -x) } }{ \sqrt { cot(\frac { \pi }{ 2 } -x) } +\sqrt { tan(\frac { \pi }{ 2 } -x) } } dx } \)
= \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { tanx } }{ \sqrt { tanx } +\sqrt { cotx } } dx } \) ...(2)
(1) + (2) ⇒ 2I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { cot \ x } +\sqrt { tan \ x } }{ \sqrt { cot \ x } +\sqrt { tan \ x } } dx } \)
=\(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ dx } ={ \left[ x \right] }_{ 0 }^{ \frac { \pi }{ 2 } }=\frac { \pi }{ 2 } -0=\frac { \pi }{ 2 } \)
∴ I = \(\frac { \pi }{ 4 } \)

y = sin²x
\(\frac { dy }{ dx } \) = 2 sin x cos x = sin 2x
∴ m = \(\left( \frac { dy }{ dx } \right) _{ \left( \frac { \pi }{ 3 } ,\frac { 3 }{ 4 } \right) }=sin\frac { 2\pi }{ 3 } =\frac { \sqrt { 3 } }{ 2 } \)
∴ Slope of the normal = \(-\frac { 1 }{ m } =-\frac { 2 }{ \sqrt { 3 } } \)
∴ Equation of normal is y-y₁ = \(-\frac { 1 }{ m } \)(x-x₁)
⇒ \(y-\frac { 3 }{ 4 } =-\frac { 2 }{ \sqrt { 3 } } \left( x-\frac { \pi }{ 3 } \right) \)
⇒ 12\(\sqrt { 3 } \)y-9\(\sqrt { 3 } \) = -24x + 8π [ multiply 12\(\sqrt { 3 } \)]
∴ 24x + 12\(\sqrt { 3 } \)y = 8π + 9\(\sqrt { 3 } \).

\(S=\left\{\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right\}\)
(i) The sample space
S = {1, 2, 3, 4, 5, 6}\(\times\){1, 2, 3, 4, 5, 6}
consists of 36 ordered pairs (α, β) where α and β can take any integer value between 1 and 6 as shown. X is assigned to each point (α, β) the sum of the numbers on the dice .
That is X (α, β) = α + β
Therefore
X (1,1) = 1+1 = 2
X (1, 2) = X (2,1) = 3
X (1,3) = X (2,2) = X (3,1)= 4
X (1, 4) = X (2,3) = X (3, 2)  X (4,1) = 5
X (1,5) = X (2,4) = X (3,3) = X (4, 2) = X (5,1) = 6
X (1,6) = X (2,5) = X (3, 4) = X (4,3 = X (5, 2) X (6,1) = 7
X (2,6) = X (3,5) = X (4,4) = X (5,3) = X (6,2) = 8
X (3,6) = X (4,5) = X (5,4)  X (6,3) = 9
X (4,6) = X (5,5) X (6,4) = 10
X (5,6) = (6,5) = 11
X (6,6) = 12
(ii) Then the random variable X takes on the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
(iii) The inverse images of 10 is {(4, 6), (5, 5), (6, 4)}.
(iv) The number of inverse images are given below

Let X be the random variable of number of black cards occur.
X = {0,1,2}
Sample space ⁵²C₂ = 1326
Let X denote the number of black cards drawn.
X = 0, X (both are red cards) = ²⁶C₂ = 325  
X = 1 (1 black card and 1 red card) = ²⁶C₁ \(\times\) ²⁶C₁ = 676  
X = 2 (both are black cards) = ²⁶C₂ = 325  
∴ X takes the values 0, 1, 2

Given y² = 2a\(\left( x+a\frac { 2 }{ 3 } \right) \)
\(\Rightarrow { y }^{ 2 }=2ax+2a\frac { 2 }{ 3 } \)
Differentiating with respect to 'x' we get,
\( 2 y \frac{d y}{d x}=2 \mathrm{a}+0 \Rightarrow \frac{d y}{d x}=\frac{2 a}{2 y} \\ \frac{d y}{d x}=\frac{a}{y} \Rightarrow a=y \frac{d y}{d x} \)
Substituting the values of a in equation (1), we get
\( y^2=2\left(y \frac{d y}{d x}\right) x+2\left(y \frac{d y}{d x}\right)^{\frac{5}{3}} \\ y^2=2 x y \frac{d y}{d x}+2\left(y \frac{d y}{d x}\right)^{\frac{5}{3}} \)
 \(y^2-2 x y \frac{d y}{d x}=2\left(y \frac{d y}{d x}\right)^{\frac{5}{3}}\)
Raising to cubical power on both sides, we get
\(\left(y^2-2 x y \frac{d y}{d x}\right)^3=8\left(y \frac{d y}{d x}\right)^5\)
Hence \(y^2=2 a x+2 a^{\frac{5}{3}}\)  is a solution of the differential equation
\(\left(y^2-2 x y \frac{d y}{d x}\right)^3=8\left(y \frac{d y}{d x}\right)^5\)

Let z = -2i = 2(-i)
= 2\(\left[ cos\left( -\frac { \pi }{ 2 } \right) +isin\left( -\frac { \pi }{ 2 } \right) \right] \)
[∵ cos(-θ) = cos θ and sin(-θ) = -sinθ]
Principal value of -2i c is\(\frac { \pi }{ 2 } \).

Given line in Ix + my + n = 0 ....(1)
tangent at (x₁, y₁) to the circle x² + y² = 9² is
xx₁ + yy₁= a₂ ...(2)
Comparing the co-efficients of like terms in (1)
and (2), we get, \(\frac { { x }_{ 1 } }{ l } =\frac { { y }_{ 1 } }{ m } =\frac { -{ { a }^{ 2 } } }{ n } \)
\({ x }_{ 1 }=\frac { -{ a }^{ 2 }l }{ n } \), and \({ y }_{ 1 }=\frac { -{ a }^{ 2 }m }{ n } \)
Since (x₁ , y₁) is a point on the circle, x²₁ + y²₁ = a²
\(\left( \frac { -{ a }^{ 2 }l }{ n } \right) +\left( \frac { -{ a }^{ 2 }m }{ n } \right) ={ a }^{ 2 }\)
\(\frac { -{ a }^{ 4 }{ l }^{ 2 } }{ { n }^{ 2 } } +\frac { { a }^{ 4 }{ m }^{ 2 } }{ { n }^{ 2 } } ={ a }^{ 2 }\)
⇒ \(-{ a }^{ 4 }{ l }^{ 2 }+{ a }^{ 4 }{ m }^{ 2 }={ a }^{ 2 }\)
\({ a }^{ 2 }({ l }^{ 2 }+{ m }^{ 2 })=1\)

L. H. S = \(\left[ \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } ,\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } ,\overset { \rightarrow }{ c } \right] \)
\(=\left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) .\left\{ \left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) \times \overset { \rightarrow }{ c } \right\} \)
\(=\left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) .\left\{ \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \times \overset { \rightarrow }{ c } \right\} \)
\(=\left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) .\left\{ \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right\} \ \left[ \because \overset { \rightarrow }{ c } \times \overset { \rightarrow }{ c } =\overset { \rightarrow }{ 0 } \right] \)
\(=\overset { \rightarrow }{ a } .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) +\overset { \rightarrow }{ b } .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) +\overset { \rightarrow }{ c } .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) \)
\(=\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] +0+0\)
\(\left[ \because \left[ \overset { \rightarrow }{ b } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] =\left[ \overset { \rightarrow }{ c } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] =0 \right] \)
\(=\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] \)= R. H. S
Hence proved

Let A =\(\left[ \begin{matrix} 4 \\ -2 \\ 1 \end{matrix}\begin{matrix} 4 \\ 3 \\ 4 \end{matrix}\begin{matrix} 0 \\ -1 \\ 8 \end{matrix}\begin{matrix} 3 \\ 5 \\ 7 \end{matrix} \right] \)
A\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ -2 \\ 4 \end{matrix}\begin{matrix} 4 \\ 3 \\ 4 \end{matrix}\begin{matrix} 8 \\ -1 \\ 0 \end{matrix}\begin{matrix} 3 \\ 5 \\ 7 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }+2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-4{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 4 \\ 11 \\ -12 \end{matrix}\begin{matrix} 8 \\ 15 \\ -32 \end{matrix}\begin{matrix} 7 \\ 19 \\ -25 \end{matrix} \right] \)
A is in row - echelon form and it has 3 non-zero rows.
∴ \(\rho\) (A) = 3

Given adj A =\(\left[ \begin{matrix} 2 & -4 & 2 \\ -3 & 12 & -7 \\ -2 & 0 & 2 \end{matrix} \right] \)
We know that A = \(\pm \frac { 1 }{ \sqrt { |adjA| } } \) adj (adj A)..(1)
|adj A| = \(2\left| \begin{matrix} 12 & -7 \\ 0 & 2 \end{matrix} \right| +4\left| \begin{matrix} -3 & -7 \\ -2 & 2 \end{matrix} \right| +2\left| \begin{matrix} -3 & 12 \\ -2 & 0 \end{matrix} \right| \)
[Expanded along R₁]
= 2(24-0)+4(-6-14)+2(0+24)
= 2(24)+4(-20)+2(24) = 48-80+48
= 96-80 = 16
Now, adj (adj A)
=\(\left[ \begin{matrix} +\left| \begin{matrix} 12 & -7 \\ 0 & 2 \end{matrix} \right| & -\left| \begin{matrix} -3 & -7 \\ -2 & 2 \end{matrix} \right| & +\left| \begin{matrix} -3 & 12 \\ -2 & 0 \end{matrix} \right| \\ -\left| \begin{matrix} -4 & 2 \\ 0 & 2 \end{matrix} \right| & +\left| \begin{matrix} 2 & 2 \\ -2 & 2 \end{matrix} \right| & -\left| \begin{matrix} 2 & -4 \\ -2 & 0 \end{matrix} \right| \\ +\left| \begin{matrix} -4 & 2 \\ 12 & -7 \end{matrix} \right| & -\left| \begin{matrix} 2 & 2 \\ -3 & -7 \end{matrix} \right| & +\left| \begin{matrix} 2 & -4 \\ -3 & 12 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(24-0)-(6-14)+(0+24) \\ -(-8-0)+(4+4)-(0-8) \\ +(28-24)-(-14+6)+(24-12) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 24 & 20 & 24 \\ 8 & 8 & 8 \\ 4 & 8 & 12 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 24 & 8 & 4 \\ 20 & 8 & 8 \\ 24 & 8 & 12 \end{matrix} \right] \)
= \(4\left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] \)
Substituting (2) and (3) in (1) we get,
A = \(\frac { 1 }{ \sqrt { 16 } } .4\left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] \)
A = \(\pm \frac { 4 }{ 4 } \left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] =\pm \left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] \)

Given v = 3-4i, w = 4+3i and \(\frac { 1 }{ u } =\frac { 1 }{ v } +\frac { 1 }{ w } \)
∴ \(\frac { 1 }{ u } =\frac { 1 }{ 3-4i } +\frac { 1 }{ 4+3i } \)
= \(\frac { 3+4i }{ (3-4i)(3+4i) } +\frac { 4-3i }{ (4+3i)(4-3i) } \)
= \(\\ \frac { 3+4i }{ 9-(4i)^{ 2 } } +\frac { 4-3i }{ 16-(3i)^{ 2 } } =\frac { 3+4i }{ 9+16 } +\frac { 4-3i }{ 16+9 } \)
= \(\frac { 3+4i }{ 25 } +\frac { 4-3i }{ 25 } =\frac { 3+4i+4-3i }{ 25 } \)
\(\frac { 1 }{ u } =\frac { 7+i }{ 25 } \)
∴ u = \(\frac { 25 }{ 7+i } \times \frac { 7-i }{ 7-i } =\frac { 25(7-i }{ 7^{ 2 }-({ i }^{ 2 }) } \) 
= \(\frac { 25(7-i) }{ 49+1 } =\frac { 25(7-i) }{ 50 } =\frac { 1 }{ 2 } \)(7-i)
∴ u = \(\frac { 1 }{ 2 } \)(7-i) or \(\frac { 7 }{ 2 } \) - \(\frac { i }{ 2 } \)

Since this is neither a tautology not a contradiction
( p ⟶ q) ↔️ (~p ⟶ q) is a contingency.

Consider the set of all even integers Z_e = {2k | k ∈ Z} = {...,−6, −4, −2, 0, 2, 4, 6,...}.
Let us verify the properties satisfied by + on Z_e.
(i) The sum of any two even integers is also an even integer.
Because x, y∈Z_e, ⇒ x = 2m and y = 2n , m,n∈Z.
So (x + y) = 2m + 2n = 2(m+n)∈Z_e. Hence + is a binary operation on Z_e.
(ii) ∀ x, y∈Z_e, (x + y) = 2(m + n) = 2(m + n) = 2(n + m) = (2n + 2m) = (y + x).
So + has commutative property
(iii) Similarly it can be seen that ∀x, y, z∈Z_e, (x + y) + z = x + ( y + z).
Hence the associative property is true.
(iv) Now take x = 2k , then 2k + e = e + 2k = 2k ⇒ e = 0.
Thus ∀ x ∈ Z_e, ヨ0∈Z_e, ⋺x+0 = 0+x = x.
So, 0 is the identity element.
(v) Taking x = 2k and x′ as its inverse, we have 2k+x' = 0 = x'+2k ⇒ x' = −2k. i.e., x' = −x.
Thus ∀x ∈ Z_e, ヨ-x∈Z_e ⋺x + (−x) = (−x) + x = 0
Hence -x is the inverse of x ∈Z_e.

The equation of the parabola is ( y + 2)² = −(x−9). The parabola crosses the y- axis at (0, −5) and (0, 1). The vertex is at (9, −2) and the axis of the parabola is y = −2. The required area is sketched
Viewing in the positive direction of x-axis, and making horizontal strips, the required area A is given by

\(A=\int _{ -5 }^{ 1 }{ xdy } =\int _{ -5 }^{ 1 }{ (5-4y-{ y }^{ 2 })dy={ \left[ 5y-2{ y }^{ 2 }-\frac { { y }^{ 3 } }{ 3 } \right] }_{ -5 }^{ 1 }=\frac { 8 }{ 3 } -\left( -\frac { 100 }{ 3 } \right) =36 } \)

Dividing by x we get,
\(\frac { dy }{ dx } +\frac { 1 }{ x } y=xlogx\)
This is a linear differential equation 
\(\therefore P=\frac { 1 }{ x } ;Q=logx\)
\(\int { pdx } =\int { \frac { 1 }{ x } } dx=logx\)
\(I.F={ e }^{ \int { pdx } }={ e }^{ log\quad x }=x\)
\(\therefore\) The solution is \({ e }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }=dx+c } \)
\(u=cos\quad x;dv=x\)
\(du=\frac { 1 }{ x } ,v=\frac { { x }^{ 2 } }{ x } \)
\(\int { udv } =uv-\int { vdu } \)
\(yx=\int { xlogxdx+c } \)
\(\Rightarrow xy=\frac { { x }^{ 2 } }{ x } logx-\int { \frac { { x }^{ 2 } }{ 2 } } .\frac { 1 }{ x } dx\)
\(\Rightarrow xy=\frac { { x }^{ 2 } }{ x } logx-\frac { 1 }{ 2 } \int { xdx } \)
\(\Rightarrow xy=\frac { { x }^{ 2 } }{ x } logx-\frac { 1 }{ 2 } .\frac { { x }^{ 2 } }{ 2 } +c\)
\(\Rightarrow xy=\frac { { 2x }^{ 2 }logx-{ x }^{ 2 }+4c }{ 4 } \)
\(\Rightarrow 4xy=2{ x }^{ 2 }logx-{ x }^{ 2 }+4c\)

\({ tan }^{ -1 }\left( { \frac { x-1 }{ x-2 } } \right) +{ tan }^{ -1 }\left( \frac { x+1 }{ x+2 } \right) =\frac { \pi }{ 4 } \)
\(\Rightarrow { tan }^{ -1 }\left( \cfrac { \frac { x-1 }{ x-2 } +\frac { x+1 }{ x+2 } }{ 1-\left( \frac { x-1 }{ x-2 } \right) \left( \frac { x+1 }{ x+2 } \right) } \right) =\frac { \pi }{ 4 } \)

\(\Rightarrow \frac { 2{ x }^{ 2 }-4 }{ { x }^{ 2 }-4-{ x }^{ 2 }+1 } =1\)
\(\Rightarrow\) 2x²- 4 = -3
\(\Rightarrow \) 2x²- 4 = -3
\(\Rightarrow\) 2x² = -3 + 4 = 1
\(\Rightarrow\) \({ x }^{ 2 }=\frac { 1 }{ 2 } \)
\(\Rightarrow\) \(x=\frac { 1 }{ \sqrt { 2 } } \)

Given (x₁ + iy₁)(x₂ + iy₂) ....(x_n + iy_n) = a + ib
Taking modulus
|(x₁+ iy₁)(x₂ + iy₂) ....(x_n + iy_n)| = |a + ib|
|x₁ + iy₁| + |x₂ + iy₂|+...+ |x_n + iy_n| = |a + ib|
\( \sqrt{x_{1}^{2}+y_{1}^{2}} \sqrt{x_{2}^{2}+y_{2}^{2}} \sqrt{x_{3}^{2}+y_{3}^{2}} \cdots \sqrt{x_{n}{ }^{2}+y_{n}^{2}} =\sqrt{a^{2}+b^{2}} \)
Squaring on both sides
\( \left(x_{1}^{2}+\mathrm{y}_{1}^{2}\right)\left(x_{2}^{2}+\mathrm{y}_{2}^{2}\right)\left(x_{3}^{2}+\mathrm{y}_{3}^{2}\right) \ldots\left(x_{\mathrm{n}}^{2}+\mathrm{y}_{\mathrm{n}}^{2}\right) =\mathrm{a}^{2}+\mathrm{b}^{2} \)
Hence proved.

Let the time by one man alone be x days and one woman alone be y days
∴ By the given data,
\(\frac { 4 }{ x } +\frac { 4 }{ y } =\frac { 1 }{ 3 } \)
and \(\frac { 2 }{ x } +\frac { 5 }{ y } =\frac { 1 }{ 4 } \)
put \(\frac { 1 }{ x } \) = s and \(\frac { 1 }{ y } \) = t
∴ 4s + 4t = \(\frac { 1 }{ 3 } \)
and 2s + 5t = \(\frac { 1 }{ 4 } \)
The matrix form of the system of equation is
\(\left[ \begin{matrix} 4 & 4 \\ 2 & 5 \end{matrix} \right] \left[ \begin{matrix} s \\ t \end{matrix} \right] =\left[ \begin{matrix} \frac { 1 }{ 3 } \\ \frac { 2 }{ 4 } \end{matrix} \right] \) ⇒ AX = B where
A = \(\left[ \begin{matrix} 4 & 4 \\ 2 & 5 \end{matrix} \right] \) and B =\(\left[ \begin{matrix} \frac { 1 }{ 3 } \\ \frac { 2 }{ 4 } \end{matrix} \right] \)
X = A^-1B
Now |A| = \(\left| \begin{matrix} 4 & 4 \\ 2 & 5 \end{matrix} \right| \) = 20 - 8 =12 ≠ 0
∴ A^-1 =\(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 12 } \left[ \begin{matrix} 5 & -4 \\ -2 & 4 \end{matrix} \right] \)
∴ X = A^-1B = \(\frac { 1 }{ 12 } \left[ \begin{matrix} 5 & -4 \\ -2 & 4 \end{matrix} \right] \left[ \frac { \begin{matrix} 1 \\ 3 \end{matrix} }{ \begin{matrix} 1 \\ 4 \end{matrix} } \right] \)
=\(\frac { 1 }{ 12 } \left[ \begin{matrix} \frac { 5 }{ 3 } & -1 \\ \frac { -2 }{ 3 } & +1 \end{matrix} \right] \)
= \(\frac { 1 }{ 12 } \left[ \frac { \begin{matrix} 2 \\ 3 \end{matrix} }{ \begin{matrix} 1 \\ 3 \end{matrix} } \right] =\left[ \begin{matrix} \frac { 2 }{ 3 } \times \frac { 1 }{ 12 } \\ \frac { 1 }{ 3 } \times \frac { 1 }{ 12 } \end{matrix} \right] =\left[ \begin{matrix} \frac { 1 }{ 18 } \\ \frac { 1 }{ 36 } \end{matrix} \right] \)
∴ \(\frac { 1 }{ 18 } \Rightarrow \frac { 1 }{ x } =\frac { 1 }{ 18 } \Rightarrow \)x = 18
t = \(\frac { 1 }{ 36 } \Rightarrow \frac { 1 }{ y } =\frac { 1 }{ 36 } \Rightarrow \)y = 36
one man can do 18 days
one woman can do 36 days.

We observe that the sum of the coefficients of the odd powers and that of the even powers are equal.
Hence −1 is a root of the equation.
To find other roots, we divide 2x³+11x²-9x-18 by x+1 and get 2x²+9x-18 as the quotient.
Solving this we get \(\frac{3}{2}\) and -6 as roots.
Thus -6, -1, \(\frac{3}{2}\) are the roots or solutions of the given equation.