Given lines 2x + y = 8, y = 2, y = 4

\(\therefore\) Required area
= Area of the region ABCDA
\(\begin{aligned} & =\int_2^4 x d y=\int_2^4\left(\frac{8-y}{2}\right) d y=\frac{1}{2} \int_2^4(8-y) d y \end{aligned}\)
\(\begin{aligned} & =\frac{1}{2}\left[8 y-\frac{y^2}{2}\right]_2^4 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{2}\left[\left(8 \times 4-\frac{16}{2}\right)-\left(8 \times 2-\frac{2^2}{2}\right)\right] \end{aligned}\)
\(=\frac{1}{2}[32-8-16+2]=\frac{1}{2}(10)\) = 5 sq units

We have, f(x) = x³
Let x₁, x₂ ∈ R such that
f(x₁) = f(x₂)
⇒ x₁³ = x₂³  
⇒ x₁ = x₂
∴ f(x) is one-one function.

Let \(\begin{aligned} I & =\int \frac{1}{5+4 x-x^2} d x \end{aligned}\)
\(\begin{aligned} =\int \frac{1}{5+2 \cdot 2 x-x^2-(2)^2+(2)^2} d x \end{aligned}\)
\(\begin{aligned} =\int \frac{1}{5+4-\left(x^2+(2)^2-2 \cdot 2 \cdot x\right)} d x \end{aligned}\)
\(\begin{aligned} =\int \frac{1}{9-(x-2)^2} d x=\int \frac{1}{(3)^2-(x-2)^2} \end{aligned}\)
\(\begin{aligned} =\frac{1}{2 \times 3} \log \left|\frac{3+x-2}{3-x+2}\right|+C \end{aligned}\)
\(\begin{aligned} \left[\because \int \frac{d x}{a^2-x^2}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C\right] \end{aligned}\)
\(=\frac{1}{6} \log \left|\frac{1+x}{5-x}\right|+C\)

A = {1, 2, 3 ...13, 14}
R = {(x, y) : 3x - y = 0}
∴ R = {(1, 3), (2, 6), (3, 9),(4, 12)}
R is not reflexive since (1, 2), (2, 2)...(14, 14) ∉ R
Also, R is not symmetric as (1, 3) ∈R, but (3, 1) ∉ R.[3(3) - 1 ≠ 0]
Also, R is not transitive as (1, 3), (3, 9) ∈R, but (1, 9) ∉ R.
[3(1) - 9 ≠ 0]
Hence R is neither reflexive, nor symmetric, nor transitive. 

Given curve is  \(\begin{equation} y=9 x^{2} \end{equation} \)
\(\Rightarrow x^{2}=\frac{1}{9} y\)  
It is a parabolic curve, which open upwards, symmetrical about y-axis and passes through the origin.

∴ Required area = Area of bounded region ABCDA
\(\begin{equation} \left.=\int_{1}^{4} x d y=\frac{1}{3} \int_{1}^{4} \sqrt{y} d y \quad \because x^{2}=\frac{1}{9} y \Rightarrow x=\frac{1}{3} \sqrt{y}\right] \end{equation}  \)
\(=\frac{1}{3}\left[\frac{y^{3 / 2}}{3 / 2}\right]_{1}^{4}=\frac{1}{3} \times \frac{2}{3}\left[(4)^{3 / 2}-(1)^{3 / 2}\right]  \)
\(=\frac{2}{9}\left[(2)^{3}-1\right]=\frac{14}{9} \text { sq units } \)
Hence, the required area is \(\frac{14}{9} \)sq units.

\(\int \frac{1}{(2-x)^{2}+1} d x=-\int \frac{1}{t^{2}+1} d t=-\tan ^{-1} t+C=-\tan ^{-1}(2-x)+C\)

Our problem is to minimise and maximise
Z = x + 2y                     ....(i)
Subject to constraints
\(\begin{aligned} & x+2 y \geq 100 \end{aligned}\)        ....(ii)
\(\begin{aligned} 2 x-y \leq 0 \end{aligned}\)            ....(iii)
\(\begin{aligned} 2 x+y \leq 200 \end{aligned}\)        ....(iv)
and \(\begin{aligned} x \geq 0, y \geq 0 \end{aligned}\)     ....(v)
Table for line x + 2y = 100 is

So, the line x + 2y = 100 is passing through the points (0, 50) and (100, 0).
On putting (0, 0) in the inequality x + 2y \(\geq\) 100, we get
0 + 2 \(\times\) 0 \(\geq\) 100
\(\Rightarrow\)     0 \(\geq\) 100 (which is false)
So, the half plane is away from the origin.
Table for line 2x - y = 0 is

So, the line 2x - y = 0 is passing through the points (0, 0) and (10, 20).
On putting (5, 0) in the inequality 2x - y \(\leq\) 0,
we get
2 \(\times\) 5 - 0 \(\leq\) 0
\(\Rightarrow\)     10 \(\leq\) 0 (which is false)
So,the half-plane is towards Y-axis.
Table for line 2x + y = 200 is

So, the line 2x + y = 200 is passing through the points (0, 200) and (100, 0).
On putting (0, 0) in the inequality 2x + y \(\leq\) 200, we get 2 \(\times\) 0 + 0 \(\leq\) 200 \(\Rightarrow\) 0 \(\leq\) 200 (which is true).
So, the half plane is towards the origin.
Also, x, y \(\geq\) 0.
So, the region lies in the lst quadrant.

Clearly, feasible region is ABCDA.
On solving equations 2x - y=0 and x + 2y=100,
we get B(20, 40).
Again, solving the equations 2x - y = 0 and 2x + y = 200, we get C(50, 100).
The corner points of the feasible region are A(0, 50), B(20, 40), C(50, 100) and D(0, 200).
The values of Z at corner points are given below

The maximum value of Z is 400 at D(0, 200) and the minimum value of Z is 100 at all the points on the line segment joining A(0, 50) and B(20, 40).

Given, Z = x + 2y
Subject to the constraints
\(2 x+y \geq 3 ; x+2 y \geq 6 \text { and } x \geq 0, y \geq 0\)
Now, considering the inequations as equations, we get
2x + y = 3             ....(i)
x + 2y = 6             ....(ii)
Table for line 2x + y = 3 is

On putting (0, 0) in the inequality \(2 x+y \geq 3\) 0 \(\geq 3\) (which is false)
So, the half plane, is away from the origin
Table for line x + 2y = 6

On putting (0, 0) in the inequality x + 2y \(\geq\) 6 0  \(\geq\) 6 (which is false)
So, the half plane is away from the origin.
Also, x \(\geq\) 0 and y \(\geq\) 0, so the feasible region lies in the Ist quadrant.
The point of intersection of Eqs. (i) and (ii) is (0, 3).
The graphical representation of the above system of inequations is given below.

\(\therefore\) Feasible region is shaded region above.

Here, the feasible region is unbounded and the open half plane determined by x + 2y < 6 has no point in common with the feasible region. Hence, Z = 6 is minimum at A(0, 3) and C(6, 0) and also on the line segment AC.

The given function is f: \(R \rightarrow R\) such that
\(f(x)=4 x^3+7\)
To show f is bijective, we have to show that f is one-one and onto.
One-One function Let \(x_1, x_2 \in R\) such that
\(f\left(x_1\right)=  f\left(x_2\right)\)
\(\Rightarrow  4 x_1^3+7=4 x_2^3+7\)
\(\Rightarrow  4 x_1^3=4 x_2^3 \)
\(\Rightarrow  x_1^3-x_2^3=0\)
\(\Rightarrow  \left(x_1-x_2\right)\left(x_1^2+x_1 x_2+x_2^2\right)=0\)
\(\Rightarrow  {\left[\because a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]} \)
\(\Rightarrow  \left(x_1-x_2\right)\left[\left(x_1+\frac{x_2}{2}\right)^2+\frac{3}{4} x_2^2\right]=0\)
⇒ Either \(x_1-x_2=0\)
or \(\left(x_1+\frac{x_2}{2}\right)^2+\frac{3}{4} x_2^2=0\)
But Eq. (ii) gives complex roots as \(x_1, x_2 \in R\)
\(\therefore x_1-x_2  =0 \)
\(\Rightarrow  x_1 =x_2 \)
\(\text { Thus, }  f\left(x_1\right)  =f\left(x_2\right) \)
\(\Rightarrow  x_1 =x_2, \forall x_1, x_2 \in R\)
Therefore, f(x) is a one-one function.
Onto function Let ve R (codomain) be any arbitraty number.
\(\text { Then, }  f(x)=y \)
\(\Rightarrow  4 x^3+7=y\)
\(\Rightarrow  4 x^3=y-7\)
\(\Rightarrow  x^3=\frac{y-7}{4}\)
\(\Rightarrow  x=\left(\frac{y-7}{4}\right)^{1 /3}\)
which is a real number.
[\(\because y \in R\)]
Thus, for every \(y \in R\) (codomain), there exists
\(\mathrm{x}  =\left(\frac{y-7}{4}\right)^{1 / 3} \in R \text { (domain) such that } \)
\(f(x)  =f\left[\left(\frac{y-7}{4}\right)^{1 / 3}\right]=4\left[\left(\frac{y-7}{4}\right)^{1 / 3}\right]^3+7\)
\(=4\left(\frac{y-7}{4}\right)+7=y-7+7=y\)
\(\Rightarrow f(x)\) is an onto function.
Since, f(x) is both one-one and onto, so it is a bijective.

Let \(I=\int \sqrt{x^{2}-8 x+7} d x\) 
Here, the coefficient of x² is unity.
\(\therefore I=\int \sqrt{x^{2}-8 x+7} d x=\int \sqrt{x^{2}-8 x+7+4^{2}-(4)^{2}} d x\) 
[ here a = 1and b = -8, adding and subtracting \(\left(\frac{b}{2 a}\right)^{2}=\left(\frac{-8}{2}\right)^{2}=(4)^{2}\) under, the square root]
\(=\int \sqrt{(x-4)^{2}+7-16} d x=\int \sqrt{(x-4)^{2}-(3)^{2}} d x\) 
Now, put \(x-4=t \Rightarrow d x=d t\) 
\(\therefore \ I=\int \sqrt{t^{2}-3^{2}} d t\) 
\(=\frac{t}{2} \sqrt{t^{2}-3^{2}}-\frac{(3)^{2}}{2} \log \left|t+\sqrt{t^{2}-3^{2}}\right|+C\) 
 \(\left[\because \int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|\right]\) 
\( =\frac{(x-4)}{2} \sqrt{(x-4)^{2}-(3)^{2}} -\frac{(3)^{2}}{2} \log \left|x-4+\sqrt{(x-4)^{2}-(3)^{2}}\right|+C \) 

\(I=\int { \frac { 2 }{ (1-x)(1+{ x }^{ 2 }) } } dx\)
Let  \(\frac { 2 }{ (1-x)(1+{ x }^{ 2 }) } \equiv \frac { A }{ 1-x } +\frac { Bc+C }{ 1+{ x }^{ 2 } } ....(1)\)
Multiplying by \((1-x)\quad (1+{ x }^{ 2 }),\) 
we get : \(2\equiv A(1+{ x }^{ 2 })+(Bx+C)(1-x).\)  
Putting \(x-1,2=A(2) \Rightarrow A=1.\)
Putting  \(x=0,=A+C\Rightarrow C=2-1=1.\)
Comparing coeffs. of  \({ x }^{ 2 },0=A-B\) 
\(\Rightarrow B=A-1.\)
ஃ From (1),  \(\frac { 2 }{ (1-x)(1+{ x }^{ 2 }) } =\frac { 1 }{ 1-x } +\frac { x+1 }{ 1+{ x }^{ 2 } } .\)
\(\therefore I= \int { \frac { 2 }{ (1-x)(1+{ x }^{ 2 }) } } dx\)
\(=\int { \frac { 1 }{ 1-x } } dx+\frac { 1 }{ 2 } \int { \frac { 2x }{ 1+{ x }^{ 2 } } dx } +\int { \frac { 1 }{ 1+{ x }^{ 2 } } } \)
\(=\int { \frac { \log { \left| 1-x \right| } }{ -1 } } +\frac { 1 }{ 2 } \log { \left| 1+{ x }^{ 2 } \right| } +\tan ^{ -1 }{ x+C } \)
\(=-\log { \left| 1-x \right| } +\frac { 1 }{ 2 } \log { \left| 1+{ x }^{ 2 } \right| } +\tan ^{ -1 }{ x+C } \)
\(\left[ \because { x }^{ 2 }\ge 0\Rightarrow 1+{ x }^{ 2 }>0\Rightarrow \left| 1+{ x }^{ 2 } \right| =1+{ x }^{ 2 } \right] \)

We have: f(x) = 4x + 3
Now \(f(x_{ 1 })=f(x_{ 2 })\)
\(\Rightarrow \) \(4x_{ 1 }+3=4x_{ 2 }+3\)
\(\Rightarrow \) \(x_{ 1 }=x_{ 2 }\) \(\Rightarrow \) \(f\) is one-one.
\(\Rightarrow \) \(f\) is invertible.
Let y = f(x) = 4x + 3
\(\Rightarrow \) \(x=\frac { y-3 }{ 4 } \)
Now f(x) = y \(\Rightarrow \) \(f^{ -1 }(y)=x\)
\(\Rightarrow \) \(f^{ -1 }(y)=\frac { x-3 }{ 4 } \)
\(\Rightarrow \) \(f^{ -1 }(x)=\frac { y-3 }{ 4 } \).

(i) Required area = \(\overset { 2 }{ \underset { 1 }{ \int { } } } ydx\)  [Taking vertical strips]

\(=\overset { 2 }{ \underset { 1 }{ \int { } } } { x }^{ 2 }dx=\left[ \frac { { x }^{ 3 } }{ 3 } \right] _{ 1 }^{ 2 }\)
\(=\frac { 8 }{ 3 } -\frac { 1 }{ 3 } =\frac { 7 }{ 3 } sq.units.\)
(ii) Required. area
\(=\overset { 5 }{ \underset { 1 }{ \int { } } } ydx=\overset { 5 }{ \underset { 1 }{ \int { } } } { x }^{ 4 }dx\)
\(=\left[ \frac { { x }^{ 5 } }{ 5 } \right] _{ 1 }^{ 5 }=\frac { { 5 }^{ 5 } }{ 5 } -\frac { 1 }{ 5 } =625-\frac { 1 }{ 5 } \)
\(=\frac { 3125-1 }{ 5 } =\frac { 3124 }{ 5 } \)
\(=624.8sq.units\)

\(23\over2\)

As sin x is positive in 1^st and 2^nd quadrant and negative is 3^rd and 4^th quadrant.
Area = \(\int _{ 0 }^{ 2\pi }{ |sinx|dx } \)
\(=\int _{ 0 }^{ \pi }{ sinxdx } +\int _{ \pi }^{ 2\pi }{ (-sinx)dx } \)
= 4 sq units

∫sec²(7 – 4x)dx =\(\frac { tan(7-4x) }{ -4 } +C=-\frac { 1 }{ 4 } \)tan(7 - 4x) + C

T₁ and T₃ are similar as their sides are proportional.

\(l =\int_{-2}^1 \sqrt{5-4 x-x^2} d x\)
\(=\int_{-2}^1 \sqrt{5-\left(x^2+4 x+4\right)+4} d x\)
\(=\int_{-2}^1 \sqrt{9-(x+2)^2} d x\)
Let \(x+2=t \Rightarrow d x=d t\)
Lower limit When x=-2, then t=0
Upper limit When x=1, then t=3
\(\therefore l  =\int_0^3 \sqrt{3^2-t^2} d t\)
\(=\left[\frac{1}{2} t \sqrt{3^2-t^2}+\frac{1}{2} 3^2 \sin ^{-1}\left(\frac{t}{3}\right)\right]_0^3\)
\(=0+\frac{9}{2} \sin ^{-1} 1-0=\frac{9}{2} \cdot \frac{\pi}{2}=\frac{9 \pi}{4}\)

We will solve this integral by partial fraction.
\(\text { Let } \frac{x^2}{\left(x^2+1\right)\left(3 x^2+4\right)}=\frac{A x+B}{\left(x^2+1\right)}+\frac{C x+D}{\left(3 x^2+4\right)} \)
\( \Rightarrow x^2=(A x+B)\left(3 x^2+4\right)+(C x+D)\left(x^2+1\right) \)
\(=(3 A+C) x^3+(3 B+D) x^2+(4 A+C) x+4 B+D\)
Equating similar terms, we get
3 A+C=0,3 B+D=1,4 A+C=0 and 4 B+D=0
On solving, we get A=0, B =-1, C=0 and D=4
Thus, \(I  =\int \frac{-d x}{\left(x^2+1\right)}+\int \frac{4 d x}{\left(3 x^2+4\right)}\)
\(=\int-\frac{d x}{\left(x^2+1\right)}+\frac{4}{3} \int \frac{d x}{\left(x^2+\frac{4}{3}\right)}\)
\(=\int-\frac{d x}{x^2+1}+\frac{4}{3} \int \frac{d x}{x^2+\left(\frac{2}{\sqrt{3}}\right)^2}\)
\(=-\tan ^{-1} x+\frac{4}{3} \cdot \frac{\sqrt{3}}{2} \tan ^{-1}\left(\frac{\sqrt{3} x}{2}\right)+C\)

Our problem is to maximise
Z = 3x + 4y                 ...(i)
Subject to the constraints are
\(\begin{aligned} x+y & \leq 4 \end{aligned}\)           ...(ii)
\(\begin{aligned} x & \geq 0, y \geq 0 \end{aligned}\)       ...(iii)
Table for the line x + y = 4 is

On putting (0, 0) in the inequality x+ y \(\leq\) 4, we have
0 + 0 \(\leq\) 4
\(\Rightarrow\)    0 \(\leq\) 4 (which is true)
So, the half plane is towards the origin

So, the feasible region lies in the Ist quadrant.
\(\therefore\) Feasible region is OABO.
The corner points of the feasible region are O(0, 0), A(4, 0) and B(0, 4). The values of Z at these points are as follows

Therefore, the maximum value of Z is 16 at the point B(0, 4).

Maximise Z = 70x + 40y
Subject to the constraints
         \(\begin{array}{r} 3 x+2 y \leq 9 \end{array}\),
         \(\begin{array}{r} 3 x+y \leq 9 \end{array}\)
  and   \(\begin{array}{r} x \geq 0, y \geq 0 \end{array}\)
Now, considering the inequations as equations, we get
3x + 2y = 9         .....(i)
3x + y  =9           ....(ii)
Table for line 3x + 2y = 9 is

So, it passes through the points (3, 0) and (0, 9/2) on putting (0, 0) in the inequality 3x + 2y \(\leq\) 9.
0 \(\leq\) 9 (which is true)
So, the half plane is towards the origin.
Table for line 3x + y = 9 is

On putting (0, 0) in the inequality 3x + y \(\leq\) 9
0\(\leq\)9 (which is true)
So, the half plane is towards the origin.
Also, x \(\geq\) 0,y \(\geq\) 0, so the feasible region lies in the Ist quadrant.
The point of intersection of Eqs. (i) and (ii) is (3, 0).
The graphical representation of the above system of inequations is given below

Clearly, feasible region is OABO, whose corner points are O(0, 0), A(3, 0) and B (0, 9/2).

In the table, we find that maximum value of Z is 210 at the point A(3, 0).

The graph of y = sin x and y = cos x are given below

It is clear from the graph that the common region bounded by the X-axis, y = cos x, y = sin x and \(0 \leq x \leq \frac{\pi}{2}\) is the shaded region.
For point of intersection, consider cos x = sin x
\(\Rightarrow \tan x=1 \)
\(\Rightarrow x=\frac{\pi}{4} \) 
\(\therefore\) Required bounded area = Area of shaded region \(\left(A_{1}+A_{2}\right)\) 
\(=\int_{0}^{\pi / 4} \sin x d x+\int_{\pi / 4}^{\pi / 2} \cos x d x\) 
\(=[-\cos x]_{0}^{\pi / 4}+[\sin x]_{\pi / 4}^{\pi / 2}\) 
\(=-\left[\cos \frac{\pi}{4}-\cos 0\right]+\left[\sin \frac{\pi}{2}-\sin \frac{\pi}{4}\right]\) 
\(=-\left[\frac{1}{\sqrt{2}}-1\right]+\left[1-\frac{1}{\sqrt{2}}\right]\) 
\(=-\frac{1}{\sqrt{2}}+1+1-\frac{1}{\sqrt{2}}\) 
\(=2-\frac{2}{\sqrt{2}}\) 
\(=(2-\sqrt{2}) \text { sq units }\) 
Hence, the required area is \((2-\sqrt{2}) \text { sq units }\).

We .have a relation R on N x N defined by (a, b)R(c, d), if
ad(b + c) = bc(a + d) .
Reflexive Let (a, b) ∈ N x N be any arbitrary element.
We have to show that (a, b) R (a, b), i.e.
ab(b + a) = ba(a + b), which is always true, as natural numbers are commutative under usual multiplication and addition .•
Since, (a, b) ∈ N x N was arbitrary, so R is reflexive.
Symmetric Let (a, b), (c, d) ∈ N x N
such that (a, b) R (c, d), i.e. ad(b + c) = bc(a + d)
We have to show that (c, d) R (a, b),
i.e. cb(d + a) = da(c + b)
From Equation (i), we have ad( b + c) = bc( a + d)
⇒ da(c + b) = eb(d + a)
[natural numbers are commutative under
usual addition and multiplication] [1]
⇒ cb(d+a) = da(c+b)
⇒ (c, d)R(a, b)
So, R is symmetric. 
Transitive Let (a, b), (c, d) and (e, f) ∈N x N such that (a, b) R (c, d) and (c, d) R (e, f).
⇒ (a, b)R(c, d) ⇒ ad(b+c) = be(a+d)
\( \Rightarrow \frac{b+c}{b c}=\frac{a+d}{a d} \)
\(\Rightarrow \frac{1}{c}+\frac{1}{b}=\frac{1}{d}+\frac{1}{a}\) 
and \((c, d) R(e, f) \Rightarrow c f(d+e)=d e(c+f)\) 
\(\Rightarrow \frac{d+e}{d e}=\frac{c+f}{c f} \)
\( \Rightarrow \frac{1}{e}+\frac{1}{d}=\frac{1}{f}+\frac{1}{c}\) 
Now, adding Equation (ii) and (iii), we get
\(\left(\frac{1}{c}+\frac{1}{b}\right)+\left(\frac{1}{e}+\frac{1}{d}\right)=\left(\frac{1}{d}+\frac{1}{a}\right)+\left(\frac{1}{f}+\frac{1}{c}\right) \) 
\(\Rightarrow \frac{1}{b}+\frac{1}{e}=\frac{1}{a}+\frac{1}{f}\) 
\(\Rightarrow \frac{e+b}{b e}=\frac{f+a}{a f}\) 
\(\Rightarrow a f(e+b)=b e(f+a)\) 
\(\Rightarrow a f(b+e)=b e(a+f) \)
\(\Rightarrow (a, b) R(e, f)\) 
So, R is transitive.
Hence, R is an equivalence relation.

(i) b : Curves y = cos x and y = x + 1 meet at point C(O, 1).

(ii) C : curve y=cosx meet the x axis at \(A^{\prime}\left(\frac{-\pi}{2}, 0\right)\) and \(A\left(\frac{\pi}{2}, 0\right)\) .
(iii) (a) : \(\int_{-1}^{0}(x+1) d x=\left[\frac{x^{2}}{2}+x\right]_{-1}^{0}=0-\left(\frac{1}{2}-1\right)=\frac{1}{2}\) 
(iv) (d) : \(\int_{0}^{\pi / 2} \cos x d x=[\sin x]_{0}^{\pi / 2}=\sin \frac{\pi}{2}-\sin 0=1\) 
(v) (b) : Required area \(\int_{-1}^{0}(x+1) d x+\int_{0}^{\pi / 2} \cos x d x\) 
\(=\frac{1}{2}+1=\frac{3}{2} \text { sq. units }\)

(i) (b): Objective function is a linear function (involve variable) whose maximum or minimum value is to be found.
(ii) (c): If optimal solution is obtained at two distinct points A and B (corners of the feasible region), then optimal solution is obtained at every point of segment [AB].
(iii) (b): When x ≥ 6 and y ≥ 2, then
2x + y ≥ 2 x 6 + 2, i.e., 2x + y ≥ 14
Hence, x ≥ 0, Y ≥ 0 and 2x + y ≥ 10 are automatically satisfied by every point of the region.
\(\{(x, y): x \geq 6\} \cap\{(x, y): y \geq 2\}\)
(iv) (b): Construct the following table of values of the objective function:

Minimum of Z = - 32 at (0, 8)
(v) (a): Construct the following table of values of the objective function F

Hence, maximum of F = 0

(i) (a) : For f(x) to be defined \(x-2 \neq 0\) i.e.,\(x \neq 2\) 
\(\therefore\) Domain of f = R - {2}
(ii) (b) : Let y =J(x), then \(y=\frac{x-1}{x-2}\) 
\(\Rightarrow x y-2 y=x-1 \Rightarrow x y-x=2 y-1 \Rightarrow x=\frac{2 y-1}{y-1}\) 
Since, \(x \in R-\{2\}\),therefore \(y \neq 1\)
Hence, range of f = R-{1} 
(iii) (d): We have,g(x) = 2f(x) - 1
\(=2\left(\frac{x-1}{x-2}\right)-1=\frac{2 x-2-x+2}{x-2}=\frac{x}{x-2}\)
(iv) (a) : We have, \(g(x)=\frac{x}{x-2}\) ,
Let  \(g\left(x_{1}\right)=g\left(x_{2}\right) \Rightarrow \frac{x_{1}}{x_{1}-2}=\frac{x_{2}}{x_{2}-2}\) 
 \(\Rightarrow x_{1} x_{2}-2 x_{1}=x_{1} x_{2}-2 x_{2} \Rightarrow 2 x_{1}=2 x_{2} \Rightarrow x_{1}=x_{2}\) 
Thus, \(g\left(x_{1}\right)=g\left(x_{2}\right) \Rightarrow x_{1}=x_{2}\) 
Hence, g(x) is one-one.
(v) (c) 

(d) R is correct; A is incorrect

(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
Let \(I =\int_2^8 \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x\)
\(=\int_2^8 \frac{\sqrt{10-(10-x)}}{\sqrt{10-x}+\sqrt{10-(10-x)}} d x\)
\(=\int_2^8 \frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}} d x\)
On adding Eqs. (i) and (ii), we get
\(2 I  =\int_2^8 \frac{\sqrt{10-x}+\sqrt{x}}{\sqrt{10-x}+\sqrt{x}} d x=\int_2^8 1 d x\)
\(\Rightarrow \quad 2 I =[x]_2^8=8-2=6 \Rightarrow I=3 \)
Thus, Assertion and Reason both are correct and Reason is the correct explanation of Assertion.