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64. The given eqⁿ of the three lines are

y = m₁ x + c₁ ______ (1)

y = m₂ x + c₂ ______ (2)

y = m₃ x + c₃ ______ (3)

The point of intersection of (2) and (3) is given by.

y - y = (m₂x + c₂) - (m₃ x + c₃)

(m₂ - m₃) x = c₃ - c₂

$\Rightarrow x=\frac{c_3-c_2}{m_2-m_3}.$

Hence, y = $\frac{m_2\left(c_3-c_2\right)}{\left(m_2-m_3\right)}+c_2$

$=\frac{m_2\left(c_3-c_2\right)+c_2\left(m_2-m_3\right)}{m_2-m_3.}$

$=\frac{m_2{c}_3-m_3{c}_2}{m_2-m_3}.$

$ie,\left(\frac{c_3-c_2}{m_2-m_3},\frac{m_2{c}_3-m_3{c}_2}{m_2-m_3}\right)$

As the three lines are concurrent, the point of intersection of (2) and (3) lies on line (1) also

i e, $\frac{m_2{c}_3-m_3{c}_2}{m_2-m_3}=m_1\left(\frac{c_3-c_2}{m_2-m_3}\right)+c_1$

$\Rightarrow \frac{-m_1\left(c_2-c_3\right)+c_1\left(m_2-m_3\right)}{m_2-m_3}=\frac{-m_2{c}_3-m_3{c}_2}{m_2-m_3}.$

m₁ (c₂ - c₃) - c₁ (m₂ - m₃) + m₂ c₃ - m₃ c₂ = 0

m₁ (c₂ - c₃) - m₂ c₁ + m₃ c₁ + m₂ c₃ - m₃ c₂ = 0

m₁ (c₂ - c₃) + m₂ (c₃ - c₁) + m₃ (c₁ - c₂) = 0

A binary operation * on {a, b} is a function from {a, b} × {a, b} → {a, b}

i.e., * is a function from { (a, a), (a, b), (b, a), (b, b)} → {a, b}.

Hence, the total number of binary operations on the set {a, b} is 2⁴ i.e., 16.

The correct answer is B.

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It is given that,

$f:R\to R$ is defined as $f\left(x\right)=\{\begin{array}{ccc}\hfill 1\hfill & \hfill x>0\hfill & \hfill \hfill \\ \hfill 0\hfill & \hfill x=0\hfill & \hfill \hfill \\ \hfill -1\hfill & \hfill x<0\hfill & \hfill \hfill \end{array}$

Also, $g:R\to R$ is defined as $g\left(x\right)=\left[x\right]$ , where [x] is the greatest integer less than or equal to x.

Now, let $x\in \left(0,1\right)$

Then, we have:

$\left[x\right]=1$ if $x=1$ and $\left[x\right]=0$ if $0<x<1$

$\begin{array}{l}\therefore fog\left(x\right)=f\left(g\left(x\right)\right)=f\left(\left[x\right]\right)=\{\begin{array}{cc}\hfill f\left(1\right)\hfill & \hfill if,x=1\hfill \\ \hfill f\left(0\right)\hfill & \hfill if,x\in \left(0,1\right)\hfill \end{array}=\left\{\left(1,"if,x=1"\right),\left(0,:if,x\in \left(0,1\right)"\right):\right\}\\ gof\left(x\right)=g\left(f\left(x\right)\right)\\ =g\left(1\right)\left[x>0\right]\\ =\left[1\right]=1\end{array}$

Thus, when $x\in \left(0,1\right)$ , we have $fog\left(x\right)=0and,gof\left(x\right)=1.$

Hence, fog and gof do not coincide in (0, 1).

Therefore, option (B) is correct.

2, Therefore, option (B) is correct.

63. The given eqⁿ of the lines are.

3x + y - 2 = 0 _____ (1)

Px + 2y - 3 = 0 ______ (2)

2x - y - 3 = 0 _____ (3)

Point of intersection of (1) and (3) is given by,

(3x + y - 2) + (2x - y - 3) = 0

=> 5x - 5 = 0

=> x = $\frac{5}{5}$

=> x = 1

So, y = 2 - 3x = 2 -3 (1) = 2 - 3 = 1.

i e, (x, y) = (1, -1).

As the three lines interests at a single point, (1, -1) should line on line (2)

i e, P * 1 + 2 * (-1)- 3 = 0

P - 2 - 3 = 0

P = 5

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It is clear that 1 is reflexive and symmetric but not transitive.

Therefore, option (A) is correct.

It is given that A = {–1, 0, 1, 2}, B = {–4, –2, 0, 2}

Also, it is given that $f,g:A\to B$ are defined by $f\left(x\right)=x^2-x,x\in A$ and $g\left(x\right)=2x-\frac{1}{2}-1,x\in A$ .

It is observed that:

$\begin{array}{l}f\left(-1\right)=\left(1^2\right)-\left(-1\right)=1+1=2\\ g\left(-1\right)=2\left(-1\right)-\frac{1}{2}-1=2\left(\frac{3}{2}\right)-1=3-1=2\\ \Rightarrow f\left(-1\right)=g\left(-1\right)\\ f\left(0\right)={\left(0\right)}^{\wedge }2-0=0\\ g\left(0\right)=2\left(0\right)-\frac{1}{2}-1=2\left(\frac{1}{2}\right)-1=1-1=0\\ \Rightarrow f\left(0\right)=g\left(0\right)\\ f\left(1\right)={\left(1\right)}^{\wedge }2-1=1-1=0\\ g\left(1\right)=2a-\frac{1}{2}-1=2\left(\frac{1}{2}\right)-1=1-1=0\\ \Rightarrow f\left(1\right)=g\left(1\right)\\ f\left(2\right)={\left(2\right)}^{\wedge }2-2=4-2=2\\ g\left(2\right)=2\left(2\right)-\frac{1}{2}-1=2\left(\frac{3}{2}\right)-1=3-1=2\\ \Rightarrow f\left(2\right)=g\left(2\right)\\ \therefore f\left(a\right)=g\left(a\right)\forall a\in A\end{array}$

Hence, the functions f and g are equal.

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Let X={0, 1, 2, 3, 4, 5}.

The operation* on X is defined as:

$a*b=\{\begin{array}{cc}\hfill a+b\hfill & \hfill if,a+b<6\hfill \\ \hfill a+b-6\hfill & \hfill if,a+b\ge 6\hfill \end{array}$

An element $e\in X$ is the identity element for the operation*, if $a*e=a=e*a\forall a\in X$

For $a\in X$ we observed that

$\begin{array}{l}a*0=a+0=a\left[a\in X\Rightarrow a+0<6\right]\\ 0*a=0+a=a\left[a\in X\Rightarrow 0+a<6\right]\\ \therefore a*0=0*a\forall a\in X\end{array}$

Thus, 0 is the identity element for the given operation*.

An element $a\in X$ is invertible if there exists $b\in X$ such that $a*0=0*a.$

$ie\{\begin{array}{cc}\hfill a+b=0=b+a\hfill & \hfill if,a+b<6\hfill \\ \hfill a+6-6=0=b+a-6\hfill & \hfill if,a+b\ge 6\hfill \end{array}$

i.e.,

$a=-b,or,b=6-a$

But, X={0, 1, 2, 3, 4, 5} and $a,b\in X$ . Then, $a\ne -b$ .

$\therefore b=6-a$ is the inverse of $a\&mnForE;a\in X.$

Hence, the inverse of an element $a\in X,a\ne 0$ is 6-a i.e., $a^{-1}=6-a.$

29. Given, f(x)=|x – 1|.

The given function is defined for all real number x.

Hence, domain of f(x)=R.

As f(x)=|x – 1|, x $\in$ R is a non-negative no.

Range of f(x)=[0, ?), if positive real numbers.

It is given that*: $R\times R\to$ and $o:R\times R\to R$ is defined as

$a*b=\left|a-b\right|and,aob=a,\&mnForE;a,b\in R.$

For $a,b\in R$ , we have:

$\begin{array}{l}a*b=\left|a-b\right|\\ b*a=\left|b-a\right|=\left|-\left(a-b\right)\right|=\left|a-b\right|\\ \therefore a*b=b*a\end{array}$

$\therefore$ The operation* is commutative.

It can be observed that,

$\begin{array}{l}\left(1.2\right).3=\left(\left|1-2\right|\right).3=1.3=\left|1-3\right|=2\\ 1*\left(2*3\right)=1*\left(\left|2-3\right|\right)=1*1=\left|1-1\right|=0\\ \therefore \left(1*2\right)*3=1*\left(2*3\right)\left(where,1,2,3\in R\right)\end{array}$

$\therefore$ The operation* is not associative.

Now, consider the operation o:

It can be observed that $1o2=1,and,2o1=2.$

$\therefore 1o2\ne 2o1\left(where,1,2\in R\right)$

$\therefore$ The operation o is not commutative.

Let, $a,b,c\in R$ . Then we have:

$\begin{array}{l}\left(aob\right)oc=aoc=a\\ ao\left(boc\right)=aob=a\\ \Rightarrow \left(aob\right)oc=ao\left(boc\right)\end{array}$

$\therefore$ The operation o is associative.

Now, $a,b,c\in R$ . Then we have:

$\begin{array}{l}a*\left(boc\right)=a*b=\left|a-b\right|\\ \left(a*b\right)o\left(a*c\right)=\left(\left|a-b\right|\right)o\left(\left|a-c\right|\right)=\left|a-b\right|\\ Hence,a*\left(boc\right)=\left(a*b\right)o\left(a*c\right).\\ Now,\\ 1o\left(2o3\right)=1o\left(\left|2-3\right|\right)=1o1=1\\ \left(1o2\right)*\left(1o3\right)=1*1=\left|1-1\right|=0\\ \therefore 1o\left(2o3\right)\ne \left(1o2\right)*\left(1o3\right)\left(where,1,2,3\in R\right)\end{array}$

$\therefore$ The operation o does not distribute over*.