Maths

35. Equation of line with intercept form is

$\frac{x}{a}+\frac{y}{b}=1\left(1\right)$

As R (h, x) divides line segment joining point A (a, 0) and B (0, b) in the ratio 1 : 2 we can write,

$\left(h,k\right)=\left(\frac{1*0+2*a}{1+2},\frac{1*b+2*0}{1+2}\right)$

So,  $h=\frac{0+2a}{3}k=\frac{b+0}{3}$

$\Rightarrow \mathrm{3h}=2ab=3k$

$\Rightarrow a=\frac{3h}{2}b=3k$

Hence, putting value of a and b in equation (1) we get,

$\frac{x}{3h/2}+\frac{y}{\mathrm{3k}}=1$

$\Rightarrow \frac{x}{\mathrm{3h}}+\frac{y}{\mathrm{3k}}=1$

6. Given,

A *B = { (a, x), (a, y), (b, x), (b, y)}

We know that,

A *B = { (p, q); p ∈ A and q ∈ B}

So, A = {a, b} and B = {x, y}.

34.

Since P (a, b) is the mid-point of the line segment say AB with points A (0, y) and B (x, 0) we can write,

$\left(a,b\right)=\left(\frac{x+0}{2},\frac{y+0}{2}\right)$

$\Rightarrow a=\frac{x}{2}\text{?}b=\frac{y}{2}$

$\Rightarrow x=2a\text{?}y=2b$

So, the equation of line with x and y intercept 2a and 2b using intercept form is

$\frac{x}{\mathrm{2a}}+\frac{y}{\mathrm{2b}}=1$

$\Rightarrow \frac{x}{a}+\frac{y}{b}=2$

Hence, proved

5. Given, A = {1,1}

So, A* A = { (1,1), (1,1), (1,1), (1,1)}

A *A *A = { (1,1), (1,1), (1,1), (1,1)} * {1,1}

= { (1,1.1), (1, 1), (1, ), (1,1,1), (1,1,1), (1,1), (1,1), (1,1,1)}

The $fx$ n is $f(x)=\frac{1}{x}$ , which is a $f:R_{*}$ $\to$ $R_{*}$ and $R_{*}$ is set of all non-zero real numbers

For, $x_1,x_2\in R_{*},f(x_1)=f(x_2)$

$\Rightarrow \frac{1}{x_1}=\frac{1}{x_2}$

$\Rightarrow x_1=x_2$ So, $f$ is one-one

For, $y\in R_{*},$ $x=\frac{1}{f(x)}=\frac{1}{y}$ such that

So, $f(x)=y$

So, every element in the co-domain has a pre-image in $f$

So, $f$ is onto

If $f:N\to R_{*}$ such that $f(x)=\frac{1}{x}$

For, $x_1,x_2\in N,$ $f(x_1)=f(x_2)$

$\Rightarrow \frac{1}{x_1}=\frac{1}{x_2}$

$\Rightarrow x_1=x_2$ So, $f$ is one-one

For, $y\in R_{*}$ and $f(x)=y$ we have $x=\frac{1}{y}\notin N$

Eg., $3\in R_{*}$ so $x=\frac{1}{3}\notin N$

So, $f$ is not onto

4. (i) False. Here P = {m, n}, n (p)=2

Q = {n, m}, n (Q)=2

n (P* Q) = n (P)* n (Q) = 2* 2 = 4.

So, P *Q = { (m, n), (m, m), (n, n), (n, m)}

(ii) True.

(iii) True. { A * (B ∩ ?) = A* ? . {∴ B ∩ ? = ?  }

= n (A) *0 {? is empty set}

= ? 

33. Assuming the price per litre say P in x-axis and the corresponding demand say D in y-axis, we have two point (14, 980) and (16, 1220) in xy plane. Then the points (P, D) will satisfy the equation.

$\Rightarrow D-980=\frac{1220-980}{16-14}(P-14)$

$\Rightarrow D-980=\frac{240}{2}(P-14)$

$\Rightarrow D-120(P-14)+980$

$\Rightarrow D-12\mathrm{0P}-1680+980$

$\Rightarrow D=12\mathrm{0P}-700$

Which is the required relation

Where P = 17, we have

D = 120 * 17 – 700

D = 1340

Hence, the owner can sell 1340 litres of milk weekly at? 17/litre

The given relation in set N defined by

$R=\left\{\left(a,b\right):a=b-2,b>6\right\}$

For (2,4),        4>6 is not true

For (3,8),     8>6  but  3= 8-2 ⇒3=6 is not true

For (6,8),      8>6 and 6= 8-2 ⇒6=6 is true

And for (8,7), 7>6 but 8= 7-2 ⇒8=5 is not true

Hence, option (C) is correct

The set in $A=\left\{1,2,3,4\right\}$

The relation in this set $A$ is given by

$R=\left\{\left(1,2\right),\left(2,2\right),\left(1,1\right),\left(4,4\right),\left(1,3\right),\left(3,3\right),\left(3,2\right)\right\}$

$R$ is reflexive as $\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right)\in R$

As, $\left(1,2\right)\in R$ but $\left(2,1\right)\notin R$

$R$ is not symmetric

For $\left(1,2\right)\in R$ and $\left(2,2\right)\in R;\left(1,2\right)\in R$

And for $\left(1,3\right)\in R$ and $\left(3,2\right)\in R;\left(1,3\right)\in R$

∴ $R$ is transitive

Hence, option (B) is correct

The given relation in the set $L=$ all lines in $XY-$ plane is defined as

$R=\{\left(L_1,L_2\right):L_1$ is parallel to $L_2\}$

Let $L_1\in A$ then as $L_1$ is parallel to $L_1$ ,

$\left(L_1,L_1\right)\in R$

So, $R$ is reflexive

Let $L_1,L_2\in A$ and $\left(L_1,L_1\right)\in R$

Then, $L_1$ is parallel to $L_2$

$L_2$ is parallel to $L_1$

So, $\left(L_2,L_1\right)\in R$

i.e., $R$ is symmetric

Let $L_1,L_2,L_3\in A$ and $\left(L_1,L_2\right)$ and $\left(L_2,L_3\right)\in R$

Then, $L_1?L_2$ and $L_2?L_3$

So, $L_1?L_3$

i.e., $(L_1,L_2)\in R$

So, $R$ is transitive

Hence, $R$ is an equivalence relation

The set of lines related to $y=2x+4$ is given by the equation $y=2x+C$ where $C$ is some constant.