Class 11th

7. Given,

A= {1, 2}, B = {1,2,3,4}, C= {5,6} and D= {5,6,7,8}

(i) L.H. S = A * (B∩ C) = {1,2} [ {1,2,3,4} ∩ {5,6}]

= {1,2}* 

=  .

R.H.S = (A* B)∩ (A *C)= [ {1,2}* {1,2,3,4}]∩ [ {1,2} {5,6}]

= [ { (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)]∩ [ {1,5), (1,6), (2,5), (2,6)}]

= .

Hence, L.H.S= R.H.S.

(ii) A* C = {1, 2}* {5,6}

= { (1,5), (1,6), (2,5), (2,6)}

B* D = {1,2,3,4} * {5,6,7,8}

= { (1,5), (1,6), (1,7), (1,8), (2,5), (2,6), (2,7), (2,8), (3,5), (3,6), (3,7), (3,8), (4,5), (4,6), (4,7), (4,8)}

As every element of A C is also an element of B* D.

A *C $\subset$ B *D

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)}

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

3. Given, G = {7, 8} and H = {5, 4, 2}

By the definition of the Cartesian product,

G *H = { (x, y): x∈G and y = ∈ H}

= { (7, 5), (7, 4), (7, 2), (8, 5), (8,4), (8,2)}

H* G = { (x, y): x∈ H and y ∈G}

= { (5, 7), (5, 8), (4,7), (4, 8), (2, 7), (2,8)}

2. Given, n (A) = 3

n (B) = 3 or B = {3,4,5}

So, number of elements in A* B = n (A* B) = n (A)* n (B) = 3 *3 = 9.

33. (i) False, as {2,3,4,5} ∩ {3,6} = {3} ≠ $?$ .Hence sets are not disjoint.

(ii) False as {a, e, i, o, u} ∩ {a, b, c, d} = {a} ≠ $?$ Hence sets are not disjoint.

(iii) True as {2,6,10,14} ∩ {3,7,11,15} = $?$ . Hence sets are disjoint.

(iv) True as {2,6,10} ∩ {3,7,11}= $?$ . Hence sets are disjoint.