Ncert Solutions Maths class 11th

10. Given, n (A * A)=9

n (A) *n (A) = 9.

n (A)² = 3².

n (A) = 3 .

And (–1,0), (0,1) $\in$ A * A i.e., A * A = { (x, y), x $\in$ A, y $\in$ B}

? A= {–1,0,1}

And A * A= {–1,0,1} * {–1,0,1}

= { (–1, –1), ( –1,0), ( –1,1), (0, –1), (0,0), (0,1), (1, –1), (1,0), (1,1)}

9. Given, n (A)=3

n (B)= 2

So, n (Ax B)=n (A).n (B)=3x 2=6

as (x, 1), (y, 2), (z, 1) ∈Ax B= { (x, y), x∈Aand y∈B}.

A= {x, y, z} and B= {1,2}

As n (A) = 3as n (B) = 2

36. 

Let the given points be A (3, 0), B (–2, –2) and C (8, 2). Then by two point form we can write equation of line passing point A (3, 0) and B (–2, –2) as

$y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$

$\Rightarrow y-0=\frac{-2-0}{-2-3}\left(x-3\right)$

$\Rightarrow y=\frac{-2}{-5}\left(x-3\right)$

$\Rightarrow 5y=2\left(x-3\right)$

$\Rightarrow 5y=2x-6$

$\Rightarrow 2x-5y-6=0\left(1\right)$

If the three points A, B and C are co-linear, C will also lieonm the line formed by AB or satisfies equation (1).

Hence, putting x = 8 and y = 2 we have

L.H.S. = 2 * 8 – 5 * 2 – 6

= 16 – 10 – 6

= 0 = R.H.S.

The given three points are collinear.

8. A Given, A= {1,2}

B= {3,4}

So, A* B= { (1,3), (1,4), (2,3), (2, 4)}

i.e., n (A *B)=4

A *B will have subset =2⁴=16.They are,

Φ, { (1,3)}, { (1,4)}, { (2, 3)}, { (2,4)}, { (1,3), (1,4)}, { (1,3), (2,3)},

{ (1,3), (2, 4)}, { (1,4), (2, 3)}, { (1,4), (2, 4)}, { (2,3), (2, 4)},

{ (1,3), (1,4), (2, 3)}, { (1,3), (1,4), (2,4)}, { (1,3), (2,3), (2, 4)}, { (1,4), (2,3), (2,4)},

and { (1,3), (1,4), (2,3), (2,4)}

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}

= ?