Maths

Kindly go through the solution

17. GivenR= { (x, x³) : x is a prime number less than 10}

R = { (x, x³) : x = 2,3,5,7}

= { (2,2³), (3,3³), (5,5³), (7,7³)}

= { (2,8), (3,27), (5,125), (7,343)}

40.

The given equation of the line is.

12 (x + 6) = 5 (y- 2)

⇒ 12x + 72 = 5y- 9

⇒ 12x- 5y + 72 + 9 = 0

⇒ 12x- 5y + 82 = 0

The perpendicular distance of point (-1, 1) from the line is given by

16. Given, R = { (x, x+5): x $\in$  {0,1,2,3,4,5}

= { (0,0+5), (1,1+5), (2,2+5), (3,3+5), (4,4+5), (5,5+5)}

= { (0,5), (1,6), (2,7), (3,8), (4,9), (5,10)}

So, domain of R= {0,1,2,3,4,5}

range of R= {5,6,7,8,9,10}

15. Given, A= {1,2,3,4,6}

R= { (a, b): a, b $\in$ A, b is exactly divisible by a}

(i) R= { (1,1), (1,2), (1,3), (1,4), (1,6), (2,2), (2,4), (2,6), (3,3), (3,6), (4,4), (6,6)}

(ii) Domain of R= {1,2,3,4,6}

(iii) Range of R= {1,2,3,4,6}

14. As R is a relation from set P to Q.

(i) R = { (x, y): x – 2 = y ; 5 ≤ x ≤ 7}

(ii) R = { (5,3), (6,4), (7,5)}

Domain of R= {5,6,7}

range of R= {3,4,5}

38. (i) Given, 3x + 2y 12 = 0.

3x + 2y = 12

Dividing both sides by 12 we get,

$\Rightarrow \frac{3x}{12}+\frac{2y}{12}=\frac{12}{12}$

$\Rightarrow \frac{x}{4}+\frac{y}{6}=1.$

Comparing the above equation with $\frac{x}{a}+\frac{y}{b}=1$ = we get, x-intercept, a = 4 and y-intercept b = 6.

(ii) Given, 4x - 3y = 6

Dividing the both sides by 6.

$\frac{4x}{6}-\frac{3y}{6}=\frac{6}{6}$

$\Rightarrow \frac{2x}{3}-\frac{y}{2}=1$

$\Rightarrow \frac{x}{3/2}+\frac{y}{(-2)}=1.$

Comparing above equation by $\frac{x}{a}+\frac{y}{b}=1$ we get, x-intercept a = $\frac{3}{2}$ and y-intercept, b = -2

(iii) Given, 3y + 2 = 0.

3y = -2

$y=-\frac{2}{3}$

As the equation of line is of form y = constant, it is parallel to x-axis and has no x-intercept.

y-intercept = -$\frac{2}{3}$

13. Given, A= {1,2,3,5}

B= {4,6,9}

R= { (x, y) : the difference of x & y is odd; x $\in$ A, y $\in$ B}.

= { (x, y):|x – y| is odd and x $\in$ A, y $\in$ B}

= { (1,4), (1,6), (2,9), (3,4), (3,6), (5,4), (5,6)}.

12. Given, R = { (x, y): y = x + 5, x is a natural number less than 4; x, y $\in$ N}

= { (x, y): y = x + 5; x, y $\in$ N and x < 4}.

= { (1,1+5), (2,2+5), (3,3+5)}

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

So, domain of R = {1,2,3}

range of R = {6,7,8}