Answer : Given: A = {3, 4} and B = {7, 9} R = {(a, b): a ϵ A, b ϵ B and (a – b) is odd} So, R = {(4, 7), (4, 9)} An empty relation means there is no elements in the relation set. Here we get two...
Answer : Given: A = {2, 3} and B= {3, 5} (i) (A × B) = {(2, 3), (2, 5), (3, 3), (3, 5)} Therefore, n(A × B) = 4 (ii) No. of relation from A to B is a subset of Cartesian product of (A × B)....
Answer : Given: R = {(x, y): x, y ϵ Z and x2 + y2 ≤ 4} (i) R is Foster Form is, R = {(-2, 0), (-1, -1), (-1, 0), (-1, 1), (0, -2), (0, -1), (0, 0), (0, 1), (0, 2), (1, -1), (1, 0), (1, 1), (2, 0)}...
Answer : Given: R = {(a, b): a, b ϵ Z and (a – b) is an integer The condition satisfies for all the values of a and b to be any integer. So, R = {(a, b): for all a, b ϵ (-∞, ∞)} Dom(R) = {-∞, ∞}...
Answer : Given: A = {1, 2, 3, 4, 6} (i) R = {(a, b) : a, b ϵ A, and a divides b} R is Foster Form is, R = {(1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}...
Answer : Given: R = {(x, x + 5): x ϵ {9, 1, 2, 3, 4, 5}} (i) R is Foster Form is, R = {(9, 14), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)} (ii) Dom(R) = {1, 2, 3, 4, 5, 9} Range(R) = {6, 7, 8, 9, 10,...
Answer : Given: A = {1, 2, 3, 4, 5, 6} (i) R = {(x, y): y = x + 1} So, R is Roster Form is, R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)} (ii) Dom(R) = {1, 2, 3, 4, 5} Range(R) = {2, 3, 4, 5, 6}...
Answer : Given: A = {(x, y): x + 3y = 12, x ϵ N and y ϵ N} (i) So, R in Roster Form is, R = {(3, 3), (6, 2), (9, 1)} (ii) Dom(R) = {3, 6, 9} Range(R) = {1, 2, 3}
Answer : Given: A = {1, 2, 3, 5} AND B = {4, 6, 9} R = {(x, y): x ϵ A, y ϵ B and (x – y) is odd} Therefore, R in Roster Form is, R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}
Answer : Given: A = {2, 3, 4, 5} and B = {3, 6, 7, 10} (i) R = {(x, y), : x ϵ A, y ϵ B and x is relatively prime to y} So, R in Roster Form, R = {(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5,...
Answer : Given: A = {2, 4, 5, 7} and b = {1, 2, 3, 4, 5, 6, 7, 8} (i) R = {(x, y) x ϵ A, y ϵ B and x divides y} So, R in Roster Form, R = {(2, 2), (2, 4), (2, 6), (2, 8), (4, 4), (4, 8), (5, 5), (7,...
Answer : Given: A = {1, 3, 5, 7} and B = {2, 4, 6, 8} (i) R = {(x, y), : x ϵ A, y ϵ B and x > y} So, R in Roster Form, R = {(3, 2), (5, 2), (5, ), (7, 2), (7, 4), (7, 6)} (ii) Dom(R) = {3, 5, 7}...
Answer : (i) Given: R = {(–1, 1), (1, 1), (–2, 4), (2, 4), (2, 4), (3, 9)} Dom(R) = {x: (x, y) R} = {-2, -1, 1, 2, 3} Range(R) = {y: (x, y) R} = {1, 4, 9} (ii) Given: R = {(x, y): x +...
are 
, a and 
, find the values of a and b.using the relationship between the zeroes of he quadratic polynomial. Sum of zeroes $=\frac{-\left(\text { coef ficient of } x^{2}\right)}{\text { coefficient of } x^{3}}$ $\therefore...
are the zeroes of the polynomial 
, then 
.using the relationship between the zeroes of the quadratic polynomial. We have Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes...
are the zeroes of the polynomial 
, then 
using the relationship between the zeroes of he quadratic polynomial. Sum of zeroes $=\frac{-\text { (coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text {...
and 
are the zeros of the polynomial 
find the value of 
using the relationship between the zeroes of the quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text {...
are the zeroes of the polynomial 
such that 
, find the value of 
using the relationship between the zeroes of the quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text {...
are , a and , find the values of a and b.using the relationship between the zeroes of he quadratic polynomial. Sum of zeroes $=\frac{-\left(\text { coef ficient of } x^{2}\right)}{\text { coefficient of } x^{3}}$ $\therefore...
are the zeroes of the polynomial , then .using the relationship between the zeroes of the quadratic polynomial. We have Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes...
are the zeroes of the polynomial , then using the relationship between the zeroes of he quadratic polynomial. => Sum of zeroes $=\frac{-\text { (coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text...
and are the zeros of the polynomial find the value of using the relationship between the zeroes of the quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text {...
are the zeroes of the polynomial such that , find the value of using the relationship between the zeroes of the quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text {...
.For finding the zeroes of the quadratic polynomial we will equate $f(x)$ to 0 Hence, the zeroes of the quadratic polynomial $f(x)=4 \sqrt{3} x^{2}+5 x-2 \sqrt{3}$ are $-\frac{2}{\sqrt{3}}$ or...
.To find the zeroes of the quadratic polynomial we will equate $\mathrm{f}(\mathrm{x})$ to 0 Hence, the zeroes of the quadratic polynomial $f(x)=6 x^{2}-3$ are...
and 
respectively. Write the polynomial.We can find the quadratic polynomial if we know the sum of the roots and product of the roots by using the formula $\mathrm{x}^{2}-($ sum of the zeroes $) \mathrm{x}+$ product of zeroes $\Rightarrow...
"If $\mathrm{f}(\mathrm{x})$ and $\mathrm{g}(\mathrm{x})$ are two polynomials such that degree of $\mathrm{f}(\mathrm{x})$ is greater than degree of $\mathrm{g}(\mathrm{x})$ where...
and be the zeroes of the polynomial 
write the value of 
.using the relationship between the zeroes of he quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text {...
is divisible by 
, write the value of a and 
.Equating $\mathrm{x}^{2}-\mathrm{x}$ to 0 to find the zeroes, we will get x(x-1)=0 x(x-1)=0 ⇒x=0 or x-1=0 \Rightarrow \mathrm{x}=0 \text { or } \mathrm{x}-1=0...
and are zeros of the polynomial 
write the value of a.using the relationship between the zeroes of the quadratic polynomial.$$ \begin{aligned} &\text { Sum of zeroes }=\frac{-\left(\text { coefficient of } x^{2}\right)}{\text { coefficient of } x^{3}}...
is a factor of 
, then find the value of 
$(x+a)$ is a factor of $2 x^{2}+2 a x+5 x+10$ $x+a=0$ $\Rightarrow \mathrm{X}=-\mathrm{a}$ Since, $(x+a)$ is a factor of $2 x^{2}+2 a x+5 x+10$ Hence, It will satisfy the above polynomial...
is 3 . Find the value of 
.using the relationship between the zeroes of he quadratic polynomial. Product of zeroes $=\frac{\text { constant term }}{\text { coefficient of } x^{2}}$ $\Rightarrow 3=\frac{k}{1}$ $\Rightarrow...
is 1 write the value of 
.using the relationship between the zeroes of the quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ $\Rightarrow 1=\frac{-(-3)}{k}$...
.$f(x)=x^{2}-x-6$ $=x^{2}-3 x+2 x-6$ $=x(x-3)+2(x-3)$ $=(x-3)(x+2)$ $f(x)=0 \Rightarrow(x-3)(x+2)=0$ $$ \begin{aligned} &\Rightarrow(x-3)=0 \text { or }(x+2)=0 \\ &\Rightarrow x=3 \text { or } x=-2...
is a zero of the polynomial 
then find the value of .$x=-2$ is one zero of the polynomial $3 x^{2}+4 x+2 k$ Therefore, it will satisfy the above polynomial. Now, we have $3(-2)^{2}+4(-2) 1+2 k=0$ $\Rightarrow 12-8+2 k=0$ $\Rightarrow...
is 1 , then find the value of a.$x=1$ is one zero of the polynomial $a x^{2}-3(a-1) x-1$ Therefore, it will satisfy the above polynomial. Now, we have $a(1)^{2}-(a-1) 1-1=0$ $\Rightarrow a-3 a+3-1=0$ $\Rightarrow-2 \mathrm{a}=-2$...
is a zero of the polynomial 
is , then find the value of .$x=-4$ is one zero of the polynomial $x^{2}-x-(2 k+2)$ Therefore, it will satisfy the above polynomial. Now, we have $(-4)^{2}-(-4)-(2 k+2)=0$ $\Rightarrow 16+4-2 \mathrm{k}-2=0$ $\Rightarrow 2...
, find the value of .$x=3$ is one zero of the polynomial $2 x^{2}+x+k$ Therefore, it will satisfy the above polynomial. Now, we have $2(3)^{2}+3+k=0$ $\Rightarrow 21+\mathrm{k}=0$ $\Rightarrow...
is 2 , then find the value of .$x=2$ is one zero of the quadratic polynomial $k x^{2}+3 x+k$ Therefore, it will satisfy the above polynomial. $k(2)^{2}+3(2)+k=0$ $\Rightarrow 4 \mathrm{k}+6+\mathrm{k}=0$ $\Rightarrow 5...
are the zeros of polynomial 
and 
then write the polynomial.If the zeroes of the quadratic polynomial are $\alpha$ and $\beta$ then the quadratic polynomial can be found as $\mathrm{x}^{2}-(\alpha+\beta) \mathrm{x}+\alpha \beta$ $\ldots \ldots(1)$...
$f(x)=x^{2}-3 x-m(m+3)$ adding and subtracting $\mathrm{mx}$, $f(x)=x^{2}-m x-3 x+m x-m(m+3)$ $=x[x-(m+3)]+m[x-(m+3)]$ $=[x-(m+3)](x+m)$ $f(x)=0 \Rightarrow[x-(m+3)](x+m)=0$...
$f(x)=x^{2}+x-p(p+1)$ adding and subtracting $\mathrm{px}$, we get $f(x)=x^{2}+p x+x-p x-p(p+1)$ $=x^{2}+(p+1) x-p x-p(p+1)$ $=x[x+(p+1)]-p[x+(p+1)]$ $=[x+(p+1)](x-p)$ $f(x)=0$...
Is 
, write the other zero.Let the other zeroes of $x^{2}-4 x+1$ be a (using the relationship between the zeroes of the quadratic polynomial) sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coef ficient of }...