Answer : N is the set of all the natural numbers. N = {1, 2, 3, 4, 5, 6, 7…..} R = {(a, b) : a, b, ϵ N and a < b} R = {(1, 2), (1, 3), (1, 4) …. (2, 3), (2, 4), (2, 5) ……} For reflexivity, A...
Answer : Any relation on A is a subset of A×A. A×A = {(a, a), (a, b), (b, a), (b, b)} The subsets are. {} empty set {(a, a)} {(a, b)} {(a, a), (a, b)} {(b, a)} {(b, b)} {(b, a), (b, b)} {(a, a), (b,...
Answer : (i) R = {(1, 2), (1, 3), (2, 3), (3, 2), (4, 5)} R–1 = {(2, 1), (3, 1), (3, 2), (2, 3), (5, 4)} (ii) R = {(x, y) : x, y ϵ N, x + 2y = 8}. Put x = 2, y = 3 Put x = 4, y = 2 Put x = 6, y = 1...
Answer : x2 + y2 = 25 Put x = 0, y = 5, 02 + 52 = 25 Put x = 3, y = 4, 32 + 42 = 25 R = {(0, 5), (0, -5), (5, 0), (-5, 0), (3, 4), (-3, 4), (-3, -4), (3, -4)} Since, x and y get interchanged in the...
Answer : A = {1, 2, 3, 4, 5} Since, x ≤ y R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3) ,(2, 4), (2, 5), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5), (5, 5) } The domain of R is the set of...
Answer : x2 + y2 = 9 We can have only integral values of x and y. Put x = 0 , y = 3 , 02 + 32 = 9 Put x = 3 , y = 0 , 32 + 02 = 9 R = {(0, 3) , (3, 0) , (0 , -3) , (-3 , 0)} The domain of R is the...
Answer : b = 2a – 4 Put b = -2 , a = 1 Put a = 4 , b = 4 a = 1 , b = 4
Answer : b = 2a – 4 Put b = -2 , a = 1 Put a = 4 , b = 4 a = 1 , b = 4
Answer : Put a = 1 , b = 1 |12 – 12| ≤ 5, (1, 1) is an ordered pair. Put a = 1 , b = 2 |12 – 22| ≤ 5, (1, 2) is an ordered pair. Put a = 1 , b = 3 |12 – 32| > 5, (1, 3) is not an ordered pair....
Answer : (i) Reflexivity: Let a є Z, a - a = 0 є Z which is also even. Thus, (a, a) є R for all a є Z. Hence, it is reflexive Symmetry: Let (a, b) є R (a, b) є R è a - b is even -(b - a) is even (b...
Answer : An equivalence relation is one which possesses the properties of reflexivity, symmetry and transitivity. Reflexivity: A relation R on A is said to be reflexive if (a, a) є R for all a є...
Answer : 3a + 2b = 15 a=1 è b=6 a=3 è b=3 a=5 è b=0 R = {(1, 6), (3, 3), (5, 0)} ????−1 = {(6, 1), (3, 3), (0, 5)} The domain of R is the set of first co-ordinates of R Dom(R) = {1, 3, 5} The range...
Answer : A = {0, 1, 2, 3, 4, 5, 6, 7, 8} 2a + 3b = 12 a=0 è b=4 a=3 è b=2 a=6 è b=0 R = {(0, 4), (3, 2), (6, 0)} Since, R is a subset of A × A, it a relation to A. The domain of R is the set...
Answer : First, calculate A×A. A×A = {(2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5)} Since, R is a subset of A × A, it’s a binary relation on A. The domain of R is the set...
Answer : Any subset of (A × A) is called a binary relation to A. Here, (A × A) is the cartesian product of A with A. Let A = {4, 5, 6) and R = {(4, 5), (6, 4), (5, 6)} Here, R is a binary relation...
The correct option is option (d) $5 \mathrm{x}^{2}$ is a monomial. $5 \mathrm{x}^{2}$ consists of one term only. So, it is a monomial.
The correct option is (c) either $\mathrm{r}(\mathrm{x})=0$ or $\operatorname{deg} \mathrm{r}(\mathrm{x})<\operatorname{deg} \mathrm{g}(\mathrm{x})$ By division algorithm on polynomials, either...
The correct option is option (d) 2 $\alpha$ and $\beta$ are the zeroes of $2 \mathrm{x}^{2}+5 \mathrm{x}+\mathrm{k}$, we have: $\alpha+\beta=\frac{-5}{2}$ and $\alpha \beta=\frac{k}{2}$ it is given...
The correct option is option (c) $1-\mathrm{a}+\mathrm{b}$ $-1$ is a zero of $x^{3}+a x^{2}+b x+c$, we have $(-1)^{3}+a \times(-1)^{2}+b \times(-1)+c=0$ $\Rightarrow a-b+c+1=0$ $\Rightarrow...
The correct option is option (b) $\frac{c}{a}$ $\alpha, \beta$ and 0 be the zeroes of $a x^{3}+b x^{2}+c x+d$. Then, sum of the products of zeroes taking two at a time is given by $(\alpha...
The correct option is option (a) $\frac{-b}{a}$ $\alpha, 0$ and 0 be the zeroes of $a x^{3}+b x^{2}+c x+d=0$ Then the sum of zeroes $=\frac{-b}{a}$ $\Rightarrow \alpha+0+0=\frac{-b}{a}$ $\Rightarrow...
The correct option is option (c) $x^{3}-3 x^{2}-10 x+24$ $\alpha, \beta$ and $\gamma$ are the zeroes of polynomial $p(x)$. $(\alpha+\beta+\gamma)=3,(\alpha \beta+\beta \gamma+\gamma \alpha)=-10$ and...
The correct option is option (a) $-3$ $\alpha, \beta$ and $\gamma$ are the zeroes of $2 \mathrm{x}^{3}+\mathrm{x}^{2}-13 \mathrm{x}+6$, we have: $\alpha \beta \gamma=\frac{-(\text { constant term...
The correct option is option (a) $-1$ It is given that $\alpha, \beta$ and $\gamma$ are the zeroes of $x^{3}-6 x^{2}-x+30$ $\therefore(\alpha \beta+\beta \gamma+\gamma \alpha)=\frac{\text {...
The correct option is option (b) $-3$ $\alpha$ and $\beta$ be the zeroes of $x^{2}+6 x+2$, we have: $\alpha+\beta=-6$ and $\alpha \beta=2$...
The correct option is option (d) $\frac{-2}{3}$ $\alpha$ and $\beta$ be the zeroes of $\mathrm{kx}^{2}+2 \mathrm{x}+3 \mathrm{k}$. Then $\alpha+\beta=\frac{-2}{k}$ and $\alpha \beta=3$ $\Rightarrow...
The correct option is option (a) $\mathrm{k}=3$ $\alpha$ and $\frac{1}{\alpha}$ be the zeroes of $3 x^{2}-8 x+k$. Then the product of zeroes $=\frac{k}{3}$ $\Rightarrow \alpha \times...
The correct option is option (c) $a=-2, b=-6$ $-2$ and 3 are the zeroes of $x^{2}+(a+1) x+b$. $(-2)^{2}+(a+1) \times(-2)+b=0 \Rightarrow 4-2 a-2+b=0$ $\Rightarrow b-2 a=-2$ ….(1) $3^{2}+(a+1) \times...
The correct option is option (b) $\frac{5}{4}$ Since $-4$ is a zero of $(k-1) x^{2}+k x+1$ $(k-1) \times(-4)^{2}+k \times(-4)+1=0$ $\Rightarrow 16 \mathrm{k}-16-4 \mathrm{k}+1=0$ $\Rightarrow 12...
The correct option is option (d) $\frac{-6}{5}$ Since 2 is a zero of $k x^{2}+3 x+k$, we have: $\mathrm{k} \times(2)^{2}+3(2)+\mathrm{k}=0$ $\Rightarrow 4 \mathrm{k}+\mathrm{k}+6=0$ $\Rightarrow 5...
The correct option is option (c) $\frac{-9}{2}$ $\alpha$ and $\beta$ be the zeroes of $2 \mathrm{x}^{2}+5 \mathrm{x}-9$. If $\alpha+\beta$ are the zeroes, then $\mathrm{x}^{2}-(\alpha+\beta)...
The correct option is option (b) both negative $\alpha$ and $\beta$ be the zeroes of $\mathrm{x}^{2}+88 \mathrm{x}+125$ Then $\alpha+\beta=-88$ and $\alpha \times \beta=125$ This can only happen...
The correct option is option (d) $x^{2}-\frac{1}{10} x-\frac{3}{10}$ Since, the zeroes are $\frac{3}{5}$ and $\frac{-1}{2}$ $\alpha=\frac{3}{5}$ and $\beta=\frac{-1}{2}$ sum of the zeroes,...
(c) $x^{2}-2 x-15$ Since, the zeroes are 5 and $-3$. $\alpha=5$ and $\beta=-3$ Therefore, sum of the zeroes, $\alpha+\beta=5+(-3)=2$ product of the zeroes, $\alpha \beta=5 \times(-3)=-15$ The...
The correct option is option (c) $x^{2}-3 x-10$ Sum of zeroes, $\alpha+\beta=3$ Also, product of zeroes, $\alpha \beta=-10$ $\therefore$ Required polynomial $=\mathrm{x}^{2}-(\alpha+\beta)+\alpha...
The correct option is option (a) $\frac{2}{3}, \frac{-1}{7}$ $f(x)=7 x^{2}-\frac{11}{3} x-\frac{2}{3}=0$ $\Rightarrow 21 \mathrm{x}^{2}-11 \mathrm{x}-2=0$ $\Rightarrow 21 \mathrm{x}^{2}-14...
The correct option is option (b) $\frac{-3}{2}, \frac{4}{3}$ $f(x)=x^{2}+\frac{1}{6} x-2=0$ $\Rightarrow 6 \mathrm{x}^{2}+\mathrm{x}-12=0$ $\Rightarrow 6 x^{2}+9 x-8 x-12=0$ $\Rightarrow 3 x(2...
The correct option is option (c) $\frac{3}{\sqrt{2}}, \frac{\sqrt{2}}{4}$ $f(x)=4 x^{2}+5 \sqrt{2} x-3=0$ $\Rightarrow 4 x^{2}+6 \sqrt{2} x-\sqrt{2} x-3=0$ $\Rightarrow 2 \sqrt{2}...
The correct option is option (b) $3 \sqrt{2},-2 \sqrt{2}$ $f(x)=x^{2}-\sqrt{2} x-12=0$ $\Rightarrow x^{2}-3 \sqrt{2} x+2 \sqrt{2} x-12=0$ $\Rightarrow x(x-3 \sqrt{2})+2 \sqrt{2}(x-3 \sqrt{2})=0$...
The correct option is option (c) $3,-1$ Let $f(x)=x^{2}-2 x-3=0$ $=x^{2}-3 x+x-3=0$ $=x(x-3)+1(x-3)=0$ $=(x-3)(x+1)=0$ $\Rightarrow \mathrm{x}=3$ or $\mathrm{x}=-1$
The correct option is option (d) $x+\frac{3}{x}$ is not a polynomial. It is because in the second term, the degree of $x$ is $-1$ and an expression with a negative degree is not a polynomial.
The correct option is option (d) none of these A polynomial in $x$ of degree $n$ is an expression of the form $p(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots \ldots+$ $a_{n} x^{n}$, where $a_{n} \neq 0$.