Solution: (i) One-One but not Onto $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{N}$ be a mapping given by $\mathrm{f}(\mathrm{x})=\mathrm{x} 2$ For one-one $\begin{array}{l} f(x)=f(y) \\ x_{2}=y z \\...
Solution: (i) Into Function: It is is a function where there is atleast one element is Set B who is not the image of any element in set A. For Example: $f(x) = 2x - 1$ from the set of Integers to...
Solution: (i)Bijective function: It is, also known as one-one onto function and is a function where for every element of set A, there is exactly one image in set B, such that no element is set B is...
Solution: (i) Injective function: It is, also known as one-one function and is a type of function where every element in set A has an image in set B. Hence, f: A → B is one-one or injection function...
Solution: A function is stated as the relation between the two sets, where there is exactly one element in set B, for every element of set A. A function is represented as f: A → B, which means ‘f’...
Solution: $\begin{array}{l} \mathrm{A}=\{1,2,3,4\} \text { and } \mathrm{R}=\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,3) \\ (3,2)\} \text { (Given) } \end{array}$ $\mathrm{R}$ is reflexive if $\mathrm{a}...
Solution: $R=\{(a, b): a>b\}$ on $N$ (given) Non-Reflexivity: Assume $a$ be an arbitrary element of $\mathrm{N}$ a cannot be greater than $a$ $\Rightarrow(\mathrm{a}, \mathrm{a}) \notin...
Solution: $\mathrm{R}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}_{2}+\mathrm{b}_{2}=1\right\}$ and $\mathrm{a}, \mathrm{b} \in \mathrm{S}$ (As given) Non-Reflexivity: Assume $a$ be an arbitrary...
Solution: $\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathrm{A}$ and $|\mathrm{a}-\mathrm{b}|$ is even $\}$ where $\mathrm{A}=$ (As given) If $\mathrm{R}$ is Reflexive,...
Solution: (i) Non-transitive: If $p, q$ and $r \in A$ such that $(p, q) \in R$ and $(q, r) \in R \Rightarrow(p, r) \in R$, then $\mathrm{R}$ is transitive. Here, $(1,2) \in \mathrm{R}$ and $(2,3)...
Solution: (i) Non - Symmetric: Suppose $\mathrm{p}$ and $\mathrm{q} \in \mathrm{S}$, such that $(\mathrm{p}, \mathrm{q}) \in \mathrm{R}$ $\Rightarrow \mathrm{p} \leq \mathrm{q}$ $\Rightarrow...
Solution: Suppose $A$ be the set of all points in a plane and $O$ be the origin. (As given) Therefore, $\mathrm{R}=\{(\mathrm{P}, \mathrm{Q}): \mathrm{P}, \mathrm{Q} \in \mathrm{A}$ and...
Solution: (i) Suppose $A, B, C \in S$ such that $(A, B) \in R$ and $(B, C) \in R$ $\Rightarrow \mathrm{A}$ is a proper subset of $\mathrm{B}$ and $\mathrm{B}$ is a proper subset of $\mathrm{C}$...
Solution: Since 1<a< 5, $a = 2, 3, 4$ Therefore, R $= {\{(2,{\frac12}), (3,{\frac13}), (4,{\frac14})}\}$ As a result, Domain(R) = {2, 3, 4} and Range(R) =...
Solution: As $|a| < 3$, $a = −2$, $−1$, $0$, $1$, $2$ Therefore, R = {(−2, 3), (−1, 2), (0, 1), (1, 0), (2, 1)} As a result, Domain(R) = {-2, -1, 0, 1, 2} and Range(R) = {3, 2, 1, 0}
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 : (i) R = {(3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6)} (ii) The domain of R is the set of first co-ordinates of R Dom(R) = {3, 4, 5} The range of R is the set of second...
Answer : i) {(x, x2) : x is a prime number less than 10}. Roster form: R = {(1, 1), (2, 4), (3, 9), (5, 25), (7, 49)} (ii) The domain of R is the set of first co-ordinates of R Dom(R) = {1, 2, 3, 5,...
Answer : A = {5} B = {5, 6} A × B = {(5, 5), (5, 6)} All the possible subsets of A × B are, {(5, 5)} {(5, 6)} {(5, 6), (5, 6)}
Answer : Let A and B be any two sets such that A × B = {(a, b): a ϵ A, b ϵ B} Now, B × A = {(b, a): a ϵ A, b ϵ B} A × B = B × A (a, b) = (b, a) We can see that this is possible only when the ordered...
Answer : Given: A ⊆ B Then, A = B at some value Multiplying by C both sides, we get, A × C = B × C Hence, Proved.
Answer : A = {3, 4}, B = {4, 5} and C = {5, 6} B × C = {(4, 5), (4, 6), (5, 5), (5, 6)} A × (B × C) = {(3, 4, 5), (3, 4, 6), (3, 5, 5), (3, 5, 6), (4, 4, 5), (4, 4, 6), (4, 5, 5), (4, 5, 6)}...
Answer : A = {1, 2} A × A = {1, 2} × {1, 2} = {(1, 1), (1, 2), (2, 1), (2, 2)} A × A × A = {1, 2} × {(1, 1), (1, 2), (2, 1), (2, 2)} Therefore A × A × A = {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2,...
Answer : (i) n(A ???? B) = n(A) + n(B) - n(A ∩ B) = 5 + 3 – 2 =6 (ii) n(A × B) = n(A) × n(B) = 5 × 3 = 15
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...
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 +...
Answer : (i) If A and B are two nonempty sets, then any subset of the set (A × B) is said to a relation R from set A to set B. That means, if R be a relation from A to B then R ⊆ (A × B). Therefore,...
Answer : Given: A = {a, b, c, d,}, B = {c, d, e} and C = {d, e, f, g} Need to prove: A × (B ∩ C) = (A × B) ∩ (A × C) Left hand side, (B ∩ C) = {d, e} ⇒ A × (B ∩ C) = {(a, d), (a, e), (b, d), (b, e),...
Answer : Given: A = {1, 2} and B = {2, 3} Need to write: All possible subsets of A × B A = {1, 2} and B = {2, 3} So, all the possible subsets of A × B are: (A × B) = {(x, y): x A and y B} =...
Answer : We know, (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A) Here A and B have an element in common i.e., n(A ∩ B) = 1 = (B ∩ A) So, n((A × B) ∩ (B × A)) = n((A ∩ B) × (B ∩ A)) = n(A ∩ B) × n(B ∩ A) = 1...
Answer : Given: n(A) = 3, n(B) = 4 and n(A ∩ B) = 2 n(A × B) = n(A) × n(B) ⇒ n(A × B) = 3 × 4 ⇒ n(A × B) = 12 n(B × A) = n(B) × n(A) ⇒ n(B × A) = 4 × 3 ⇒ n(B × A) = 12 (iii) n((A × B) ∩ (B × A)) =...
Answer : Given: A × B ⊆ C × D and A × B ≠ ϕ Need to prove: A ⊆ C and B ⊆ D Let us consider, (x, y) (A × B)---- (1) ⇒ (x, y) (C × D) [as A × B ⊆ C × D]---- (2) From (1) we can say that, x A...
Answer : (i) Given: A ⊆ B Need to prove: A × C ⊆ B × C Let us consider, (x, y) (A × C) That means, x A and y C Here given, A ⊆ B That means, x will surely be in the set B as A is the subset of...
Answer : Given: A = B, where A and B are nonempty sets. Need to prove: A × B = B × A Let us consider, (x, y) (A × B) That means, x A and y B As given in the problem A = B, we can write, ⇒...
Answer : Given: A and B two sets are given. Need to prove: (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A) Let us consider, (x, y) (A × B) ∩ (B × A) ⇒ (x, y) (A × B) and (x, y) (B × A) ⇒ (x A...
Answer : Given: A, B and C three sets are given. Need to prove: A × (B – C) = (A × B) – (A × C) Let us consider, (x, y) A × (B – C) ⇒ x A and y (B – C ) ⇒ x A and (y B and y ∉ C) ⇒ (x A and y B)...
Answer : Given: A, B and C three sets are given. Need to prove: A × (B ∩ C) = (A × B) ∩ (A × C) Let us consider, (x, y) A × (B ∩ C) ⇒ x A and y (B ∩ C) ⇒ x ⇒ (x A and (y...
Answer : Given: A, B and C three sets are given. Need to prove: A × (B ???? C) = (A × B) ???? (A × C) Let us consider, (x, y) A × (B ???? C) ⇒ x A and y (B ???? C) ⇒ x A and (y B or...
(iii) Given: B = {1, 3} and C = {3, 5} To find: B × C By the definition of the Cartesian product, Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs of...
Answer : (i) Given: A = {-3, -1} and B = {1, 3} To find: A × B By the definition of the Cartesian product, Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered...
Answer : We have, A = {5, 7} So, By the definition of the Cartesian product, Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs of elements from P and Q,...
(iii) Given: A = {-2, 2} To find: A × A By the definition of the Cartesian product, Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs of elements from P...
Answer : (i) Given: A = {-2, 2} and B = {0, 3, 5} To find: A × B By the definition of the Cartesian product, Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered...
Answer : Since, (a, 0), (b, 1), (c, 0) are the elements of A × B. ∴ a, b, c Є A and 0, 1 Є B It is given that n(A) = 3 and n(B) = 2 ∴ a, b, c Є A and n(A) = 3 ⇒ A = {a, b, c} and 0, 1 Є B and n(B) =...
Answer : Given: A × B = {(a, b): b = 3a – 2} and {(x, -5), (2, y)} Є A × B For (x, -5) Є A × B b = 3a – 2 ⇒ -5 = 3(x) – 2 ⇒ -5 + 2 = 3x ⇒ -3 = 3x ⇒ x = -1 For (2, y) Є A × B b = 3a – 2 ⇒ y = 3(2) –...
Answer : Given: A = {2, 3} and B = {4, 5} To find: A × B By the definition of the Cartesian product, Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs...
Answer : Here, A × B = {(–2, 3), (–2, 4), (0, 4), (3, 3), (3, 4)} To find: A and B Clearly, A is the set of all first entries in ordered pairs in A × B ∴ A = {-2, 0, 3} and B is the set of all...
Answer : (i) Given: A = {x ϵ W : x < 2} Here, W denotes the set of whole numbers (non – negative integers). ∴ A = {0, 1} [∵ It is given that x < 2 and the whole numbers which are less than 2...
Answer : (i) Given: A = {1, 3, 5}, B = {3, 4} and C = {2, 3} H. S = A × (B ⋃ C) By the definition of the union of two sets, (B ⋃ C) = {2, 3, 4} = {1, 3, 5} × {2, 3, 4} Now, by the definition of the...
Answer : Given: A = {x ϵ N: x ≤ 3} Here, N denotes the set of natural numbers. ∴ A = {1, 2, 3} [∵ It is given that the value of x is less than 3 and natural numbers which are less than 3 are 1 and...
(iii) Given: A = {2, 3, 5} and B = {2, 3, 5} To find: A × A By the definition of the Cartesian product, Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered...
Answer : (i) Given: A = {2, 3, 5} and B = {5, 7} To find: A × B By the definition of the Cartesian product, Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered...
Answer : Given: P = {a, b} and Q = {x, y, z} To show: P × Q ≠ Q × P Now, firstly we find the P × Q and Q × P By the definition of the Cartesian product, Given two non – empty sets P and Q. The...
Answer : Given: A = {9, 1} and B = {1, 2, 3} To show: A × B ≠ B × A Now, firstly we find the A × B and B × A By the definition of the Cartesian product, Given two non – empty sets P and Q. The...
Since, the ordered pairs are equal, the corresponding elements are ∴, a – 2 = b – 1 …(i) & 2b + 1 = a + 2 …(ii) Solving eq. (i), we get a – 2 = b – 1 ⇒ a – b = -1 + 2 ⇒ a – b = 1 … (iii) Solving...
Answer : Since, the ordered pairs are equal, the corresponding elements are equal. ∴, a + 3 = 5 …(i) and b – 2 = 1 …(ii) Solving eq. (i), we get a + 3 = 5 ⇒ a = 5 – 3 ⇒ a = 2 Solving eq. (ii), we...
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 $a_{n} \neq 0$.
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...
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...
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 {...
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 {...
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 {...
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...
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...
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...
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 {...
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...
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...
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 {...
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...
using the relationship between the zeroes of the quadratic polynomial.$$ \begin{aligned} &\text { Sum of zeroes }=\frac{-\left(\text { coefficient of } x^{2}\right)}{\text { coefficient of } x^{3}}...
$(x+a)$ is a factor of $2 x^{2}+2 a x+5 x+10$ $x+a=0$ $\Rightarrow \mathrm{X}=-\mathrm{a}$ Since, $(x+a)$ is a factor of $2 x^{2}+2 a x+5 x+10$ Hence, It will satisfy the above polynomial...
using the relationship between the zeroes of he quadratic polynomial. Product of zeroes $=\frac{\text { constant term }}{\text { coefficient of } x^{2}}$ $\Rightarrow 3=\frac{k}{1}$ $\Rightarrow...
using the relationship between the zeroes of the quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ $\Rightarrow 1=\frac{-(-3)}{k}$...
$f(x)=x^{2}-x-6$ $=x^{2}-3 x+2 x-6$ $=x(x-3)+2(x-3)$ $=(x-3)(x+2)$ $f(x)=0 \Rightarrow(x-3)(x+2)=0$ $$ \begin{aligned} &\Rightarrow(x-3)=0 \text { or }(x+2)=0 \\ &\Rightarrow x=3 \text { or } x=-2...
$x=-2$ 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...
$x=1$ is one zero of the polynomial $a x^{2}-3(a-1) x-1$ Therefore, it will satisfy the above polynomial. Now, we have $a(1)^{2}-(a-1) 1-1=0$ $\Rightarrow a-3 a+3-1=0$ $\Rightarrow-2 \mathrm{a}=-2$...
$x=-4$ 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...
$x=3$ 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...
$x=2$ 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...
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 $f(x)=x^{2}+p x+x-p x-p(p+1)$ $=x^{2}+(p+1) x-p x-p(p+1)$ $=x[x+(p+1)]-p[x+(p+1)]$ $=[x+(p+1)](x-p)$ $f(x)=0$...
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 }...
$f(x)=2 x^{4}-11 x^{3}+7 x^{2}+13 x-7$. Since $(3+\sqrt{2})$ and $(3-\sqrt{2})$ are the zeroes of $f(x)$ it follows that each one of $(x+3+\sqrt{2})$ and $(x+3-\sqrt{2})$ is a factor of $f(x)$...
The given polynomial is $f(x)=x^{4}+4 x^{3}-2 x^{2}-20 x-15$. Since $(x-\sqrt{5})$ and $(x+\sqrt{5})$ are the zeroes of $f(x)$ it follows that each one of $(x-\sqrt{5})$ and $(x$ $+\sqrt{5})$ is a...
The given polynomial is $f(x)=2 x^{4}-3 x^{3}-5 x^{2}+9 x-3$ Since $\sqrt{3}$ and $-\sqrt{3}$ are the zeroes of $f(x)$, it follows that each one of $(x-\sqrt{3})$ and $(x+\sqrt{3})$ is a factor of...
Let f(x)=x4+x3-23x2-3x+60 \text { Let } f(x)=x^{4}+x^{3}-23 x^{2}-3 x+60 Since $\sqrt{3}$ and $-\sqrt{3}$ are the zeroes of $f(x)$, it follows that each one of $(x-\sqrt{3})$ and...
Let $f(x)=x^{4}+x^{3}-34 x^{2}-4 x+120$ Since 2 and $-2$ are the zeroes of $f(x)$, it follows that each one of $(x-2)$ and $(x+2)$ is a factor of $f(x)$ Consequently,...
Since 3 and $-3$ are the zeroes of $f(x)$, it follows that each one of $(x+3)$ and $(x-3)$ is a factor of $f(x)$ Consequently, $(x-3)(x+3)=\left(x^{2}-9\right)$ is a factor of $f(x)$. On dividing...
Let $f(x)=x^{3}+2 x^{2}-11 x-12$ Since $-1$ is a zero of $f(x),(x+1)$ is a factor of $f(x)$. On dividing $\mathrm{f}(\mathrm{x})$ by $(\mathrm{x}+1)$, we get $$ \begin{aligned} &f(x)=x^{3}+2...
$-6 x^{3}+x^{2}+20 x+8$ and $g(x)$ as $-3 x^{2}+5 x+2$ Quotient $=2 \mathrm{x}+3$ Remainder $=x+2$ By using division rule, we have Dividend $=$ Quotient $\times$ Divisor $+$ Remainder $\therefore-6...
using division rule, Dividend $=$ Quotient $\times$ Divisor $+$ Remainder $\therefore 3 x^{3}+x^{2}+2 x+5=(3 x-5) g(x)+9 x+10$ $\Rightarrow 3 x^{3}+x^{2}+2 x+5-9 x-10=(3 x-5) g(x)$ $\Rightarrow 3...
Let $f(x)=2 x^{4}+3 x^{3}-2 x^{2}-9 x-12$ and $g(x)$ as $x^{2}-3$
$f(x)$ as $x^{4}+0 x^{3}+0 x^{2}-5 x+6$ and $g(x) a s-x^{2}+2$ Quotient $q(x)=-x^{2}-2$ Remainder $\mathrm{r}(\mathrm{x})=-5 \mathrm{x}+10$
Quotient $q(x)=x^{2}+x-3$ Remainder $r(x)=8$
Quotient $q(x)=x-\overline{3}$ Remainder $r(x)=7 x-9$ 7. If $f(x)=x^{4}-3 x^{2}+4 x+5$ is divided by $g(x)=x^{2}-x+1$
sum of the product of the zeroes taken two at a time and the product of the zeroes of a cubic polynomial then the cubic polynomial can be found as $x^{3}-($ sum of the zeroes $) x^{2}+($ sum of the...
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as $x^{3}-(a+b+c) x^{2}+(a b+b c+c a) x-a b c$ Let $a=\frac{1}{2}, b=1$ and $c=-3$ Substituting the values...
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as $x^{3}-(a+b+c) x^{2}+(a b+b c+c a) x-a b c$ Let $a=2, b=-3$ and $c=4$ Substituting the values in 1 , we...
p(x)=3x3-10x2-27x+10 p(x)=\left(3 x^{3}-10 x^{2}-27 x+10\right) p(5)=3×53-10×52-27×5+10=(375-250-135+10)=0 p(5)=\left(3 \times 5^{3}-10 \times 5^{2}-27 \times...
$x=\frac{2}{3}$ is one of the zero of $3 x^{3}+16 x^{2}+15 x-18$ Now, we have $\mathrm{x}=\frac{2}{3}$ $\Rightarrow \mathrm{x}-\frac{2}{3}=0$ Now, we divide $3 x^{3}+16 x^{2}+15 x-18$ by...
$(x+a)$ is a factor of $2 x^{2}+2 a x+5 x+10$ $x+a=0$ $\Rightarrow \mathrm{x}=-\mathrm{a}$ Since, it satisfies the above polynomial. => $2(-a)^{2}+2 a(-a)+5(-a)+10=0$ $\Rightarrow 2 a^{2}-2...
$a x^{2}+7 x+b=0$ Since, $x=\frac{2}{3}$ is the root of the above quadratic equation Hence, it will satisfy the above equation. => $a\left(\frac{2}{3}\right)^{2}+7\left(\frac{2}{3}\right)+b=0$...
Quadratic equation can be found if we know the sum of the roots and product of the roots by using the formula: $\mathrm{x}^{2}-($ Sum of the roots) $\mathrm{x}+$ Product of roots $=0$ $\Rightarrow...
Let $\alpha$ and $\beta$ be the zeroes of the required polynomial $\mathrm{f}(\mathrm{x})$. =>$(\alpha+\beta)=\frac{5}{2}$ and $\alpha \beta=1$ $\therefore...
Let $\alpha$ and $\beta$ be the zeroes of the required polynomial $\mathrm{f}(\mathrm{x})$. Then $(\alpha+\beta)=8$ and $\alpha \beta=12$ $\therefore f(x)=x^{2}-(\alpha+\beta) x+\alpha \beta$...
Let $\alpha=\frac{2}{3}$ and $\beta=\frac{-1}{4}$. Sum of the zeroes $=(\alpha+\beta)=\frac{2}{3}+\left(\frac{-1}{4}\right)=\frac{8-3}{12}=\frac{5}{12}$ Product of the zeroes, $\alpha...
Let $\alpha=2$ and $\beta=-6$ Sum of the zeroes, $(\alpha+\beta)=2+(-6)=-4$ Product of the zeroes, $\alpha \beta=2 \times(-6)=-12$ $\therefore$ Required polynomial $=\mathrm{x}^{2}-(\alpha+\beta)...
$$ \begin{aligned} &3 x^{2}-x-4=0 \\ &\Rightarrow 3 x^{2}-4 x+3 x-4=0 \\ &\Rightarrow x(3 x-4)+1(3 x-4)=0 \\ &\Rightarrow(3 x-4)(x+1)=0 \\ &\Rightarrow(3 x-4) \text { or }(x+1)=0 \\ &\Rightarrow...
f(u)=5u2+10u \mathrm{f}(\mathrm{u})=5 \mathrm{u}^{2}+10 \mathrm{u} It can be written as $5 \mathrm{u}(\mathrm{u}+2)$ ∴f(u)=0⇒5u=0 or u+2=0 \therefore \mathrm{f}(\mathrm{u})=0...
$$ \begin{aligned} &f(x)=x^{2}-5 \\ &\text { It can be written as } x^{2}+0 x-5 . \\ &=\left(x^{2}-(\sqrt{5})^{2}\right) \\ &=(x+\sqrt{5})(x-\sqrt{5}) \\ &\therefore f(x)=0...
$$ \begin{aligned} &4 x^{2}-4 x+1=0 \\ &\Rightarrow(2 x)^{2}-2(2 x)(1)+(1)^{2}=0 \end{aligned} $$ $$ \begin{aligned} &\Rightarrow(2 \mathrm{x}-1)^{2}=0 \quad\left[\because \mathrm{a}^{2}-2...
$$ \begin{aligned} &2 \sqrt{3} x^{2}-5 x+\sqrt{3} \\ &\Rightarrow 2 \sqrt{3} x^{2}-2 x-3 x+\sqrt{3} \\ &\Rightarrow 2 x(\sqrt{3} x-1)-\sqrt{3}(\sqrt{3} x-1)=0 \\ &\Rightarrow(\sqrt{3} x-1) \text {...
$f(x)=5 x^{2}-4-8 x$ $=5 x^{2}-8 x-4$ $=5 x^{2}-(10 x-2 x)-4$ $=5 x^{2}-10 x+2 x-4$ $=5 x(x-2)+2(x-2)$ $=(5 x+2)(x-2)$ $\therefore \mathrm{f}(\mathrm{x})=0 \Rightarrow(5...
$f(x)=4 x^{2}-4 x-3$ $=4 x^{2}-(6 x-2 x)-3$ $=4 x^{2}-6 x+2 x-3$ $=2 x(2 x-3)+1(2 x-3)$ $=(2 x+1)(2 x-3)$ $\therefore \mathrm{f}(\mathrm{x})=0 \Rightarrow(2 \mathrm{x}+1)(2 \mathrm{x}-3)=0$...
$f(x)=x^{2}+3 x-10$ $=x^{2}+5 x-2 x-10$ $=x(x+5)-2(x+5)$ $=(x-2)(x+5)$ $\therefore \mathrm{f}(\mathrm{x})=0 \Rightarrow(\mathrm{x}-2)(\mathrm{x}+5)=0$ $\Rightarrow x-2=0$ or $x+5=0$ $\Rightarrow...
$x^{2}-2 x-8=0$ $\Rightarrow \mathrm{x}^{2}-4 \mathrm{x}+2 \mathrm{x}-8=0$ $\Rightarrow x(x-4)+2(x-4)=0$ $\Rightarrow(x-4)(x+2)=0$ $\Rightarrow(x-4)=0$ or $(x+2)=0$ $\Rightarrow x=4$ or $x=-2$ Sum...
$x^{2}+7 x+12=0$ $\Rightarrow x^{2}+4 x+3 x+12=0$ $\Rightarrow x(x+4)+3(x+4)=0$ $\Rightarrow(x+4)(x+3)=0$ $\Rightarrow(x+4)=0$ or $(x+3)=0$ $\Rightarrow \mathrm{x}=-4$ or $\mathrm{x}=-3$ Sum of...