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 \\...
Give an example of a function which is
Define each of the following:
(i) into function
Give an example of each type of functions.
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...
Define each of the following:
(i) bijective function
(ii) many – one function
Give an example of each type of functions.
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...
Define each of the following:
(i) injective function
(ii) surjective function
Give an example of each type of functions.
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...
Define a function. What do you mean by the domain and range of a function? Give examples.
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’...
Let A = (1, 2, 3, 4) and R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (1, 3), (3, 2)}. Show that R is reflexive and transitive but not symmetric.
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}...
Show that the relation R = {(a, b) : a > b} on N is transitive but neither reflexive nor symmetric.
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...
Let be the set of all real numbers. Show that the relation is symmetric but neither reflexive nor transitive.
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...
Show that the relation R defined on the set A = (1, 2, 3, 4, 5), given by
R = {(a, b) : |a – b| is even} is an equivalence relation.
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,...
Let A = {1, 2, 3, 4, 5, 6) and let R = {(a, b) : a, b ∈ A and b = a + 1}. Show that R is
(i) not transitive.
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)...
On the set S of all real numbers, define a relation R = {(a, b) : a ≤ b}. Show that R is
(i) not symmetric.
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...
Let A be the set of all points in a plane and let O be the origin. Show that the relation R = {(P, Q) : P, Q ∈ A and OP = OQ) is an equivalence relation.
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...
Let S be the set of all sets and let R = {(A, B) : A ⊂ B)}, i.e., A is a proper subset of B. Show that R is
(i)transitive
(ii) not reflexive
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 $\mathrm{C}$...
Let and . Find the domain and range of .
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) =...
Let R = {(a, b) : b = |a – 1|, a ∈ Z and la| < 3}. Find the domain and range of 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}
Let R = {(a, b) : a, b, ϵ N and a < b}. Show that R is a binary relation on N, which is neither reflexive nor symmetric. Show that R is transitive.
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...
Let A = {3, 4, 5, 6} and R = {(a, b) : a, b ϵ A and a
(i)Write R in roster form.
(ii)Find: dom (R) and range (R)
(iii) Write R–1 in roster form
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...
Let R = {(x, x2) : x is a prime number less than 10}.
(i) Write R in roster form.
(ii)Find dom (R) and range (R).
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,...
If A = {5} and B = {5, 6}, write down all possible subsets of A × B.
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)}
Prove that A × B = B × A ⇒ A = B.
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...
If A ⊆ B, prove that A × C = B × C
Answer : Given: A ⊆ B Then, A = B at some value Multiplying by C both sides, we get, A × C = B × C Hence, Proved.
If A = {3, 4}, B = {4, 5} and C = {5, 6}, find A × (B × C).
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)}...
If A = {1, 2}, find A × A × A.
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,...
Let A and B be two sets such that n(A) = 5, n(B) = 3 and n(A ∩ B) = 2.
(i)n(A ???? B)
(ii) n(A × B)
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
Let R = {(a, b) : a, b, ϵ N and a < b}. Show that R is a binary relation on N, which is neither reflexive nor symmetric. Show that R is transitive.
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...
Let A = {a, b}. List all relation on A and find their number.
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,...
Find R–1, when
(i) R = {(1, 2), (1, 3), (2, 3), (3, 2), (4, 5)}
(ii) R = {(x, y) : x, y ϵ N, x + 2y = 8}.
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...
Let R = (x, y) : x, y ϵ Z and x2 + y2 = 25}. Express R and R–1 as sets of ordered pairs. Show that R = R–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...
Let A be the set of first five natural numbers and let R be a relation on A, defined by (x, y) ϵ R ↔ x ≤ y. Express R and R–1 as sets of ordered pairs. Find: dom (R–1) and range (R).
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...
Let R be a relation on Z, defined by (x, y) ϵ R ↔ x2 + y2 = 9. Then, write R as a set of ordered pairs. What is its domain?
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...
Let R = {(a, b) : a, b ϵ Z and b = 2a – 4}. If (a, –2} ϵ R and (4, b2) ϵ R. Then, write the values of a and b.
Answer : b = 2a – 4 Put b = -2 , a = 1 Put a = 4 , b = 4 a = 1 , b = 4
Let R = {(a, b) : a, b ϵ Z and b = 2a – 4}. If (a, –2} ϵ R and (4, b2) ϵ R. Then, write the values of a and b.
Answer : b = 2a – 4 Put b = -2 , a = 1 Put a = 4 , b = 4 a = 1 , b = 4
Let A = {1, 2, 3} and R = {(a, b) : a, b ϵ A and |a2 – b2| ≤ 5. Write R as a set of ordered pairs. Mention whether R is
(i) reflexive
(ii) symmetric
(iii) transitive. Give reason in each case.
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....
Let R = {(a, b) : a, b ϵ Z and (a – b) is even}. Then, show that R is an equivalence relation on Z.
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...
What is an equivalence relation? Show that the relation of ‘similarity’ on the set S of all triangles in a plane is an equivalence relation.
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...
If R is a binary relation on a set A define R–1 on A. Let R = {(a, b) : a, b ϵ W and 3a + 2b = 15} and 3a + 2b = 15}, where W is the set of whole numbers. Express R and R–1 as sets of ordered pairs. Show that
(i) dom (R) = range (R–1)
(ii) range (R) = dom (R–1)
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...
Let A = {0, 1, 2, 3, 4, 5, 6, 7, 8} and let R = {(a, b) : a, b ϵ A and 2a + 3b = 12}. Express R as a set of ordered pairs. Show that R is a binary relation on A. Find its domain and 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...
Let A = {2, 3, 5} and R = {(2, 3), (2, 5), (3, 3), (3, 5)}. Show that R is a binary relation on A. Find its domain and range.
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...
What do you mean by a binary relation on a set A? Define the domain and range of relation on A.
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...
Let A = {3, 4} and B = {7, 9}. Let R = {(a, b): a ϵ A, b ϵ B and (a – b) is odd}. Show that R is an empty relation from A to B.
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...
Let A = {2, 3} and B= {3, 5}
(i)Find (A × B) and n(A × B).
(ii) How many relations can be defined from A to B?
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)....
Let R = {(x, y): x, y ϵ Z and x2 + y2 ≤ 4}.
(i)Write R in roster form.
(ii)Find dom (R) and range (R).
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)}...
Define a relation R from Z to Z, given by R = {(a, b): a, b ϵ Z and (a – b) is an integer. Find dom (R) and range (R).
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) = {-∞, ∞}...
Let A = {1, 2, 3, 4, 6} and R = {(a, b) : a, b ϵ A, and a divides b}.
(i)Write R in roster form.
(ii)Find dom (R) and range (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)}...
Let R = {(x, x + 5): x ϵ {9, 1, 2, 3, 4, 5}}.
(i)Write R in roster form.
(ii)Find dom (R) and range (R).
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,...
Let A = {1, 2, 3, 4, 5, 6}. Define a relation R from A to A by R = {(x, y): y = x + 1}.
(i)Write R in roster form.
(ii)Find dom (R) and range (R).
(iii) What is its co-domain?
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}...
Let A = {(x, y): x + 3y = 12, x ϵ N and y ϵ N}.
(i)Write R in roster form.
(ii)Find dom (R) and range (R).
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}
Let A = {1, 2, 3, 5} AND B = {4, 6, 9}. Let R = {(x, y): x ϵ A, y ϵ B and (x – y) is odd}. Write R in roster form.
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)}
Let A = {2, 3, 4, 5} and B = {3, 6, 7, 10}. Let R = {(x, y): x ϵ A, y ϵ B and x is relatively prime to y}.
(i)Write R in roster form.
(ii)Find dom (R) and range (R).
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,...
Let A = {2, 4, 5, 7} and b = {1, 2, 3, 4, 5, 6, 7, 8}. Let R = {(x, y) x ϵ A, y ϵ B and x divides y}.(i) Write R in roster form. (ii) Find dom (R) and range (R).
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,...
Let A = {1, 3, 5, 7} and B = {2, 4, 6, 8}. Let R = {(x, y), : x ϵ A, y ϵ B and x > y}.
(i) Write R in roster form.
(ii)Find dom (R) and range (R).
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}...
Find the domain and range of each of the relations given below: (i) R = {(–1, 1), (1, 1), (–2, 4), (2, 4), (2, 4), (3, 9)}
(ii)R ={(x, y) : x + 2y = 8 and x, y ϵ N}
(iii) R = {(x, y), : y = |x – 1|, x ϵ Z and |x| ≤ 3}
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 +...
Let A and B be two nonempty sets.
(i) What do you mean by a relation from A to B?
(ii) What do you mean by the domain and range of a relation?
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,...
Let A = {a, b, c, d}, B = {c, d, e} and C = {d, e, f, g}. Then verify each of the following identities: (i) A × (B ∩ C) = (A × B) ∩ (A × C) (ii) A × (B – C) = (A × B) – (A × C) (iii) (A × B) ∩ (B × A) = (A ∩ B) × (A ∩ B)
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),...
Let A = {1, 2} and B = {2, 3}. Then, write down all possible subsets of A × B.
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} =...
For any two sets A and B, show that A × B and B × A have an element in common if and only if A and B have an element in common.
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...
If A and B be two sets such that n(A) = 3, n(B) = 4 and n(A ∩ B) = 2 then find.
(i)n(A × B)
(ii)n(B × A)
(iii) n(A × B) ∩ (B × A)
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)) =...
If A × B ⊆ C × D and A × B ≠ ϕ, prove that A ⊆ C and B ⊆ D.
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...
(i) If A ⊆ B, prove that A × C ⊆ B × C for any set C.
(ii) If A ⊆ B and C ⊆ D then prove that A × C ⊆ B × D.
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...
If A and B are nonempty sets, prove that A × B = B × A ⇔ A = B
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, ⇒...
For any sets A and B, prove that (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A).
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...
C. For any sets A, B and C prove that: A × (B – C) = (A × B) – (A × C)
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)...
B. For any sets A, B and C prove that: A × (B ∩ C) = (A × B) ∩ (A × C)
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...
A. For any sets A, B and C prove that: A × (B ???? C) = (A × B) ???? (A × C)
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...
Let A = {–3, –1}, B = {1, 3) and C = {3, 5). Find:
(iii)B × C
(iv)A × (B × C)
(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...
Let A = {–3, –1}, B = {1, 3) and C = {3, 5). Find:
(i) A × B
(ii) (A × B) × C
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...
If A = {5, 7), find (i) A × A × A.
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,...
Let A = {–2, 2} and B = (0, 3, 5). Find:
(iii)A × A
(iv) B × B
(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...
Let A = {–2, 2} and B = (0, 3, 5). Find:
(i) A × B
(ii) B × A
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...
Let A and B be two sets such that n(A) = 3 and n(B) = 2. If a ≠ b ≠ c and (a, 0), (b, 1), (c, 0) is in A × B, find A and B.
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) =...
Let A × B = {(a, b): b = 3a – 2}. if (x, –5) and (2, y) belong to A × B, find the values of x and y.
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) –...
Let A = {2, 3} and B = {4, 5}. Find (A × B). How many subsets will (A × B) have?
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...
If A × B = {(–2, 3), (–2, 4), (0, 4), (3, 3), (3, 4), find A and B.
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...
Let A = {x ϵ W : x < 2}, B = {x ϵ N : 1 < x ≤ 4} and C = {3, 5}. Verify that: (i) A × (B ???? C) = (A × B) ???? (A × C) (ii) A × (B ∩ C) = (A × B) ∩ (A × C)
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...
If A = {1, 3, 5) B = {3, 4} and C = {2, 3}, verify that:
(i) A × (B ???? C) = (A × B) ???? (A × C)
(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
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...
If A = {x ϵ N : x ≤ 3} and {x ϵ W : x < 2}, find (A × B) and (B × A). Is (A × B) = (B × A)?
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...
If A = {2, 3, 5} and B = {5, 7}, find:
(iii)A × A
(iv)B × B
(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...
If A = {2, 3, 5} and B = {5, 7}, find:
(i)A × B
(ii)B × A
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...
If P = {a, b} and Q = {x, y, z}, show that P × Q ≠ Q × P.
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...
If A = {9, 1} and B = {1, 2, 3}, show that A × B ≠ B × A.
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...
Find the values of a and b, when:(a – 2, 2b + 1 = (b – 1, a + 2)
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...
Find the values of a and b, when:
(i) (a + 3, b –2) = (5, 1)
(ii) (a + b, 2b – 3) = (4, –5)
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...
Which of the following is a true statement? (a) is a linear polynomial. (b) is a binomial (c) is a monomial (d) is a monomial
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.
On dividing a polynomial by a non-zero polynomial be the quotient and be the remainder, then , where (a) always (b) always (c) either or (d)
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...
If be the zeroes of the polynomial such that , then ? (a) 3 (b) (c) -2 (d) 2
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...
If one of the zeroes of the cubic polynomial is , then the product of the other two zeroes is (a) (b) (c) (d)
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...
If one of the zeroes of the cubic polynomial is 0 , then the product of the other two zeroes is (a) (b) (c) 0 (d)
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...
If two of the zeroes of the cubic polynomial are 0 , then the third zero is (a) (b) (c) (d)
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...
If be the zeroes of the polynomial such that and , then (a) (b) (c) (d) none of these
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...
If be the zeroes of the polynomial , then ? (a) (b) 3 (c) (d)
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...
If be the zeroes of the polynomial , then (a) (b) 1 (c) (d) 30
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 {...
If are the zeroes of the polynomial , then ? (a) 3 (b) (c) 12 (d)
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 $\alpha \beta=2$...
If the sum of the zeroes of the quadratic polynomial is equal to the product of its zeroes, then ? (a) (b) (c) (d)
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...
If one zero of be the reciprocal of the other, then ? (a) 3 (b) (c) (d)
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...
If and 3 are the zeroes of the quadratic polynomial , then (a) (b) (c) (d)
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...
If one zero of the quadratic polynomial is , then the value of is (a) (b) (c) (d)
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...
If one zero of the quadratic polynomial is 2 , then the value of is (a) (b) (c) (d)
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...
If and are the zeroes of , then the value of is (a) (b) (c) (d)
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 zeroes of the quadratic polynomial are (a) both positive (b) both negative (c) one positive and one negative (d) both equal
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...
A quadratic polynomial whose zeroes are and , is (a) (b) (c) (d)
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,...
A quadratic polynomial whose zeroes are 5 and -3, is (a) (b) (c) (d) none of these
(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 sum and product of the zeroes of a quadratic polynomial are 3 and respectively. The quadratic polynomial is (a) (b) (c) (d)
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 zeros of the polynomial are (a) (a) (c) (d) none of these
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 zeros of the polynomial are (a) (b) (c) (d) none of these
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 zeroes of the polynomial are (a) (b) (c) (d) none of these
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 zeroes of the polynomial are (a) (b) (c) (d)
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 Zeroes of the polynomial are (a) (b) (c) (d) 3,1
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 $\mathrm{x}=-1$
Which of the following is not a polynomial? (a) (b) (c) (d)
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.
Which of the following is a polynomial? (a) (b) (c) (d) None of these
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 $a_{n} \neq 0$.
If the zeroes of the polynomial 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...
If 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...
If 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 {...
If 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 {...
If 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 {...
If the zeroes of the polynomial 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...
If 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...
If 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...
If 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 {...
If 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 {...
Find the zeroes of the quadratic polynomial .
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...
Find the zeroes of the quadratic polynomial .
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...
Find the sum of the zeros and the product of zeros of a quadratic polynomial, 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...
State Division Algorithm for Polynomials.
"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...
If 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 {...
If 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...
If and are zeros of the polynomial write the value of a.
using the relationship between the zeroes of the quadratic polynomial.$$ \begin{aligned} &\text { Sum of zeroes }=\frac{-\left(\text { coefficient of } x^{2}\right)}{\text { coefficient of } x^{3}}...
If is a factor of , then find the value of
$(x+a)$ is a factor of $2 x^{2}+2 a x+5 x+10$ $x+a=0$ $\Rightarrow \mathrm{X}=-\mathrm{a}$ Since, $(x+a)$ is a factor of $2 x^{2}+2 a x+5 x+10$ Hence, It will satisfy the above polynomial...
If the product of the zero of the polynomial is 3 . Find the value of .
using the relationship between the zeroes of he quadratic polynomial. Product of zeroes $=\frac{\text { constant term }}{\text { coefficient of } x^{2}}$ $\Rightarrow 3=\frac{k}{1}$ $\Rightarrow...
If the sum of the zeros of the quadratic polynomial is 1 write the value of .
using the relationship between the zeroes of the quadratic polynomial. Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ $\Rightarrow 1=\frac{-(-3)}{k}$...
Write the zeros of the polynomial .
$f(x)=x^{2}-x-6$ $=x^{2}-3 x+2 x-6$ $=x(x-3)+2(x-3)$ $=(x-3)(x+2)$ $f(x)=0 \Rightarrow(x-3)(x+2)=0$ $$ \begin{aligned} &\Rightarrow(x-3)=0 \text { or }(x+2)=0 \\ &\Rightarrow x=3 \text { or } x=-2...
If 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...
If 1 is a zero of the quadratic polynomial 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 $a(1)^{2}-(a-1) 1-1=0$ $\Rightarrow a-3 a+3-1=0$ $\Rightarrow-2 \mathrm{a}=-2$...
If 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...
If 3 is a zero of the polynomial , 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...
If one zero of the quadratic polynomial 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...
Find 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 $\mathrm{x}^{2}-(\alpha+\beta) \mathrm{x}+\alpha \beta$ $\ldots \ldots(1)$...
Find the zeroes of the polynomial
$f(x)=x^{2}-3 x-m(m+3)$ adding and subtracting $\mathrm{mx}$, $f(x)=x^{2}-m x-3 x+m x-m(m+3)$ $=x[x-(m+3)]+m[x-(m+3)]$ $=[x-(m+3)](x+m)$ $f(x)=0 \Rightarrow[x-(m+3)](x+m)=0$...
Find the zeroes of the polynomial
$f(x)=x^{2}+x-p(p+1)$ adding and subtracting $\mathrm{px}$, we get $f(x)=x^{2}+p x+x-p x-p(p+1)$ $=x^{2}+(p+1) x-p x-p(p+1)$ $=x[x+(p+1)]-p[x+(p+1)]$ $=[x+(p+1)](x-p)$ $f(x)=0$...
If one zero of the polynomial 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 }...
Find all the zeroes of polynomial , it being given that two of its zeroes are and .
$f(x)=2 x^{4}-11 x^{3}+7 x^{2}+13 x-7$. 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 $f(x)$...
Obtain all other zeroes of if two of its zeroes are and .
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...
Find all the zeroes of , it is being given that two of its zeroes are and .
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...
Find all the zeroes of , if it is given that two of its zeroes are and
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...
If 2 and are two zeroes of the polynomial , find all the zeroes of the given polynomial.
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,...
If 3 and are two zeroes of the polynomial , find all the zeroes of the given polynomial.
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...
It is given that is one of the zeroes of the polynomial . Find all the zeroes of the given polynomial.
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...
Verify division algorithm for the polynomial by
$-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...
On dividing is divided by a polynomial , the quotient and remainder are and respectively. Find
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...
By actual division, show that is a factor of .
Let $f(x)=2 x^{4}+3 x^{3}-2 x^{2}-9 x-12$ and $g(x)$ as $x^{2}-3$
If is divided by .
$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$
If f(x) =
is divided by g(x)=
Quotient $q(x)=x^{2}+x-3$ Remainder $r(x)=8$
If is divided by
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$
Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and the product of its zeroes as and respectively.
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...
Find a cubic polynomial whose zeroes are and .
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...
Find a cubic polynomial whose zeroes are and
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...
Verify that and are the zeroes of the cubic polynomial and verify the relation between its zeroes and coefficients.
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...
One zero of the polynomial is . Find the other zeros of the polynomial.
$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...
If is a factor of the polynomial , 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, it satisfies the above polynomial. => $2(-a)^{2}+2 a(-a)+5(-a)+10=0$ $\Rightarrow 2 a^{2}-2...
If and are the roots of the quadratic equation then find the value of a and .
$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$...
Find the quadratic polynomial, sum of whose zeroes is and their product is .
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...
Find the quadratic polynomial, sum of whose zeroes is and their product is 1 . Hence, find the zeroes of the polynomial.
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...
Find the quadratic polynomial, sum of whose zeroes is 8 and their product is Hence, find the zeroes of the polynomial.
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$...
Find the quadratic polynomial whose zeroes are and . Verify the relation between the coefficients and the zeroes of the polynomial.
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...
Find the quadratic polynomial whose zeroes are 2 and Verify the relation between the coefficients and the zeroes of the polynomial.
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)...
Find the zeroes of the quadratic polynomial and verify the relation between the zeroes and the coefficients.
$$ \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...
Find the zeroes of the quadratic polynomial (5y and verify the relation between the zeroes and the coefficients.
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...
Find the zeroes of the quadratic polynomial and verify the relation between the zeroes and the coefficients.
$$ \begin{aligned} &f(x)=x^{2}-5 \\ &\text { It can be written as } x^{2}+0 x-5 . \\ &=\left(x^{2}-(\sqrt{5})^{2}\right) \\ &=(x+\sqrt{5})(x-\sqrt{5}) \\ &\therefore f(x)=0...
Find the zeroes of the quadratic polynomial and verify the relation between the zeroes and the coefficients.
$$ \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...
Find the zeroes of the polynomial and verify the relation between its zeroes and coefficients.
$$ \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 {...
Find the zeroes of the quadratic polynomial and verify the relationship between the zeroes and coefficients of the given polynomial.
$f(x)=5 x^{2}-4-8 x$ $=5 x^{2}-8 x-4$ $=5 x^{2}-(10 x-2 x)-4$ $=5 x^{2}-10 x+2 x-4$ $=5 x(x-2)+2(x-2)$ $=(5 x+2)(x-2)$ $\therefore \mathrm{f}(\mathrm{x})=0 \Rightarrow(5...
Find the zeroes of the quadratic polynomial and verify the relation between its zeroes and coefficients.
$f(x)=4 x^{2}-4 x-3$ $=4 x^{2}-(6 x-2 x)-3$ $=4 x^{2}-6 x+2 x-3$ $=2 x(2 x-3)+1(2 x-3)$ $=(2 x+1)(2 x-3)$ $\therefore \mathrm{f}(\mathrm{x})=0 \Rightarrow(2 \mathrm{x}+1)(2 \mathrm{x}-3)=0$...
Find the zeroes of the quadratic polynomial and verify the relation between its zeroes and coefficients.
$f(x)=x^{2}+3 x-10$ $=x^{2}+5 x-2 x-10$ $=x(x+5)-2(x+5)$ $=(x-2)(x+5)$ $\therefore \mathrm{f}(\mathrm{x})=0 \Rightarrow(\mathrm{x}-2)(\mathrm{x}+5)=0$ $\Rightarrow x-2=0$ or $x+5=0$ $\Rightarrow...
Find the zeroes of the polynomial and verify the relation between its zeroes and coefficients.
$x^{2}-2 x-8=0$ $\Rightarrow \mathrm{x}^{2}-4 \mathrm{x}+2 \mathrm{x}-8=0$ $\Rightarrow x(x-4)+2(x-4)=0$ $\Rightarrow(x-4)(x+2)=0$ $\Rightarrow(x-4)=0$ or $(x+2)=0$ $\Rightarrow x=4$ or $x=-2$ Sum...
Find the zeros of the polynomial and verify the relation between its zeroes and coefficients.
$x^{2}+7 x+12=0$ $\Rightarrow x^{2}+4 x+3 x+12=0$ $\Rightarrow x(x+4)+3(x+4)=0$ $\Rightarrow(x+4)(x+3)=0$ $\Rightarrow(x+4)=0$ or $(x+3)=0$ $\Rightarrow \mathrm{x}=-4$ or $\mathrm{x}=-3$ Sum of...