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...
, 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 $f(x)$...
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...
, 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...
, 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...
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,...
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...
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...
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...
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...
is a factor of 
.Let $f(x)=2 x^{4}+3 x^{3}-2 x^{2}-9 x-12$ and $g(x)$ as $x^{2}-3$
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 $\mathrm{r}(\mathrm{x})=-5 \mathrm{x}+10$
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Quotient $q(x)=x^{2}+x-3$ Remainder $r(x)=8$
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 $g(x)=x^{2}-x+1$
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...
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...
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...
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...