The total number of reflexive relations on a finite set having n elements is

Question: The total number of reflexive relations on a finite set havingnelements is __________. Solution: The total number of reflexive relations on a finite set havingnelements is _________. Consider a setAwithnelements SayA={1, 2, .......n1,n} out ofn2elementsnelements are compulsory for relation to be reflexive. i.e (1, 1) (2, 2) (3, 3) .... (n,n) and for remainingn2nelements, we have choice of filling i.e either they are present or absent. Hence, Total number of reflexive relation are $2^{n...

The total number of reflexive relations on a finite set having n elements is

Question: The total number of reflexive relations on a finite set havingnelements is __________. Solution: The total number of reflexive relations on a finite set havingnelements is _________. Consider a setAwithnelements SayA={1, 2, .......n1,n} out ofn2elementsnelements are compulsory for relation to be reflexive. i.e (1, 1) (2, 2) (3, 3) .... (n,n) and for remainingn2nelements, we have choice of filling i.e either they are present or absent....

Show that

Question: $\frac{x}{\left(x^{2}+1\right)(x-1)}$ Solution: Let $\frac{x}{\left(x^{2}+1\right)(x-1)}=\frac{A x+B}{\left(x^{2}+1\right)}+\frac{C}{(x-1)}$ $x=(A x+B)(x-1)+C\left(x^{2}+1\right)$ $x=A x^{2}-A x+B x-B+C x^{2}+C$ Equating the coefficients ofx2,x, and constant term, we obtain $A+C=0$ $-A+B=1$ $-B+C=0$ On solving these equations, we obtain $A=-\frac{1}{2}, B=\frac{1}{2}$, and $C=\frac{1}{2}$ From equation (1), we obtain $\therefore \frac{x}{\left(x^{2}+1\right)(x-1)}=\frac{\left(-\frac{1}...

In figure 9.36, AC ⊥ CE and ∠A:∠B: ∠C = 3: 2: 1, find

Question: In figure 9.36, AC CE and A:B: C = 3: 2: 1, find Solution: A: B: C = 3: 2: 1 Let the angles be 3x, 2x and x ⇒ 3x + 2x + x = 180 [Angle sum property] ⇒ 6x = 180 ⇒ x = 30 = ACB ECD = 180 ACB 90 [Linear pair] = 180 30 90 = 60 ECD = 60...

If (1, 3), (2, 5) and (3, 3) are three elements of

Question: If (1, 3), (2, 5) and (3, 3) are three elements ofABandn(AB) = 6, then the remaining three elements of are __________. Solution: Given (1, 3), (2, 5) and (3, 3) ABandn(AB) = 6 According to given collection, A= {1, 2, 3} (since from (x,y)xAandyB) andB= {3, 5} i.e AB ={(1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5)} Remaining elements ofABare (1, 5) (2, 3)and(3, 5) i.e Remaining elements ofABare (1, 5) (2, 3) and (3, 5)...

ABC is a triangle. The bisector of the exterior angle at B and the bisector of ∠C intersect each other at D.

Question: ABC is a triangle. The bisector of the exterior angle at B and the bisector of C intersect each other at D. Prove that D = 1/2A. Solution: Let ABE = 2x and ACB = 2y ABC = 180 2x [Linear pair] A = 180 ABC ACB [Angle sum property] = 180 180 + 2x + 2y = 2(x y) ..... (i) Now, D = 180 DBC DCB ⇒ D = 180 (x + 180 2x) y ⇒ D = 180 x 180 + 2x y = (x y) $=\frac{1}{2} \angle \mathrm{A} \ldots .$ from(i) Hence, $\angle \mathrm{D}=\frac{1}{2} \angle \mathrm{A}$....

In figure 9.35, AB divides ∠DAC in the ratio 1: 3 and AB = DB. Determine the value of x.

Question: In figure 9.35, AB divides DAC in the ratio 1: 3 and AB = DB. Determine the value of x. Solution: Let BAD = Z, BAC = 3Z ⇒ BDA = BAD = Z (∵ AB = DB) Now BAD + BAC + 108 = 180 [Linear pair] ⇒ Z + 3Z + 108 =180 ⇒ 4Z = 72 ⇒ Z = 18 Now, In ΔADC ADC + ACD = 108 [Exterior angle property] ⇒ x + 18 = 180 ⇒ x = 90...

If R and S are two equivalence relations on a set A,

Question: IfRandSare two equivalence relations on a setA, thenRSis __________. Solution: IfRandSare two equivalence relations on a Let A, SincexA $(x, x) \in R \quad$ and $\quad(x, x) \in S(\because R$ and $S$ are reflexive $)$ $\Rightarrow(x, x) \in R \cap S \quad \forall x \in A$ i.e $R \cap S$ is Reflexive $\rightarrow$ Let $(x, y) \in R \cap S$ $\Rightarrow(x, y) \in R$ and $(x, y) \in S$ $\Rightarrow(y, x) \in R$ and $(y, x) \in S(\because R$ and $S$ are symmetric $)$ $\Rightarrow(y, x) \in...

Solve each of the following systems of equations by the method of cross-multiplication :

Question: Solve each of the following systems of equations by the method of cross-multiplication : $x\left(a-b+\frac{a b}{a-b}\right)=y\left(a+b-\frac{a b}{a+b}\right)$ $x+y=2 a^{2}$ Solution: GIVEN: $x\left((a-b)+\frac{a b}{a-b}\right)=y\left((a+b)-\frac{a b}{a+b}\right)$ $x+y=2 a^{2}$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation $x\left((a-b)+\frac{a b}{a-b}\right)-y\left((a+b)-\frac{a b}{a+b}\right)=0$ $...

If R and S are two equivalence relations on a set A,

Question: IfRandSare two equivalence relations on a setA, thenRSis __________. Solution: IfRandSare two equivalence relations on a Let A, SincexA $(x, x) \in R \quad$ and $\quad(x, x) \in S(\because R$ and $S$ are reflexive $)$ $\Rightarrow(x, x) \in R \cap S \quad \forall x \in A$ i.e $R \cap S$ is Reflexive $\rightarrow$ Let $(x, y) \in R \cap S$ $\Rightarrow(x, y) \in R$ and $(x, y) \in S$ $\Rightarrow(y, x) \in R$ and $(y, x) \in S(\because R$ and $S$ are symmetric $)$ $\Rightarrow(y, x) \in...

Show that

Question: $\frac{1-x^{2}}{x(1-2 x)}$ Solution: It can be seen that the given integrand is not a proper fraction. Therefore, on dividing $\left(1-x^{2}\right)$ by $x(1-2 x)$, we obtain $\frac{1-x^{2}}{x(1-2 x)}=\frac{1}{2}+\frac{1}{2}\left(\frac{2-x}{x(1-2 x)}\right)$ Let $\frac{2-x}{x(1-2 x)}=\frac{A}{x}+\frac{B}{(1-2 x)}$ $\Rightarrow(2-x)=A(1-2 x)+B x$ ...(1) Substituting $x=0$ and $\frac{1}{2}$ in equation $(1)$, we obtain $A=2$ and $B=3$ $\therefore \frac{2-x}{x(1-2 x)}=\frac{2}{x}+\frac{3}{...

Compute the value of x in each of the following figures:

Question: Compute the value of x in each of the following figures: Solution: (i) BAC = 180 120 = 60 [Linear pair] ACB = 180 112 = 68 [Linear pair] x = 180 BAC ACB = 180 60 68 = 52 [Sum of all angles of a triangle] (ii) ABC = 180 120 = 60 [Linear pair] ACB = 180 110 = 70 [Linear pair] eBAC = x = 180 ABC ACB = 180 60 70 = 50 [Sum of all angles of a triangle] (iii) BAE = EDC = 52 [Alternate angles] DEC = x = 180 40 EDC = 180 40 52 = 180 92 = 88 [Sum of all angles of a triangle] (iv) CD is produced ...

A relation R on a set A is a symmetric relation if

Question: A relationRon a setAis a symmetric relation if __________. Solution: A relationRon a setAis symmetric relation i.e if $(x, y) \in R$ for $x, y \in A$ $\Leftrightarrow(y, x) \in R$ i.e if $(x, y) \in R$ $\Leftrightarrow(x, y) \in R^{-1}$ $\therefore R=R^{-1}$ Hence relationRon a setAis symmetric If $R=R^{-1}$...

Show that

Question: $\frac{2 x}{x^{2}+3 x+2}$ Solution: Let $\frac{2 x}{x^{2}+3 x+2}=\frac{A}{(x+1)}+\frac{B}{(x+2)}$ $2 x=A(x+2)+B(x+1)$ ...(1) Substitutingx= 1 and 2 in equation (1), we obtain $A=-2$ and $B=4$ $\therefore \frac{2 x}{(x+1)(x+2)}=\frac{-2}{(x+1)}+\frac{4}{(x+2)}$ $\Rightarrow \int \frac{2 x}{(x+1)(x+2)} d x=\int\left\{\frac{4}{(x+2)}-\frac{2}{(x+1)}\right\} d x$ $=4 \log |x+2|-2 \log |x+1|+\mathrm{C}$...

If A and B are two sets such that n (A) = 5 and n (B) =7,

Question: IfAandBare two sets such thatn(A) = 5 andn(B) =7, then the the total number of relations onABis ........ . Solution: n(A) = 5 n(B) = 7 Then total number of relation onABis 2n(A).n(B) $=2^{5 \times 7}$ i.e total number of relation on $A \times B=2^{35}$...

Question: $\frac{2 x}{x^{2}+3 x+2}$ Solution: Let $\frac{2 x}{x^{2}+3 x+2}=\frac{A}{(x+1)}+\frac{B}{(x+2)}$ $2 x=A(x+2)+B(x+1)$ ...(1) Substitutingx= 1 and 2 in equation (1), we obtain $A=-2$ and $B=4$ $\therefore \frac{2 x}{(x+1)(x+2)}=\frac{-2}{(x+1)}+\frac{4}{(x+2)}$ $\Rightarrow \int \frac{2 x}{(x+1)(x+2)} d x=\int\left\{\frac{4}{(x+2)}-\frac{2}{(x+1)}\right\} d x$ $=4 \log |x+2|-2 \log |x+1|+\mathrm{C}$...

In figure 9.30, the sides BC, CA and AB of a triangle ABC have been produced to D,

Question: In figure 9.30, the sides BC, CA and AB of a triangle ABC have been produced to D, E and F respectively. If ACD = 105 and EAF = 45, find all the angles of the triangle ABC. Solution: BAC = EAF = 45 [Vertically opposite angles] ABC = 105 45 = 60 [Exterior angle property] ACD = 180 105 = 75 [Linear pair]...

The smallest reflexive relation on a set A is the

Question: The smallest reflexive relation on a setAis the ___________ . Solution: Let us defineA= {1, 2, 3..........n} Then smaller possible reflexive relation on setAis{(1,1), (2, 2),....... (n 1,n 1), (n,n)}...

In a triangle ABC, the internal bisectors of ∠B and ∠C meet at P and the external bisectors of ∠B and ∠C meet at Q.

Question: In a triangle ABC, the internal bisectors of B and C meet at P and the external bisectors of B and C meet at Q. Prove that BPC + BQC = 180. Solution: Let ABD = 2x and ACE = 2y ABC = 180 2x [Linera pair] ACB = 180 2y [Linera pair] A + ABC + ACB = 180 [Sum of all angles of a triangle] ⇒ A + 180 2x + 180 2y = 180 ⇒ A + 2x + 2y = 180 $\Rightarrow \mathrm{x}+\mathrm{y}=90^{\circ}+\frac{1}{2} \angle \mathrm{A}$ Now in ΔBQC x + y + BQC = 180 [Sum of all angles of a triangle] $\Rightarrow 90^{...

Let n(A) = m and n(B) = n.

Question: Letn(A) =mandn(B) =n. Then the total number of non-empty relations that can be defined from AtoBis ___________ . Solution: The total number of non-empty relations that can be defined fromAtoBis 2mn1....

Show that

Question: $\frac{x}{(x-1)(x-2)(x-3)}$ Solution: Let $\frac{x}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}$ $x=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)$ ...(1) Substituting $x=1,2$, and 3 respectively in equation (1), we obtain $A=\frac{1}{2}, B=-2$, and $C=\frac{3}{2}$ $\therefore \frac{x}{(x-1)(x-2)(x-3)}=\frac{1}{2(x-1)}-\frac{2}{(x-2)}+\frac{3}{2(x-3)}$ $\Rightarrow \int \frac{x}{(x-1)(x-2)(x-3)} d x=\int\left\{\frac{1}{2(x-1)}-\frac{2}{(x-2)}+\frac{3}{2(x-3)}\right\} d x$ $=\...

Question: Letn(A) =mandn(B) =n. Then the total number of non-empty relations that can be defined from AtoBis (a) mn (b)nm 1 (c)mn 1 (d) 2mn 1 Solution: Letn(A) =m n(B) =n sincen(AB) =mn whereABdefinesAcartesianB. $\therefore$ Total number of relation from $A$ to $B$ $=$ number of subsets of $A \times B$ $=2^{m n}$ i.e, Total number of non-empty relations is $2^{m n}-1$ Hence, the correct answer is option D....

Solve each of the following systems of equations by the method of cross-multiplication :

Question: Solve each of the following systems of equations by the method of cross-multiplication : $(a+2 b) x+(2 a-b) y=2$ $(a-2 b) x+(2 a+b) y=3$ Solution: GIVEN: $(a+2 b) x+(2 a-b) y=2$ $(a-2 b) x+(2 a+b) y=3$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation $(a+2 b) x+(2 a-b) y-2=0$ $(a-2 b) x+(2 a+b) y-3=0$ By cross multiplication method we get $\frac{x}{((2 a-b) \times-3)-((2 a+b) \times(-2))}=\frac{-y}{(-...

Question: Letn(A) =mandn(B) =n. Then the total number of non-empty relations that can be defined from AtoBis (a) mn (b)nm 1 (c)mn 1 (d) 2mn 1 Solution: Letn(A) =m n(B) =n sincen(AB) =mn whereABdefinesAcartesianB. $\therefore$ Total number of relation from $A$ to $B$ $=$ number of subsets of $A \times B$ $=2^{m n}$ i.e, Total number of non-empty relations is $2^{m n}-1$ Hence, the correct answer is option D....

The exterior angles, obtained on producing the base of a triangle both ways are 104° and 136°. Find all the angles of the triangle.

Question: The exterior angles, obtained on producing the base of a triangle both ways are 104and 136. Find all the angles of the triangle. Solution: ACD = ABC + BAC [Exterior angle property] Now ABC = 180 136 = 44 [Linera pair] ACB = 180 104 = 76 [Linera pair] Now, In ΔABC A + ABC + ACB = 180 [Sum of all angles of a triangle] ⇒ A + 44 + 76 = 180 ⇒ A = 180 44 76 ⇒ A = 60...