If Δ 1 = x s i n θ c o s θ - s i n θ - x 1 c o s θ 1 x and Δ 2 = x s i n 2 θ c o s 2 θ - s i n 2 θ - x 1 c o s 2 θ 1 x , x ≠ 0 ; then for all θ ∈ 0 , π 2 :

Δ 1 + Δ 2 = - 2 ( x 3 + x - 1 )

Δ 1 - Δ 2 = x ( c o s 2 θ - c o s 4 θ )

If A = 1 2 4 2 then show that | 2 A | = 4 | A | .

Given A = 1 2 4 2 ⇒ A = 1 × 2 - 4 × 2 = - 6 ⇒ 4 A = 4 - 6 = - 24 Now 2 A = 2 1 2 4 2 = 2 4 8 4 2 A = 2 × 4 - 8 × 4 = - 24 Hence proved.

Let N be the set of natural numbers and R be the relation on N × N defined by ( a , b ) R ( c , d ) iff a d = b c for all a , b , c , d ∈ N . Show that R is an equivalence relation.

Given that, N is the set of natural numbers and R be the relation on N × N defined by ( a , b ) R ( c , d ) iff a d = b c for all a , b , c , d ∈ N We know that, a relation R that is reflexive, symmetric and transitive is called an equivalence relation. Reflexive: For any ( a , b ) ∈ N × N a · b = b · a ∴ ( a , b ) R ( a , b ) Thus, R is reflexive Symmetric: For any ( a , b ) , ( c , d ) ∈ N × N ( a , b ) R ( c , d ) ⇒ a · d = b · c ⇒ c · b = d · a ⇒ ( c , d ) R ( a , b ) Thus, R is symmetric Transitive: For any ( a , b ) , ( c , d ) , ( e , f ) , ∈ N × N ( a , b ) R ( c , d ) and ( c , d ) R ( e , f ) ⇒ a · d = b · c and c · f = d · e ⇒ a · d · c · f = b · c · d · e ⇒ a · f = b · e ∴ ( a , b ) R ( e , f ) Thus, R is transitive Therefore, R is an equivalence Relation.

Check whether the relation R defined in the set 1 , 2 , 3 , 4 , 5 , 6 as R = a , b : b = a + 1 is reflexive, symmetric or transitive.

Suppose A = 1 , 2 , 3 , 4 , 5 , 6 . A relation R is defined on set A as: R = { ( a , b ) : b = a + 1 } ∴ R = { ( 1, 2 ) , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 } Reflexive Property : A relation R in a set A is said to be reflexive if x , x ∈ R , for every x ∈ A Now, 1 , 1 , 2 , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 , 6 ∉ R ∴ R is not reflexive. Symmetric Property : A relation R in a set A is said to be symmetric if x , y ∈ R implies that y , x ∈ R for all x , y ∈ A . Now it can be observed that 1 , 2 ∈ R , but 2 , 1 ∉ R . Hence , R is not symmetric. Transitive Property : A relation R in a set A is said to be transitive if x , y ∈ R and y , z ∈ R implies that x , z ∈ R for all x , y , z ∈ A . Now, 1 , 2 , 2 , 3 ∈ R but, ( 1 , 3 ) ∉ R ∴ R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive.

Show that the relation R on R is defined as R = { ( a , b ) : a ≤ b } , is reflexive and transitive but not symmetric.

Given, R = a , b : a ≤ b , a , b ∈ R For the relation R to be reflexive: If the relation R is reflexive, then a , a ∈ R i.e. a ≤ a , which is always true for all the values of a , a ∈ R So, the given relation R is reflexive. For the relation R to be symmetric: If a , b ∈ R then b , a ∈ R i.e. if a ≤ b then b ≤ a , which is not possible for all the values of a , b ∈ R . For example: if a = 2 and b = 5 Here, a < b but b > a So, the given relation R is not symmetric. For the relation R to be transitive: If a , b ∈ R and b , c ∈ R then a , c ∈ R i.e.,if a ≤ b and b ≤ c then a ≤ c , which is true for all the values of a , b , c ∈ R . For example: if a = 2 , b = 3 and c = 4 Here, a < b < c So, the given relation is transitive. Hence, the given relation R is reflexive and transitive but not symmetric.

Show that the lines x + 3 - 3 = y - 1 1 = z - 5 5 and x + 1 - 1 = y - 2 2 = z - 5 5 are coplanar.

Consider any two points A ( x 1 , y 1 , z 1 ) and B ( x 2 , y 2 , z 2 ) and consider two vectors a → and b → having direction ratios as ( a 1 , b 1 , c 1 ) and ( a 2 , b 2 , c 2 ) respectively. Then the condition for them to be coplanar is A B → . ( a → × b → ) = 0 The two lines given in the question are x + 3 - 3 = y - 1 1 = z - 5 5 and x + 1 - 1 = y - 2 2 = z - 5 5 . These are in the form of x - x 1 a 1 = y - y 1 b 1 = z - z 1 c 1 . So, on comparing all the three lines, we get The points for the first line, x 1 = - 3 , y 1 = 1 , z 1 = 5 , a 1 = - 3 , b 1 = 1 , c 1 = 5 and Similarly, for the second line, x 2 = - 1 , y 2 = 2 , z 2 = 5 , a 2 = - 1 , b 2 = 2 , c 2 = 5 . Now the direction ratios of the vector A B → = ( - 1 - - 3 ) , ( 2 - 1 ) , ( 5 - 5 ) = 2 , 1 , 0 . Now applying the condition for coplanarity, we get x 2 - x 1 y 2 - y 1 z 2 - z 1 a 1 b 1 c 1 a 2 b 2 c 2 = 2 1 0 - 3 1 5 - 1 2 5 = 2 5 - 10 - 1 - 15 + 5 + 0 = 2 × - 5 - 1 × - 10 = 2 × - 5 - 1 × - 10 = - 10 + 10 = 0 Since the condition is satisfied, the two lines are coplanar.