Explanation:

Let R be the required relation A = {(1, 1) (2, 2), (3, 3)} (i) | R | = 3, when R = A (ii) | R | = 5, e.g. R = A  {(1, 2), (2, 1)} Number of R can be [3] (iii) R = {1, 2, 3} × {1, 2, 3}

Explanation:

Given: � ( � + � ) = � ( � ) ⋅ � ( � ) ⇒ � ( � ) = � � � f(x+y)=f(x)⋅f(y)⇒f(x)=e λx Also given: � ′ ( 0 ) = 4 � ⇒ � ′ ( � ) = � � � � ⇒ � ′ ( 0 ) = � = 4 � f ′ (0)=4a⇒f ′ (x)=λe λx ⇒f ′ (0)=λ=4a So, � ( � ) = � 4 � � f(x)=e 4ax Now consider the differential equation: � ′ ′ ( � ) − 3 � � ′ ( � ) − � ( � ) = 0 f ′′ (x)−3af ′ (x)−f(x)=0 We substitute � ( � ) = � � � f(x)=e λx : � ′ ′ ( � ) = � 2 � � � , � ′ ( � ) = � � � � f ′′ (x)=λ 2 e λx ,f ′ (x)=λe λx Substitute into the equation: � 2 � � � − 3 � � � � � − � � � = 0 ⇒ � � � ( � 2 − 3 � � − 1 ) = 0 λ 2 e λx −3aλe λx −e λx =0⇒e λx (λ 2 −3aλ−1)=0 Since � � � ≠ 0 e λx  =0, we get: � 2 − 3 � � − 1 = 0 λ 2 −3aλ−1=0 Now substitute � = 4 � λ=4a: ( 4 � ) 2 − 3 � ( 4 � ) − 1 = 16 � 2 − 12 � 2 − 1 = 4 � 2 − 1 = 0 ⇒ 4 � 2 = 1 ⇒ � 2 = 1 4 ⇒ � = ± 1 2 (4a) 2 −3a(4a)−1=16a 2 −12a 2 −1=4a 2 −1=0⇒4a 2 =1⇒a 2 = 4 1 ​ ⇒a=± 2 1 ​ Now, given: � ( � � ) = � � f(ax)=e x From above: � ( � ) = � 4 � � ⇒ � ( � � ) = � 4 � 2 � f(x)=e 4ax ⇒f(ax)=e 4a 2 x So: � ( � � ) = � � ⇒ � 4 � 2 � = � � ⇒ 4 � 2 = 1 ⇒ � = ± 1 2 f(ax)=e x ⇒e 4a 2 x =e x ⇒4a 2 =1⇒a=± 2 1 ​ Now, evaluate the area: ∫ 0 2 � 2 �   � � = [ � 2 � 2 ] 0 2 = � 4 − 1 2 ∫ 0 2 ​ e 2x dx=[ 2 e 2x ​ ] 0 2 ​ = 2 e 4 −1 ​ ✅ Final Answer: � ( � ) = � 4 � � f(x)=e 4ax , where � = ± 1 2 a=± 2 1 ​ � ( � � ) = � � f(ax)=e x confirms 4 � 2 = 1 4a 2 =1 Area under curve: ∫ 0 2 � 2 � � � = � 4 − 1 2 ∫ 0 2 ​ e 2x dx= 2 e 4 −1 ​ ​