3)

A1 = {0,2,4,6...}

A2 = {0,3,6,9,...}

A1 ∩ A2 ={0,6,12,18,24...} = {6k ; k ∈ N} = {(5+1)k ; k ∈ N} = A5

option c

9)

take for eg f(x)=x

[f(x)]^3 = x^3 (ie n=3)

watch the behaviour of the two functions x and x^3. we know the graph of the two functions. we also know that x^3 increases for all x ∈ R (evident from its graph)

so option A

i dont want to go into the formal proof (for the generalised case) right now.

27)

let 1<n<n+1 be 2 such integers ie a=n & b=n+1

so a^2 +b^2 = n^2 +(n+1)^2 = 2(n^2 +n) +1

lets see option a: it is always odd, so not true.

option b: <b>Always prime no ?

well it took Dr manindra agrawal a couple of decades to find an algorithm for generating prime numbers. it cant be that simple :P One can also verify by taking n=6. so false.

option c): it is prime for n=2 . so not true.

hence

**option d**
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"You don't have to believe in God, but you should believe in The Book." -Paul Erdős