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A is the answer.

Say $s_1, s_2 , s_3$ and $t$ are the states (in sequence) with $s_1$ being the start state

$s_1= \epsilon + s_1a + s_1b = \epsilon +s_1(a+b) = (a+b)^*$ [using Arden's theorem  $R= Q+RP$ , then $R= QP^*$][$\epsilon$ because of being the start state]

$s_2 = s_1a = (a+b)^*a$

$s_3= s_2a+s_2b = s_2(a+b) = (a+b)^*a(a+b)$

$t= s_3b= (a+b)^*a(a+b)b$

$t$ is final state so regular expression is $(a+b)^*a(a+b)b$

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the first S represents (a+b)*

and the next 2 states are S so they are same as S(first state) 

that is output of 2,3 states are generated by first S, that means we can combine those 3 states and finally make S as final state 

which lead to option D that is (a+b)*

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  • option b and option d genrate null so it is false.becz it is not accepted by finite autometa
  • option c genrate string ab which is also false

 so option a is true

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The given NFA accepts all the strings that contains at least 1 ‘a’ and ends with ‘b’

  1.  is the answer 
  2. it does not accept string starting with b while our NFA accepts string starting with both ‘a’ and ‘b’  (Hence False)
  3. this expression accepts strings ending with ‘a’ (Hence False)
  4.  Same as option C (Hence False)
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