Probability expectations question

[Spiti]

Please can somebody help me with this problem. I'm terrible at probability.
The other two options are 0 and none of the above.

[Noel]

Is it option A by any chance

[Spiti]

Could you please explain how you reached that answer?

[Noel]

I'm sorry but i actually guessed the answer without any calculation..i couldn't come up with a structured solution..is the answer even correct?

[Sinistral]

We know that E[X^k]= M^k(t=0)
Where M(t) is the moment generating function of the pdf associated with random variable X.
and M^k(t) is the k-th derivative of M(t)

So E(X)= M'(0) =1
and E(X^2) = M''(0) = 1.

 Let    M(t)   =  e^x.
         M'(0) = 1
         M''(0)=1    


Similarly E(X^100)= M^100(t=0) = 1
 
           

         
     

[The Villain]

@Sinstral...Do we have to do moment generrating functions too for entrances??Vaise i have a bit of knowledge on it as i had done for my acturial exams...bmut dowe really neee them for isi/dse??

[Spiti]

Thank you.

[Amit Goyal]

Since, E(X^2)= E(X) = 1. This implies variance of X, V(X) = E(X^2) - (E(X))^2 = 0.
Thus we have E(X) = 1, V(X) = 0
which together implies
Pr(X= 1) =  1.
Hence,
E(X^100) = 1.

[SoniaKapoor]

Sir....how did you get P(x=1)=1...?