For d= 1/3 or 1/5 or 1/9
1-d = 2/3 or 4/5 or 8/9 respectively
a = 2/3(k-1) ; 4/5(k-1) ; 8/9(k-1) ------> all three give composite numbers no matter what the value of "k"
Only for 1-d = 1/2 => d=1/2
we can have "a" as prime

In question 14 won't the answer would be a since it is asking us head after even number of tosses so sample space should be H,TTH,TTTTH,.......... according to which prob will be 2/3

This is a binomial distribution with p(success)= 1/2 and n=10
Mean score = mean of this binomial distribution= np = 5
so P(SCORE =5)=p (NO. OF SUCCESS=5)= C(10,5)* (1/2)^5 * (1/2)^5 = 63/256

If the function is of the form f(z)=z,
then from the given equation (let's call it it eq1): f(x+y)*f(x-y)=
(f(x) + f(y))^2 - 4*x^2*f(y),
we have (x+y)*(x-y) = (x+y)^2 - 4*x^2*y

Solving we get a quadratic equation 2*x^2-x-y=0.
Solving for x, we get,

x=(1+(1+8y)^.5) or x=(1-(1+8y)^.5)..........(2)

So for a given value of y, eq1 will not hold for any x not satisfying (2). But as per the question eq1 should hold for all x,y that are real.
Contradiction!
So a function of the form f(z)=z cannot hold. Or, f(2)=2 cannot hold.