a nice prob in differential calculus...must try

I think d ans is( b)

You can get away by just stating this: The rational numbers Q are dense in the field of real numbers R or a little less formally, between any two real numbers you will find a rational number.

The proof uses a property of real numbers known as the Archimedean property that states that given real numbers x,y with x>0, we can find a +ve integer n such that nx>y. This in turn is derived from the fact that a set of real numbers bounded above has a least upper bound in R

Now we are out to prove that if we have x<y then there is a rational number q such that x<q<y

From the archimedean property, since y-x>0, we have n(y-x)>1 for some natural number n.

Now there exist an integer m such that

m-1≤nx<m

Hence we have nx<m<1+nx<ny

So, nx<m<ny

So that x<m/n<y and since m and n are integers, q =m/n is the rational number that we are seeking.

Now coming to the question.

Let us say that f(x) is not a constant. Suppose f(x₁) = c₁ and f(x₂) = c₂ and say c₁<c₂

According to the problem statement c₁ and c₂ are both irrational. From the result above we have a rational number q such that c₁<q<c₂

Since f(x) is continuous, by Bolzano's theorem, we have x in the interval (x₁, x₂) such that f(x) = q contradicting the fact that f(x) takes on only irrational values

Hence f(x) is forced to be a constant

Now it is easy to see why option B is correct