If f(x) be a continuous function defined for 1≤x≤3. f(x) ϵ Q ∀ x ϵ [1,3] and f(2)=10 (Where Q is a set of all rational numbers). Then, f(1.8) is
10
Consider two arbitrary points x1,x2 ϵ[1,3]. Let f(x1)=q1 and f(x2)=q2, (q1,q2, ϵ Q)
Suppose q1≠q2. Since f(x) is continuous, f(x) must take all the values between q1 and q2.
There are infinite irrational numbers between any two rational numbers
∴ f(x) must take irrational values
But f(x) ϵ Q ∀ x ϵ [1,3]
This is a contradiction.
Hence our assumption q1≠q2 is false.
∴ q1=q2
Since q1 and q2 are arbitrary, f(x) is a constant for all x.
∵ f(2) =10, f(x) =10 ∀ x ϵ[1, 3]
∴ f(1.8) = 10.