Question

If f(x) be a continuous function (a function whose graph has no breaks) defined for $1\le x\le 3.f\left(x\right)ϵQ\forall xϵ[1,3]andf\left(2\right)=10$ (Where Q is a set of all rational numbers). Then, f(1.8) is

• A. 1
• B. 5
• C. 10
• D. 20

Answer 1

The correct option is C 10

Consider two arbitrary points $x_1,x_2ϵ[1,3].Letf\left(x_1\right)=q_{1}andf\left(x_2\right)=q_2,(q_1,q_2,ϵQ)$
Suppose $q_1\ne q_2$. Since f(x) is continuous, f(x) must take all the values between $q_{1}and{q}_2$.
There are infinite irrational numbers between any two rational numbers
$\therefore$ f(x) must take irrational values
But $f\left(x\right)ϵQ\forall xϵ[1,3]$
This is a contradiction.
Hence our assumption $q_1\ne q_2$ is false.
$\therefore q_1=q_2$
Since $q_{1}and{q}_2$ are arbitrary, f(x) is a constant for all x.
$\because$ f(2) =10, f(x) =10 $\forall xϵ$[1, 3]
$\therefore$ f(1.8) = 10.