Discussion Problem_(11)

[duck]

If f(x) is continuous for all real values of x and f(x) takes on only rational values, then, if f(1) = 1, the value of f(0) is:
(a) 0
(b) 1
(c) 2
(d) None of the above.

[anonymouse]

Logically, I think its (b)1, but I don't have a clear mathematical argument.

[laracroft]

option B
i think......

[Sumit]

suppose If we take the Function F(x)=-X+2,
Now,F(1)=1,f(0)=2 but as said f(x) only take rational values and domain of this function is all real number so all irrational number must belong to its domain too.....
So, as we put a irrational no. say square root of 2 then we will see that f(x) is also irrational no. which contradict given information of f(x) only take rational values...so, it can't be option c)
Also for option b) if we take F(x)=x^2-x+1 which also satisfying f(1)=1 n F(0)=1 but in this also if we put a irrational no. again we can see F(x) is also irrational which can't be true so, we can rule out option a) as well.

Now, If we take f(x)=X^2 then F(1)=1 n F(0)=0....also If we put irrational no square root of 2 in this function
we will get a rational no.
So, the answer is a) 0.

[Sinistral]

Sumit wrote

Now, If we take f(x)=X^2 then F(1)=1 n F(0)=0....also If we put irrational no square root of 2 in this function
we will get a rational no.
So, the answer is a) 0.

in ur above definition take x to be cube root of 2, this function f(x) again will give an irrational number. hence option a also discarded.

basically u r ignoring the condition that f(x) is continuous for all x & f(x) is rational for all x. so f(x) has to be a constant function. since f(1) is given to be 1 ==> f(x)=1.
hence option (b)

PS: the concepts underlying these kind of questions were discussed in detail by Amit Sir in one of the threads a few days back.

[anonymouse]

I found a rigorous proof. Have a look at the first answer here.

[Sumit]

thanks sinistral for pointing this out...

[duck]

Hi.. :)


There are three points in the question which can easily get you to the correct answer:
(i) Function is continuous.
(ii) Takes only rational values.
These two conditions imply that f(x) is a constant function.
Because if you make it a function of "x" then, f(x) can take irrational values also.
(iii) f(1) = 1
This condition imply that f(x) must be equal to 1.
Because if we take any other constant function then, (iii) wont be satisfied.

Therefore, we've f(x)=1

Correct answer:
Option(b): f(0)=1