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Question: Let \(f:R \to R\) be such that \(f(2x - 1) = f(x)\) for all \(x \in R\). If f is continuous at \(x =...

Let f:R→Rf:R \to R be such that f(2x−1)=f(x)f(2x - 1) = f(x) for all x∈Rx \in R. If f is continuous at x=1x = 1 and f(1)=1f(1) = 1, then
A) f(2)=1f(2) = 1
B) f(2)=2f(2) = 2
C) ff is continuous only at x=1x = 1
D) ff is continuous only at all points

Explanation

Solution

According to given in the question we have to determine the when we let f:R→Rf:R \to R be such that f(2x−1)=f(x)f(2x - 1) = f(x) for all x∈Rx \in R and if ff is continuous at x=1x = 1 and f(1)=1f(1) = 1. So, first of all we have to solve the given expression which is f(2x−1)=f(x)f(2x - 1) = f(x)
Now, we have to substitute the obtained value in the function expression we obtained to simplify it and to simplify the obtained expression we have to find the L.C.M and then we have to take the limit as mentioned in the question.

Complete step-by-step solution:
Step 1: First of all we have to solve the expression f(2x−1)=f(x)f(2x - 1) = f(x) to simplify it as mentioned in the solution hint. Hence,
⇒x→(x+12)\Rightarrow x \to \left( {\dfrac{{x + 1}}{2}} \right)
Step 2: Now, we have to substitute the obtained expression in the solution step 1 into the given function as mentioned in the solution hint. Hence,
⇒f(x)=f(x+12) ⇒f(x+12)=f(x+12+12) ⇒f(x+12+12)=f(x+1+222) \Rightarrow f(x) = f\left( {\dfrac{{x + 1}}{2}} \right) \\\ \Rightarrow f\left( {\dfrac{{x + 1}}{2}} \right) = f\left( {\dfrac{{\dfrac{{x + 1}}{2} + 1}}{2}} \right) \\\ \Rightarrow f\left( {\dfrac{{\dfrac{{x + 1}}{2} + 1}}{2}} \right) = f\left( {\dfrac{{x + 1 + 2}}{{{2^2}}}} \right)
Step 3: Now, we have to expand the terms in the expression as obtained just above in the step 2,
⇒f(x+1+222)=f(x+1+2+22+23+...........2n−12n)\Rightarrow f\left( {\dfrac{{x + 1 + 2}}{{{2^2}}}} \right) = f\left( {\dfrac{{x + 1 + 2 + {2^2} + {2^3} + {{...........2}^{n - 1}}}}{{{2^n}}}} \right)
On solving the expression as obtained just above,
⇒f(x2n+2n−12n) ⇒f(x2n+1−12n).................(1) \Rightarrow f\left( {\dfrac{x}{{{2^n}}} + \dfrac{{{2^n} - 1}}{{{2^n}}}} \right) \\\ \Rightarrow f\left( {\dfrac{x}{{{2^n}}} + 1 - \dfrac{1}{{{2^n}}}} \right).................(1)
Step 4: Now, we have to take the limit n→∞n \to \infty in the expression (1) as obtained in the solution step 3.
⇒f(x)=f(1)\Rightarrow f(x) = f(1) as f(x)f(x) continuous at x=1x = 1
⇒f(x)\Rightarrow f(x) is a continuous function.
Hence, f(x)=f(1)f(x) = f(1) as f(x)f(x) continuous at x=1x = 1 and f(x)f(x) is a continuous function.

Therefore option (A) and option (D) both are correct.

Note: To obtain the value or the limit of the function it is necessary that we have to solve the given expression and determine the x as mentioned in the question that f:R→Rf:R \to R then after substituting the x in the given expression we can obtain the limits.