
Let $f:R \to R$ be such that $f(2x - 1) = f(x)$ for all $x \in R$. If f is continuous at $x = 1$ and $f(1) = 1$, then
A) $f(2) = 1$
B) $f(2) = 2$
C) $f$ is continuous only at $x = 1$
D) $f$ is continuous only at all points
Answer
557.4k+ views
Hint: According to given in the question we have to determine the when we let $f:R \to R$ be such that $f(2x - 1) = f(x)$ for all $x \in R$ and if $f$ is continuous at $x = 1$ and $f(1) = 1$. So, first of all we have to solve the given expression which is $f(2x - 1) = f(x)$
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 $f(2x - 1) = f(x)$ to simplify it as mentioned in the solution hint. Hence,
$ \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,
$
\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,
$ \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,
$
\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 \to \infty $in the expression (1) as obtained in the solution step 3.
$ \Rightarrow f(x) = f(1)$ as $f(x)$ continuous at $x = 1$
$ \Rightarrow f(x)$ is a continuous function.
Hence, $f(x) = f(1)$ as $f(x)$ continuous at $x = 1$ and $f(x)$ 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 $f:R \to R$ then after substituting the x in the given expression we can obtain the limits.
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 $f(2x - 1) = f(x)$ to simplify it as mentioned in the solution hint. Hence,
$ \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,
$
\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,
$ \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,
$
\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 \to \infty $in the expression (1) as obtained in the solution step 3.
$ \Rightarrow f(x) = f(1)$ as $f(x)$ continuous at $x = 1$
$ \Rightarrow f(x)$ is a continuous function.
Hence, $f(x) = f(1)$ as $f(x)$ continuous at $x = 1$ and $f(x)$ 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 $f:R \to R$ then after substituting the x in the given expression we can obtain the limits.
Recently Updated Pages
A man running at a speed 5 ms is viewed in the side class 12 physics CBSE

The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

State and explain Hardy Weinbergs Principle class 12 biology CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Which of the following statements is wrong a Amnion class 12 biology CBSE

Differentiate between action potential and resting class 12 biology CBSE

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

How much time does it take to bleed after eating p class 12 biology CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Explain sex determination in humans with the help of class 12 biology CBSE

