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74 votes
74 votes

A function $f(x)$ is continuous in the interval $[0,2]$. It is known that $f(0) = f(2) = -1$ and $f(1) = 1$. Which one of the following statements must be true?

  1. There exists a $y$ in the interval $(0,1)$ such that $f(y) = f(y+1)$ 
  2. For every $y$ in the interval $(0,1)$,$f(y)$ = $f(2-y)$
  3. The maximum value of the function in the interval $(0,2)$ is $1$
  4. There exists a $y$ in the interval $(0,1)$ such that $f(y)$ = $-f(2-y)$

11 Answers

3 votes
3 votes

We are given that a function $f(x)$ is continuous in the interval $(0,2$) and $f(0) = f(2) =-1$,  $f(1) =1$

Consider a function $g(y)$ such that $g(y) = f(y) -f(y+1)$

From the given values we have,

$g(0) = f(0) – f(1) = -1-1 = -2 $

$g(1) = f(1) – f(2) = 1-(-1) =2 $

Notice how $g(0)$ and $g(1)$ have opposite signs, from a simple corollary of the Intermediate Value Theorem:

If a continuous function defined on an interval is sometimes positive and sometimes negative it must be 0 at some point.

Thus there will exist atleast one root in the interval $(0,1)$ such that

$g(y) = 0$

$ f(y) -f(y+1) =0$

$ f(y) = f(y+1)$

Thus, option A is true.

For every $y$ in the interval $(0,1)$,  $f(y) = f(2-y)$ need not be true. Since, we are given that $f(0) = -1$ and $f(1) =1$ the curve will be an increasing one ( need not be strictly monotonically increasing, but it will be increasing as the function needs to reach till $y = 1$ at $x=1$.). So, its not necessary that every y follows the condition in option B hence, it can be eliminated.

Just by knowing three random points of a curve we cannot determine the maxima or minima of an unknown function, be it any interval. Thus, option C is eliminated

Let us consider again that $g(y) = f(y) + f(2-y)$

We have,

$g(0) = f(0) + f(2) = -2$

$g(1) = f(1) +f(1) = 2$

With a similar argument as above we can prove $f(y) = -f(2-y)$

Correct options: $A,D$ 

3 votes
3 votes

The solution can be found by defining a new function

Option 1:

  • t(y) = f(y) – f(y+1)
Since (f) is continuous in [0, 2], t will also be continuous in [0, 1]. Given that f(0)= -1, f(2) = -1, and f(1) = 1, we can calculate t(0) and t(1):
  • t(0) = f(0) - f(1) = -1 - 1 = -2
  • t(1) = f(1)- f(2) = 1+1 = 2
Since the function changes its value from -2 to +2, it means functionm t is increasing and there exists a point where t will be 0 in (0,1).

 

  • When t=0, then f(y)= f(y+1).

so that option 1 is correct 

The same logic can be applied to option 4 by defining a new function:

Option 4: 

  • h(y) = f(y) + f(2 – y).
We know that function f is continuous in [0,2], so function h will also be continuous in [0,1]. Given that f(0)=-1, f(2)=-1, and f(1)=1, we can calculate:
  • h(0) = f (0) - f (1) = -1 - 1 = -2
  • h(1) = f (1)- f (2) = 1+1 = 2
Since the function is changing its value from -2 to +2, it means function h is increasing and there exists a point where h would be 0 in (0,1).
  • When h=0 then f (y)= -f (y+2).

Hence option 4 is also correct .

edited by
2 votes
2 votes

Given:

$f\left ( x \right )$ is continuous in the interval $\left [ 0,2 \right ]$

$f\left ( 0 \right )=-1$

$f\left ( 1 \right )=1$

$f\left ( 2 \right )=-1$

Now,

A) There exists a $y$ in the interval $\left ( 0,1 \right )$ such that $f\left ( y \right )=f\left ( y+1 \right )$

What they mean by this I think , at some point  value of function $f\left ( y \right ),f\left ( y+1 \right )$ will meet

and that is possible

B) 1st picture has shifted left , and 2nd will shift right

But as per my assumption here also both graph will meet at point x=2

C) will not be the ans as at point $f\left ( 1.5 \right ) it can take value 2$

D) cannot be answer. See the picture it can never be intersect

So, answer will be A) and B)

https://math.stackexchange.com/questions/2779216/describe-these-functions/2779249?noredirect=1#comment5731686_2779249

1 votes
1 votes
Consider this  as a new function, g(y) = f(y) -f(y+1)
Since function f is continuous in [0, 2], therefore g would be continuous in [0, 1]
g(0) = -2, g(1) = 2
Since g is continuous and goes from negative to positive value in [0,1]. Therefore at some point g would be 0 in (0,1).
g=0 ⇒ f(y) = f(y+1) for some y in (0,1).
Apply similar logic to option D, Let g(y) = f(y) + f(2 - y)
Since function f is continuous in [0, 2], therefore g would be continuous in [0, 1] (sum of two continuous functions is continuous)
g(0) = -2, g(1) = 2
Since g is continuous and goes from negative to positive value in [0, 1]. Therefore at some point g would be 0 in (0, 1).
There exists y in the interval (0, 1) such that:
g=0 ⇒ f(y) = -f(2 – y)
Both A, D are answers.
Answer:

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