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+48 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 $f(y)$ = $-f(2-y)$
in Calculus by | 7.3k views
anyone knows how to solve this??
Just think it via some random graph.
B) it means $f(x)$ is symmetric around the $x=1$which is not always true, So FALSE;

C) not necessary so False;

6 Answers

+67 votes
Let us define a new function $g:$

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

Since function $f$ is continuous in $[0,2],$  $g$ would be continuous in $[0,1]$

$g(0) = -2, g(1) = 2$

Since $g$ is continuous and goes from negative to positive value in $[0,1],$ at some point $g$ would be $0$ in $(0,1).$

$g=0 \implies f(y) = f(y+1)$ for some $y \in (0,1).$

Therefore, correct answer would be $(A).$
edited by
Why is D not correct ?
I also think both A and D are correct. But GATE key says only A.
how to think this in exam?
Why D is wrong?
Can someone add little more explanation to this question. Not able to follow it properly :(
@Dulqar , In the solution , Suraj sir has defined a new function 'g' as g(y) = f(y) - f(y+1)

Now In the question , It is given that f(x)  is continuous when  0 <= x <= 2 ... and f(0) = f(-2) = -1 and f(1) = 1

Now  do the same thing for g(y) = f(y) - f(y+1) by changing variable from x to y and  take y from 0 to 1..

Now for g(y)

If , y = 0 , then f(0) = -1 and f(0+1) = f(1) = 1 (given)

so , g(0) = f(0) - f(1) = -1-1 = -2.. so g(0) = -2

If y =1 , then f(1) = 1 and f(1+1)=f(2) = -1 (given)

so , g(1) = f(1) - f(2) = 1 - (-1) = 2..

Now , finally we get g(0) = -2 and g(1) = 2..

Since function is changing its value from -2 (ie. negative ) to +2 (i.e. positive).. It means this function g is increasing and will cut the axis when g(y) = 0..(think same as f(x) when u plot the graph)..

So, g(y) = 0 means f(y) = f(y+1) for the given interval 0 <= y <= 1..

do the same thing for other options.. we will get option A & D
@ankit, why we take y from 0 to 1 ?
@gate-17 , bcoz they are asking about interval (0,1)..
Can someone prove C is false?
Isn't it straight forward? For instance, plot a curve with the following points, f(0)=f(2)=-1, f(1)=1, f(1.5)=2 (or any value larger than 1).
It might  be trivial, but can u please tell how to calculate the interval in which function g is continuous
I may be wrong but I understand that the question was attacked from the options.

Is there any way we can get these kind of clues in exams? I think no.

Anyway I am really appreciating the way the answer was constructed.

Sai Krishna This question is testing aptitude rather than calculus 

D) cannot be answer, plz chk answer below

but why B) cannot be answer?
answer should be both A) and D)
How to define fuction g in the form f.i did not understand? How to solve this question
+21 votes
If we 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:

$g=0 => f(y) = -f(2 - y).$

So, with the same logic, both options $A$ and $D$ should be correct.
edited by
Can you prove C is false?
It might  be trivial, but can u please tell how to calculate the interval in which function g is continuous when we know the interval in which f is continuous
+18 votes
We can use a graphical approach, though we don't know the exact function but we have got some point like $f(x)=-1$ at $x=0,f(x)=1$ at $x=1$ and $f(x)=-1$ at $x=2$ and $f(x)$ is continuous between $0$ and $2.$

By $1$ unit horizontal left shift we get $f(x+1)$

$f(x+1)=-1$ at $x=-1,f(x)=1$ at $x=0$ and $f(x)=-1$ at $x=1$ and $f(x)$ is continuous between $-1$ and $1.$

So, both of them are continuous in the interval $0$ to $1,$ and $f(x)$ has to reach $-1$ to $1$ in this interval and $f(x+1)$ has to reach $1$ to $-1$ in this interval. This means they must intersect at some point between $0$ and $1.$ So, option A is TRUE.

$ -f(2-y)$ is continuous between $0$ and $2$ [reflect about $Y-$axis then horizontal right shift by 2 then reflect about $X-$ axis to get $-f(2-y)$ from $f(y)].$  So, $-f(2-y)$ goes from $1$ to $-1$ where as $f(y)$ goes from $-1$ to $1$ in the interval $0$ to $1.$ This means, they must intersect. So, D is also correct.

Correct answer: A, D
edited by
Shreya, How did you say that "-f(2-y) is continuous between -2 and -4". I believe  we can only say that -f(2-y) is also  continuous between 0 and 2.
yes d is right , I have rectified the answer..
Yeah it is continous but the Interval will (0,2)

If we use x=0 , interval is 2,

then the interval goes on decreasing ,

But in the option A the Interval is (0,1)
Shreya mam -f(2-y) is continuous in [2,4] and f(y) in [0,2] so there wont exist any common point .Hence d is wrong .How are you getting [0,2]???

This random graph may be helpful: 


@Dhananjay Is it necessary to take another function g(x)? Why can't we simply substitute value in both LHS and RHS. Hence why can't B be the correct option?

+2 votes

This random graph following given conditions will be helpful:

Can someone explain how this graph is able to differentiate between option b and d ?
+1 vote


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

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

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

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


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)

0 votes
Option A is Correct but D is Wrong ,


The difference between y and 2-y should be less than the length of the interval given .
Difference between y and 2-y is 2 and length of interval is (0,1)
Therefore D is Wrong
Why is that needed? f is defined over [0,2] and so the difference can be 2.
In option d they mentioned the range (0,1) for y but actually range is coming as 2 . SO i believe this is wrong choice
not really, it just wants 'y' in (0,1).
I am not sure enough . Agreeing with you @arjun sir .
Arjun Sir,  -f(2-y) is continuous between -2 and -4 .. [reflect about y axis then horizontal left shift by 2 then reflect about x axis to get -f(2-y)from f(y)].. so we can't say f(y)will intersect with -f(2-y) in between 0 and 1 ie. f(y)=-f(2-y) for some y in the interval 0 to 1 . so d is false
That is D option rt?
Sorry , D is right bcz -f(2-y) is continuous between 0 and 2 ,I did the mistake cz I was shifting it left by 2 unit but it will be shifted towards right by 2 unit since shifting is done wrt. -y not respect to +y ,here I committed the mistake.. Yes both a and d correct
by applying intermediate value theorem both options a and d are coming true
What is that theorem ? How to apply it here? Can you please explain ? @Kaluti


the Theorem is

if g (x) is some function which is defined over a close interval(a,b) than the value of resultant function will always pass through (a,b).

i think A & D both are correct as explained by Shreya
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