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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 $f(y)$ = $-f(2-y)$
asked in Calculus | 4k views
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anyone knows how to solve this??
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Just think it via some random graph.
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B) it means $f(x)$ is symmetric around the $x=1$which is not always true, So FALSE;

C) not necessary so False;
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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).

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Why is D not correct ?
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I also think both A and D are correct. But GATE key says only A.
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how to think this in exam?
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Why D is wrong?
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Can someone add little more explanation to this question. Not able to follow it properly :(
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@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
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@ankit, why we take y from 0 to 1 ?
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Can someone prove C is false?
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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).
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It might  be trivial, but can u please tell how to calculate the interval in which function g is continuous
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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.
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Sai Krishna This question is testing aptitude rather than calculus

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

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 option A and D should be correct.

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Can you prove C is false?
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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
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 betwwen 0 to 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 betwwen -1 to 1

So, both of them are continuous in the interval 0 and 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, so they must intersect at some point between 0 and 1 since they are continuous in the interval. so option a

-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 and 1 , so they must intersect so d is also correct.
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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.
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yes d is right , I have rectified the answer..
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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)
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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]???
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This random graph may be helpful:

This random graph following given conditions will be helpful:

+1 vote

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

Option A is Correct but D is Wrong ,

Reason

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
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Why is that needed? f is defined over [0,2] and so the difference can be 2.
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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
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not really, it just wants 'y' in (0,1).
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I am not sure enough . Agreeing with you @arjun sir .
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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
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That is D option rt?
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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
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by applying intermediate value theorem both options a and d are coming true
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What is that theorem ? How to apply it here? Can you please explain ? @Kaluti
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@xylene

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).

http://mathworld.wolfram.com/IntermediateValueTheorem.html

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

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