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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 $f(y)$ = $-f(2-y)$
in Calculus by Veteran (98.9k points) | 5.4k views
0
anyone knows how to solve this??
0
Just think it via some random graph.
0
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

+53 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 $(A).$
by Loyal (6k points)
edited by
+1
Why is D not correct ?
+8
I also think both A and D are correct. But GATE key says only A.
+9
how to think this in exam?
0
Why D is wrong?
0
Can someone add little more explanation to this question. Not able to follow it properly :(
+9
@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
+1
@ankit, why we take y from 0 to 1 ?
+1
@gate-17 , bcoz they are asking about interval (0,1)..
0
Can someone prove C is false?
0
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).
0
It might  be trivial, but can u please tell how to calculate the interval in which function g is continuous
0
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.
0

Sai Krishna This question is testing aptitude rather than calculus 

0
D) cannot be answer, plz chk answer below

but why B) cannot be answer?
0
answer should be both A) and D)
0
How to define fuction g in the form f.i did not understand? How to solve this question
+15 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.
by Active (1.2k points)
edited by
0
Can you prove C is false?
+1
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
+15 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
by Active (4.4k points)
edited by
0
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.
0
yes d is right , I have rectified the answer..
0
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)
0
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]???
0

This random graph may be helpful: 

0

@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:

by Junior (521 points)
+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

by Veteran (113k points)
0 votes
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
by Boss (21.2k points)
0
Why is that needed? f is defined over [0,2] and so the difference can be 2.
0
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
+1
not really, it just wants 'y' in (0,1).
0
I am not sure enough . Agreeing with you @arjun sir .
0
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
0
That is D option rt?
0
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
0
by applying intermediate value theorem both options a and d are coming true
0
What is that theorem ? How to apply it here? Can you please explain ? @Kaluti
0

@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

0
i think A & D both are correct as explained by Shreya
Answer:

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