GATE CSE 2014 | Set 1 | Question: 47

Question

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?

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

Comments

Detail Explanation 

This problem is based on Variation of Intermediate Value Theorem which is Bolzano's theorem.

Bolzano's theorem : link

Source :https://web.ma.utexas.edu/users/m408n/CurrentWeb/LM2-5-10.php

Variation of IVT :

1) To prove uniqueness, you must additionally show that f is strictly monotonic ( f′(x) > 0  or f′(x) < 0 on the interval using differentiability)

 

2) Let f be a real-valued function that is continuous on the closed interval [a,b].
If f(a) and f(b) have opposite signs (i.e., f(a) < 0< f(b) or f(b) < 0 < f(a), then there exists at least one point c ∈ (a,b) such that: $$f(c)=0.$$

Option A: \( \exists y \in (0,1) \) such that \( f(y) = f(y+1) \)

Proof:

Define the function:

\[

g(x) = f(x) - f(x+1) \quad \text{for } x \in [0, 1].

\]

Since \( f \) is continuous, \( g \) is continuous on \([0, 1]\).

Now  \( g \) at the endpoints:

\( g(0) = f(0) - f(1) = -1 - 1 = -2 \) (Negative)

\( g(1) = f(1) - f(2) = 1 - (-1) = 2 \) (Positive)

Since \( g(0) < 0 \) and \( g(1) > 0 \), by the Intermediate Value Theorem, there exists at least one \( y \in (0, 1) \) such that:

\[

g(y) = 0 \implies f(y) - f(y+1) = 0 \implies f(y) = f(y+1).

\]

This is  TRUE.

Option D: \( \exists y \in (0,1) \) such that \( f(y) = -f(2-y) \)

Proof:

Define the function:

\[

h(x) = f(x) + f(2-x) \quad \text{for } x \in [0, 1].

\]

Since \( f \) is continuous, \( h \) is continuous on \([0, 1]\).

Now evaluate \( h \) at the endpoints:

\( h(0) = f(0) + f(2) = -1 + (-1) = -2 \) (Negative)

\( h(1) = f(1) + f(1) = 1 + 1 = 2 \) (Positive)

Since \( h(0) < 0 \) and \( h(1) > 0 \), by the Intermediate Value Theorem, there exists at least one \( y \in (0, 1) \) such that:

\[

h(y) = 0 \implies f(y) + f(2-y) = 0 \implies f(y) = -f(2-y).

\]

This is  TRUE.

 

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If you are struggling to understand the question, here what's been asked:

Function is continuous in interval [0,2] the intervals are closed means continuity is defined for points 0 and 2 as well. But why this continuity is mentioned? Well it will help later on.

now some values are given inside this interval such as f(0) = -1, f(2) = -1, f(1) = 1

see when x is = 0 we have y = -1 and when x is 1 y is 1 as well that means the connection between these 2 points must cross the x axis. Why? because the function is continuous, it must never happen that you can say we can take a direct jump between these 2 point ie the function is broken between any 2 points you must travel in continuity. 

But what does crossing the x axis have to do? 

When the x axis is crossed at that point y is zero.

Now we use the same understanding to check the options

 

Option A: There exists a y in the interval (0,1) such that f(y) = f(y+1)

now f(y) - f(y+1) = 0 

and if we put values of 

y = 0 f(0) - f(1) = -1 - 1 = -2

and y = 1 f(1) - f(2) = 1 - -1 = 2

the connection between 0,-2 and 1,2 must cross the x axis and that is what equation f(y) - f(y+1) = 0 is saying that in the interval 0 and 1 when the connection will cut the x axis then this will be equal to zero. and that means f(y) = f(y+1)

 

Option B: For every y in the interval (0,1),f(y) = f(2−y)

f(y) - f(2-y) = 0

as same logic for option A we can say we can guarantee atleast a point where f(y) - f(2-y) = 0 is true, we can't say for every point.

 

Option C: we can't say about max value in the interval (0,2) as there are infinite points for which no max or min can be determined.

Option D: There exists a y in the interval (0,1) such that f(y) = −f(2−y)

f(y) + f(2-y) = 0

we just need to know that if in the interval (0,1) we are crossing the x axis or not.

at y = 0 f(0) + f(2) = -1 -1 = -2

at y = 1 f(1) + f(1) = 1 + 1 = 2

 so between these coordinates (0,-2) and (1,2) we must cross the x axis at least once 

therefore f(y) + f(2-y) = 0 

means f(y) = - f(2 -y)

 

please note you may think I have put y values and getting x absicca. It's not me it's the professors who have used "y" in the options. Also if open intervals are given then why would I take the boundary points? It is because these points are first of all for absicca and for which we are getting the function output inside these intervals. 

these A and D are true. both.

 

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➜ Option A (True): ➜ Let $D(y) = f(y+1) - f(y)$. ➜ At $y = 0$: $D(0) = f(1) - f(0) = 1 - (-1) = 2$ (Positive). ➜ At $y = 1$: $D(1) = f(2) - f(1) = -1 - 1 = -2$ (Negative). ➜ Since $D(y)$ changes sign from positive to negative, by IVT, $D(y) = 0 \implies f(y) = f(y+1)$.

 

➜ Option B (False):

➜ Test with asymmetric function $f(x) = -2x^2 + 4x - 1$ at $y = 0.5$:

➜ $f(0.5) = 0.5$ and $f(2 - 0.5) = f(1.5) = 0.5$.

➜ Here $f(0.5) = f(1.5)$ happens to hold, but test another point like $y = 0.2$:

➜ $f(0.2) = -0.28$ and $f(1.8) = -0.28$ (does not mean it holds universally for every $y$ in $(0, 1)$ under different curves).

 

➜ Option C (False):

➜ $f(1) = 1$ is a local maximum at $x=1$, but the function could exceed $1$ elsewhere if the curve reaches a higher peak (e.g., $f(x)$ can take values greater than $1$).

 

➜ Option D (False): ➜ Let $S(y) = f(y) + f(2-y)$. ➜ Using $f(x) = -2x^2 + 4x - 1$ at $y = 0.5$: ➜ $f(0.5) + f(1.5) = 0.5 + 0.5 = 1 \neq 0$ (fails the requirement $f(y) = -f(2-y)$).

Answer 1

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 $(A).$

Comments

You just put the cake to eat, greater would be the thing how did you come up with this intuition

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neel19 yes sum, difference, product and division (denominator function !=0) of two continuous functions is also continuous.

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How is g continuous in [0,1], since g(0) not equal to g(1)  ??

Answer 2

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

Comments

This random graph may be helpful: 

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@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?

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[Reflect about y axis then horizontal left shift by 2..

Answer 3

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.

Comments

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

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@Daddy

The graph can be as maximum as minimum as possible. It's continuous.

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See for that it is importan to consider all critical points as well the function itself, since we are not given both, it will not be righ to ceclare maximum value to be 1.

Answer 4

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: $A,D$ 

Comments

Very much. Thank you for your excellent explanation.That too options wise. I saw few explanations, but I was unable to understand.

Answer 5

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 .

Answer 6

This random graph following given conditions will be helpful:

Comments

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

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Thanks for the example. This is the easiest solution one can get.