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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?

- There exists a $y$
- For every $y$
- The maximum value of the function in the interval $(0,2)$ is $1$
- There exists a $y$

+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 $(A).$

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

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 $(A).$

+12

@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

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

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

+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.

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.

+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

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

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

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)

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

@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

+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)

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

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

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

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

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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