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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 $f(y)$ = $-f(2-y)$

M not understanding how C is false

Can anyone explain it?

@MANSI_SOMANI Think of the function f(x) graphically,

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

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

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

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

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
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.

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?

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

This random graph following given conditions will be helpful:

### 1 comment

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

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