# Which of the following statements are true $(NBHM - 2015)$?

Let $$X =\{f \in C[-5, 5] : f(-5)= f(5) = 0 \}$$

• There exist $$f \in X$$ such that $$f \equiv 2$$ on $$[-1, 0]$$ and $$f \equiv 3$$ on $$[1 , 2] \cup [3 , 4]$$

• For every $$f \in X$$, there exist distinct points $$x_1$$ and $$x_2$$ in $$]-5,5[$$ such that $$f(x_1) = f(x_2)$$

• For every $$f \in X$$ , therer exist $$x \in ]-5,5[$$ such that $$f(x) = x$$

• If $$f :\mathbb R \rightarrow \mathbb R$$ is differentiable on $$\mathbb R$$ and $$| f'| \leq M$$ for all $$t \in \mathbb R$$, then there exist $$\epsilon_o >0$$ such that for all $$0 < \epsilon \leq \epsilon_o$$, the function $$g(x) = x + \epsilon f(x)$$ is injective.

I would like to explain what all i have tried and if there is any better way to do please let Me know.

• define $$f(x) = \begin{cases} \frac{1}{2}(x+5) & x \in [-5,-1] \\ 2 & x \in [-1,0] \\ x+2 & x \in [0,1] \\ 3 & x \in [1,4] \\ -3x +15 & x \in [4,5] \end{cases}$$

Easly we can prove that $$f \in [-5,5]$$

• I think this can be proved by sketching the graph of $$f$$, but if someone have any counter example , provide me.

• Define $$g(x) = f(x) -x$$ $$\Rightarrow g(-5) = 5$$ and $$g(5) = -5$$, So by intermediate value theorem , there exist $$x \in ]-5, 5[$$ such that $$g(x) = x$$

$$\Rightarrow f(x) =x$$

• Let if possible there exist $$\epsilon_o > 0$$, for each $$\epsilon \leq \epsilon_o$$

define $$g(x) = x + \epsilon f(x)$$

$$\Rightarrow g'(x) = 1 + \epsilon f'(x)$$

$$\Rightarrow f'(x) = \frac{g'(x)- 1}{\epsilon}$$

as $$f'(x) \rightarrow \infty$$ for $$\epsilon \rightarrow 0$$ , which is a contradiction

So This is a false statement.

Please check my solution. Thank you

• all of the options are true, in last option $g'(x)=1+\epsilon f'(x)>1-\epsilon M$ Now how will you choose $\epsilon$ st $g'(x)>0$ ? Jan 27, 2015 at 16:18
• @Neeraj :what is your idea about fourth statement Jan 27, 2015 at 16:24
• but there exist an $\epsilon$ such that $f'(x) > M$ which is a contradiction Jan 27, 2015 at 16:26
• you cannot calculate $\lim f'(x)$ because you don't know anything about $lim$ of $g(x)$, so do like as I earlier commented. Jan 27, 2015 at 16:30
The second statement is true, and you can do better than "sketching the graph of $f$".
If $f$ is constantly zero in $]-5,0[$ the proof is obvious. Otherwise, there is a value $c$ in $]-5,0[$ where $f(c)\ne 0$. Then by continuity there are $x_1$ and $x_2$ such that $-5<x_1<c<x_2<5$ and $f(x_1)=f(x_2)=\frac{f(c)}2$.
About the last problem $g'(x)=1+\epsilon f'(x)>1-\epsilon M$ , so choosing $\epsilon_0\le\frac{1}{M}$ will ensure that for all $\epsilon\le\epsilon_0$ , $g'(x)>0$ thus $g$ is injective.