ISI MStat 2016 (Sample) Problem 2 | Continuous function | PSB

This is a beautiful sample problem from ISI MStat 2016 PSB Problem 2. This is based on application of continuity and integration.

Problem- ISI MStat 2016 Problem 2

Let $f:[-1,1] \rightarrow \mathbb{R}$ be a continuous function. Suppose that $f^{\prime}(x)$ exists and $f^{\prime}(x) \leq 1$ for all $x \in(-1,1)$ . If $f(1)=1$ and $f(-1)=-1,$ prove that $f(x)=x \text { for all } x \in[-1,1]$.

Solution

Given function is continuous and $f^{\prime}(x)$ exists .

Hence we can integrate $f^{\prime}(x)$ .

Another thing is given that $f^{\prime}(x) \leq 1$ for all $x \in(-1,1)$ ----(1) (say)

Now if we integrate both side of (1) from -1 to y , where $y \in(-1,1)$ then we get ,

$\int^y_{-1} f^{\prime}(x) \,dx \leq \int^y_{-1} 1\,dx$ , for all $y \in(-1,1)$

$\Rightarrow f(y)-f(-1) \leq y+1 \Rightarrow f(y) \leq y$ ,for all $y \in(-1,1)$ [ since $f(-1)=-1$ given ]---(2)

Again if we integrate both side of (1) from y to 1 ,where $y \in(-1,1)$ then we get,

$\int^1_{y} f^{\prime}(x) \,dx \leq \int^1_{y} 1\,dx$ , for all $y \in(-1,1)$

$\Rightarrow f(1)-f(y) \leq 1-y \Rightarrow f(y) \geq y$ ,for all $y \in(-1,1)$ [ since $f(1)=1$ given ] ---(3)

Hence from (2) & (3) we get $f(y)=y$ for all $y \in(-1,1)$ and $f(-1)=-1 , f(1)=1$ .

Therefore , $f(x)=x$ for all $x \in[-1,1]$ ( proved )

Challenge Problem

$f:(-1,1) \to (-1,1)$ be continuous function such that $f(x)=f(x^2)$ for every x and $f(0)=\frac{1}{2}$

Then find f(x) .

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