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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].

Prerequisites

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) .

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].

Prerequisites

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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