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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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