Let [t] denote the greatest integer less than | vedclass

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(A) Given $f(x) = x - [x]$ and $g(x) = 1 - (x - [x])$. Let ${x} = x - [x]$. Then $f(x) = {x}$ and $g(x) = 1 - {x}$.$h(x) = \min \{ {x}, 1 - {x} \}$.Fo

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continuous in $[-2, 2]$ but not differentiable at more than four points in $(-2, 2)$

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(A) Given $f(x) = x - [x]$ and $g(x) = 1 - (x - [x])$. Let ${x} = x - [x]$. Then $f(x) = {x}$ and $g(x) = 1 - {x}$.$h(x) = \min \{ {x}, 1 - {x} \}$.For $x \in [n, n+1)$,${x} = x - n$. So $h(x) = \min \{ x-n, 1-(x-n) \} = \min \{ x-n, 1-x+n \}$.The graphs of $f(x)$ and $g(x)$ intersect when ${x} = 1 - {x}$,which implies $2{x} = 1$,or ${x} = 0.5$.In each interval $[n, n+1)$,the function $h(x)$ increases from $0$ to $0.5$ and then decreases from $0.5$ to $0$.Since $h(x)$ is continuous everywhere and the graph shows sharp corners at $x = n$ (where ${x}=0$) and at $x = n + 0.5$ (where ${x}=0.5$),we count the points of non-differentiability in $(-2, 2)$.The points are $x = -1.5, -1, -0.5, 0, 0.5, 1, 1.5$. There are $7$ such points.Since $7 > 4$,option $A$ is correct.

Let $[t]$ denote the greatest integer less than or equal to $t$. Let $f(x)=x-[x]$,$g(x)=1-x+[x]$,and $h(x)=\min \{f(x), g(x)\}$ for $x \in [-2, 2]$. Then $h$ is :

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