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) = minf(x), g(x), x \in [ - 2, 2]. Then h is :

Limits Continuity and Differentiability PYQ — JEE Main 2021 · 26 Aug · Shift 2 · Q23 with Solution

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Correct answer: A

Identify the basic functions

Given $f(x)=x-[x], \qquad g(x)=1-x+[x].$

Now, for any real $x$ , the fractional part is $\{x\}=x-[x].$ So, $f(x)=\{x\}.$ Also, $g(x)=1-(x-[x])=1-\{x\}.$

Hence $h(x)=\min\{\{x\},1-\{x\}\}.$

Work on each interval $[n,n+1)$

For $x\in[n,n+1)$ , where $n\in\mathbb Z$ , $[x]=n, \qquad \{x\}=x-n.$ Thus $f(x)=x-n, \qquad g(x)=1-(x-n)=n+1-x.$ So $h(x)=\min(x-n,n+1-x).$

This is the distance of $x$ from the nearer endpoint of the interval $[n,n+1]$ .

More explicitly, on $[n,n+1)$ ,

\begin{cases} x-n, & n\le x\le n+\frac12,\\[4pt] n+1-x, & n+\frac12\le x< n+1. \end{cases}$$ Thus, on each unit interval, $h$ is a triangular graph. --- 3. **Check continuity at integers in $[-2,2]$** Potential discontinuities come from the floor function, i.e. at integers. Let $x=n$ where $n$ is an integer. - At $x=n$, since $\{n\}=0$, $$h(n)=\min(0,1)=0.$$ - Left-hand limit as $x\to n^-$: then $x\in(n-1,n)$, so $\{x\}\to 1^-$. Hence $$h(x)=\min(\{x\},1-\{x\})\to \min(1,0)=0.$$ - Right-hand limit as $x\to n^+$: then $\{x\}\to 0^+$, so $$h(x)\to \min(0,1)=0.$$ Therefore $h$ is continuous at every integer. Away from integers, it is piecewise linear, hence continuous. So **$h$ is continuous on $[-2,2]$**. --- 4. **Check differentiability** From the piecewise formula, on each interval $[n,n+1)$: - slope is $+1$ on $(n,n+\tfrac12)$, - slope is $-1$ on $(n+\tfrac12,n+1)$. So nondifferentiability can occur at: 1. the midpoints $n+\tfrac12$, 2. the integers $n$. Now restrict to $(-2,2)$. ### (a) Midpoints in $(-2,2)$ These are $$-\frac32,\,-\frac12,\,\frac12,\,\frac32.$$ At each such point, left derivative is $+1$ and right derivative is $-1$, so $h$ is **not differentiable** at these 4 points. ### (b) Integers in $(-2,2)$ These are $$-1,\,0,\,1.$$ Near an integer $n$: - from the left, slope is $-1$, - from the right, slope is $+1$. So $h$ is **not differentiable** at these 3 points. Hence total number of points in $(-2,2)$ where $h$ is not differentiable is $$4+3=7.$$ Thus it is continuous on $[-2,2]$ but not differentiable at **more than four points** in $(-2,2)$. --- 5. **Evaluate options** - **A:** continuous in $[-2,2]$ but not differentiable at more than four points in $(-2,2)$ True, since there are $7$ such points. - **B:** not continuous at exactly three points in $[-2,2]$ False, it is continuous everywhere. - **C:** continuous in $[-2,2]$ but not differentiable at exactly three points in $(-2,2)$ False, there are $7$ points. - **D:** not continuous at exactly four points in $[-2,2]$ False. --- 6. **Final answer** The correct option is $$\boxed{\text{A}}.$$

More from Limits Continuity and Differentiability

View written solutionFree

Correct answer: A

Identify the basic functions

Given $f(x)=x-[x], \qquad g(x)=1-x+[x].$

Now, for any real $x$ , the fractional part is $\{x\}=x-[x].$ So, $f(x)=\{x\}.$ Also, $g(x)=1-(x-[x])=1-\{x\}.$

Hence $h(x)=\min\{\{x\},1-\{x\}\}.$

Work on each interval $[n,n+1)$

For $x\in[n,n+1)$ , where $n\in\mathbb Z$ , $[x]=n, \qquad \{x\}=x-n.$ Thus $f(x)=x-n, \qquad g(x)=1-(x-n)=n+1-x.$ So $h(x)=\min(x-n,n+1-x).$

This is the distance of $x$ from the nearer endpoint of the interval $[n,n+1]$ .

More explicitly, on $[n,n+1)$ ,

\begin{cases} x-n, & n\le x\le n+\frac12,\\[4pt] n+1-x, & n+\frac12\le x< n+1. \end{cases}$$ Thus, on each unit interval, $h$ is a triangular graph. --- 3. **Check continuity at integers in $[-2,2]$** Potential discontinuities come from the floor function, i.e. at integers. Let $x=n$ where $n$ is an integer. - At $x=n$, since $\{n\}=0$, $$h(n)=\min(0,1)=0.$$ - Left-hand limit as $x\to n^-$: then $x\in(n-1,n)$, so $\{x\}\to 1^-$. Hence $$h(x)=\min(\{x\},1-\{x\})\to \min(1,0)=0.$$ - Right-hand limit as $x\to n^+$: then $\{x\}\to 0^+$, so $$h(x)\to \min(0,1)=0.$$ Therefore $h$ is continuous at every integer. Away from integers, it is piecewise linear, hence continuous. So **$h$ is continuous on $[-2,2]$**. --- 4. **Check differentiability** From the piecewise formula, on each interval $[n,n+1)$: - slope is $+1$ on $(n,n+\tfrac12)$, - slope is $-1$ on $(n+\tfrac12,n+1)$. So nondifferentiability can occur at: 1. the midpoints $n+\tfrac12$, 2. the integers $n$. Now restrict to $(-2,2)$. ### (a) Midpoints in $(-2,2)$ These are $$-\frac32,\,-\frac12,\,\frac12,\,\frac32.$$ At each such point, left derivative is $+1$ and right derivative is $-1$, so $h$ is **not differentiable** at these 4 points. ### (b) Integers in $(-2,2)$ These are $$-1,\,0,\,1.$$ Near an integer $n$: - from the left, slope is $-1$, - from the right, slope is $+1$. So $h$ is **not differentiable** at these 3 points. Hence total number of points in $(-2,2)$ where $h$ is not differentiable is $$4+3=7.$$ Thus it is continuous on $[-2,2]$ but not differentiable at **more than four points** in $(-2,2)$. --- 5. **Evaluate options** - **A:** continuous in $[-2,2]$ but not differentiable at more than four points in $(-2,2)$ True, since there are $7$ such points. - **B:** not continuous at exactly three points in $[-2,2]$ False, it is continuous everywhere. - **C:** continuous in $[-2,2]$ but not differentiable at exactly three points in $(-2,2)$ False, there are $7$ points. - **D:** not continuous at exactly four points in $[-2,2]$ False. --- 6. **Final answer** The correct option is $$\boxed{\text{A}}.$$

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

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Limits Continuity and Differentiability question

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