Let f( x ) = \ \beginmatrix \max \ | x |,x^2 \ & | x | \le 2 8 - 2| x | & 2 \lt | x | \le 4 \endmatrix . Let S be the set of points in the interval (– 4, 4) at which f is not differentiable. Then S

Limits Continuity and Differentiability PYQ — JEE Main 2019 · 10 Jan · Shift 1 · Q31 with Solution

View written solutionFree

Correct answer: B

Understand the piecewise function

Given

f(x)= \begin{cases} \max\{|x|,x^2\}, & |x|\le 2,\\[4pt] 8-2|x|, & 2<|x|\le 4. \end{cases}

We need the points in $(-4,4)$ where $f$ is not differentiable.

Simplify $\max\{|x|,x^2\}$ for $|x|\le 2$

We compare $|x|$ and $x^2$ .

Since both are nonnegative, $x^2\ge |x| \iff |x|^2\ge |x| \iff |x|(|x|-1)\ge 0.$ So:

if $|x|\ge 1$ , then $x^2\ge |x|$ ,
if $|x|\le 1$ , then $|x|\ge x^2$ .

Thus for $|x|\le 2$ ,

\max\{|x|,x^2\}= \begin{cases} x^2, & 1\le |x|\le 2,\\ |x|, & |x|\le 1. \end{cases}

Hence the full function becomes:

f(x)= \begin{cases} x^2, & -2\le x\le -1,\\ |x|, & -1\le x\le 1,\\ x^2, & 1\le x\le 2,\\ 8+2x, & -4< x< -2,\\ 8-2x, & 2< x< 4. \end{cases}

(Here we rewrote $8-2|x|$ separately for negative and positive $x$ .)

Check differentiability inside each smooth interval

On the open intervals $(-4,-2),\; (-2,-1),\; (-1,0),\; (0,1),\; (1,2),\; (2,4),$ $f$ is given by either a linear function, $x^2$ , or $|x|$ split as $\pm x$ , all of which are differentiable.

So possible non-differentiable points are only where formulas change: $x=-2,-1,0,1,2.$

Check at $x=-2$

For $x<-2$ , $f(x)=8+2x \implies f'_-( -2)=2.$

For $-2<x<-1$ , $f(x)=x^2 \implies f'_+( -2)=2x\big|_{x=-2}=-4.$

Since $f'_-( -2)\ne f'_+( -2),$ $f$ is not differentiable at $x=-2$ .

Also continuity holds: $8+2(-2)=4, \qquad (-2)^2=4.$

Check at $x=-1$

For $-2<x<-1$ , $f(x)=x^2 \implies f'_-( -1)=2x\big|_{x=-1}=-2.$

For $-1<x<0$ , $f(x)=|x|=-x \implies f'_+( -1)=-1.$

Since $-2\ne -1,$ $f$ is not differentiable at $x=-1$ .

Continuity holds since $(-1)^2=1=|-1|.$

Check at $x=0$

Near $0$ , $f(x)=|x|.$ So

for $x<0$ , $f(x)=-x \implies f'_-(0)=-1$ ,
for $x>0$ , $f(x)=x \implies f'_+(0)=1$ .

Since $f'_-(0)\ne f'_+(0),$ $f$ is not differentiable at $x=0$ .

Check at $x=1$

For $0<x<1$ , $f(x)=|x|=x \implies f'_-(1)=1.$

For $1<x<2$ , $f(x)=x^2 \implies f'_+(1)=2x\big|_{x=1}=2.$

Since $1\ne 2,$ $f$ is not differentiable at $x=1$ .

Continuity holds since $|1|=1=1^2.$

Check at $x=2$

For $1<x<2$ , $f(x)=x^2 \implies f'_-(2)=2x\big|_{x=2}=4.$

For $x>2$ , $f(x)=8-2x \implies f'_+(2)=-2.$

Since $4\ne -2,$ $f$ is not differentiable at $x=2$ .

Continuity holds since $2^2=4, \qquad 8-2(2)=4.$

Conclusion

Therefore,

S=\{-2,-1,0,1,2\}.

This matches Option B.

Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.

More from Limits Continuity and Differentiability

View written solutionFree

Correct answer: B

Understand the piecewise function

Given

f(x)= \begin{cases} \max\{|x|,x^2\}, & |x|\le 2,\\[4pt] 8-2|x|, & 2<|x|\le 4. \end{cases}

We need the points in $(-4,4)$ where $f$ is not differentiable.

Simplify $\max\{|x|,x^2\}$ for $|x|\le 2$

We compare $|x|$ and $x^2$ .

Since both are nonnegative, $x^2\ge |x| \iff |x|^2\ge |x| \iff |x|(|x|-1)\ge 0.$ So:

if $|x|\ge 1$ , then $x^2\ge |x|$ ,
if $|x|\le 1$ , then $|x|\ge x^2$ .

Thus for $|x|\le 2$ ,

\max\{|x|,x^2\}= \begin{cases} x^2, & 1\le |x|\le 2,\\ |x|, & |x|\le 1. \end{cases}

Hence the full function becomes:

f(x)= \begin{cases} x^2, & -2\le x\le -1,\\ |x|, & -1\le x\le 1,\\ x^2, & 1\le x\le 2,\\ 8+2x, & -4< x< -2,\\ 8-2x, & 2< x< 4. \end{cases}

(Here we rewrote $8-2|x|$ separately for negative and positive $x$ .)

Check differentiability inside each smooth interval

On the open intervals $(-4,-2),\; (-2,-1),\; (-1,0),\; (0,1),\; (1,2),\; (2,4),$ $f$ is given by either a linear function, $x^2$ , or $|x|$ split as $\pm x$ , all of which are differentiable.

So possible non-differentiable points are only where formulas change: $x=-2,-1,0,1,2.$

Check at $x=-2$

For $x<-2$ , $f(x)=8+2x \implies f'_-( -2)=2.$

For $-2<x<-1$ , $f(x)=x^2 \implies f'_+( -2)=2x\big|_{x=-2}=-4.$

Since $f'_-( -2)\ne f'_+( -2),$ $f$ is not differentiable at $x=-2$ .

Also continuity holds: $8+2(-2)=4, \qquad (-2)^2=4.$

Check at $x=-1$

For $-2<x<-1$ , $f(x)=x^2 \implies f'_-( -1)=2x\big|_{x=-1}=-2.$

For $-1<x<0$ , $f(x)=|x|=-x \implies f'_+( -1)=-1.$

Since $-2\ne -1,$ $f$ is not differentiable at $x=-1$ .

Continuity holds since $(-1)^2=1=|-1|.$

Check at $x=0$

Near $0$ , $f(x)=|x|.$ So

for $x<0$ , $f(x)=-x \implies f'_-(0)=-1$ ,
for $x>0$ , $f(x)=x \implies f'_+(0)=1$ .

Since $f'_-(0)\ne f'_+(0),$ $f$ is not differentiable at $x=0$ .

Check at $x=1$

For $0<x<1$ , $f(x)=|x|=x \implies f'_-(1)=1.$

For $1<x<2$ , $f(x)=x^2 \implies f'_+(1)=2x\big|_{x=1}=2.$

Since $1\ne 2,$ $f$ is not differentiable at $x=1$ .

Continuity holds since $|1|=1=1^2.$

Check at $x=2$

For $1<x<2$ , $f(x)=x^2 \implies f'_-(2)=2x\big|_{x=2}=4.$

For $x>2$ , $f(x)=8-2x \implies f'_+(2)=-2.$

Since $4\ne -2,$ $f$ is not differentiable at $x=2$ .

Continuity holds since $2^2=4, \qquad 8-2(2)=4.$

Conclusion

Therefore,

S=\{-2,-1,0,1,2\}.

This matches Option B.

Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

More from Limits Continuity and Differentiability

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

More from Limits Continuity and Differentiability