JEE Main 2022 · 24 Jun · Shift 2 · Mathematics Q5

← JEE Main 2022 · 24 Jun · Shift 2

MathematicsSingle correct+4 / -1Question 5 of 90

Let
$f(x)=\begin{cases}\dfrac{\sin(x-[x])}{x-[x]}, & x\in(-2,-1)\\[2mm] \max\{2x,3[|x|]\}, & |x|<1\\[1mm] 1, & \text{otherwise}\end{cases}$
where $[t]$ denotes greatest integer $\le t$ . If $m$ is the number of points where $f$ is not continuous and $n$ is the number of points where $f$ is not differentiable, then the ordered pair $(m,n)$ is:

(3,3)

(2,4)

(2,3)

(3,4)

Answer and worked solution

Official answer: C

The pieces. On $(-2,-1)$ , $x-[x]=x+2\in(0,1)$ , so $f=\dfrac{\sin(x+2)}{x+2}$ . On $|x|<1$ , $[|x|]=0$ , so $f=\max\{2x,0\}$ . Elsewhere $f=1$ .

$x=-2$ . $\dfrac{\sin u}u\to1$ as $u\to0^+$ and its derivative there is 0, matching the constant 1: continuous and differentiable.

$x=-1$ . Left limit $\sin1\ne1=f(-1)$ : discontinuous.

$x=0$ . $\max\{2x,0\}$ has a corner: continuous, not differentiable.

$x=1$ . Left limit 2, $f(1)=1$ : discontinuous.

Count. $m=2$ ( $x=\pm1$ ) and $n=3$ ( $x=-1,0,1$ ).

(C)

Topics:Differentiability Continuity Graphs of functions

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JEE Main 2022 · 24 Jun · Shift 2, Mathematics question 5