JEE Main 2022 · 26 Jun · Shift 2 · Mathematics Q5

← JEE Main 2022 · 26 Jun · Shift 2

MathematicsSingle correct+4 / -1Question 5 of 90

Let $f(x)=\min\{1,1+x\sin x\}$ , $0\le x\le2\pi$ . If $m$ is the number of points, where $f$ is not differentiable and $n$ is the number of points, where $f$ is not continuous, then the ordered pair $(m,n)$ is equal to

(2,0)

(1,0)

(1,1)

(2,1)

Answer and worked solution

Official answer: B

Where is the minimum $1+x\sin x$ ? When $x\sin x<0$ , i.e. on $(\pi,2\pi)$ ; elsewhere $f=1$ .

Continuity. At $\pi$ and $2\pi$ both pieces equal 1, so $f$ is continuous: $n=0$ .

Corners. At $x=\pi$ , $\frac d{dx}(x\sin x)=\sin x+x\cos x=-\pi\ne0$ , a corner. $x=2\pi$ is an end point. So $m=1$ .

(B)

Topics:Differentiability Continuity

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JEE Main 2022 · 26 Jun · Shift 2, Mathematics question 5