JEE Main 2022 · 29 Jun · Shift 1 · Mathematics Q4

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MathematicsSingle correct+4 / -1Question 4 of 90

Let $f:\mathbf R\to\mathbf R$ be a function defined by :
$f(x)=\begin{cases}\displaystyle\max_{t\le x}\left\{t^3-3t\right\}, & x\le2\\ x^2+2x-6, & 2<x<3\\ [x-3]+9, & 3\le x\le5\\ 2x+1, & x>5\end{cases}$
where $[t]$ is the greatest integer less than or equal to $t$ . Let $m$ be the number of points where $f$ is not differentiable and $I=\displaystyle\int_{-2}^2f(x)\,dx$ . Then the ordered pair $(m,I)$ is equal to :

\left(3,\dfrac{27}4\right)

\left(3,\dfrac{23}4\right)

\left(4,\dfrac{27}4\right)

\left(4,\dfrac{23}4\right)

Answer and worked solution

Official answer: C

First piece. The maximum of $t^3-3t$ up to $x$ is $x^3-3x$ for $x\le-1$ and 2 on $[-1,2]$ (local maximum 2 at $t=-1$ ).

Non-differentiable at $x=2$ (slopes 0 and 6), $x=3$ (8 and 0), $x=4$ and $x=5$ (jumps). At $x=-1$ both slopes are 0. So $m=4$ .

Integral. $\int_{-2}^{-1}(x^3-3x)dx+2\cdot3=\frac34+6=\frac{27}4$ .

(C)

Topics:Differentiability Properties of definite integrals

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JEE Main 2022 · 29 Jun · Shift 1, Mathematics question 4