QJEEM23Ma128a: JEE Main 2023 Mathematics - Class 12 JEEM23Ma128a Solution: Let the function be defined as where denotes the greatest…

Question

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Let the function $f:[0,2] \rightarrow \mathbb{R}$ be defined as

$f(x)= \begin{cases}e^{\min \left\{x^{2}, x-[x]\right\},} & x \in[0,1) \\ e^{\left[x-\log _{e} x\right]}, & x \in[1,2]\end{cases}$

where $[t]$ denotes the greatest integer less than or equal to $t$ . Then the value of the integral $\int_{0}^{2} x f (x) d x$ is :

Accepted Answer

\begin{aligned}\operatorname{Minimum}\left\{\mathrm{x}^2,\{\mathrm{x}\}\right\} &amp; =\mathrm{x}^2 ; \mathrm{x} \in[0,1) \\\\{\left[\mathrm{x}-\log _{\mathrm{e}} \mathrm{x}\right] } &amp; =1 ; \mathrm{x} \in[1,2)\end{aligned}

\therefore \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}\mathrm{e}^{\mathrm{x}^2} ; \mathrm{x} \in[0,1) \\\\\mathrm{e} ; \mathrm{x} \in[1,2)\end{array}\right.

\begin{aligned}&amp; \int\limits_0^2 x f(x)=\int\limits_0^1 x \cdot e^{x^2} d x+\int\limits_1^2 x \cdot e d x \\\\&amp; x^2=t \Rightarrow 2 x d x=d t\end{aligned}

=\frac{1}{2} \int\limits_0^1 e^t d t+\left.e \frac{x^2}{2}\right|_1 ^2

\begin{aligned}&amp; =\frac{1}{2}(\mathrm{e}-1)+\frac{1}{2}(4-1) \mathrm{e} \\\\&amp; =2 \mathrm{e}-\frac{1}{2}\end{aligned}

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