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QJEEM24Mac3da: JEE Main 2024 Mathematics - Class 12 JEEM24Mac3da Solution: Let denote the greatest integer less than or equal to . Let be…

Let [t][t] denote the greatest integer less than or equal to tt . Let f:[0,∞)→Rf:[0, \infty) \rightarrow \mathbf{R} be a function defined by f(x)=[x2+3]−[x]f(x)=\left[\frac{x}{2}+3\right]-[\sqrt{x}] . Let S\mathrm{S} be the set of all points in the interval [0,8][0,8] at which ff is not continuous. Then ∑aes a is equal to __________.

Step-by-Step Verified Solution

f(x)amp;=[x3+3]−[x]amp;=[x2]−[x]+3\begin{aligned}f(x) & =\left[\frac{x}{3}+3\right]-[\sqrt{x}] \\& =\left[\frac{x}{2}\right]-[\sqrt{x}]+3\end{aligned}

Critical points where f(x)f(x) might change behaviours when x2∈\frac{x}{2} \in integer and x∈\sqrt{x} \in integer

⇒\Rightarrow Critical points,

amp;f(0)=3amp;f(0+)=3amp;f(1−)=3amp;f(1+)=2amp;f(2−)=2amp;f(2+)=3amp;f(3+)=f(3−)=3=f(4+)=f(4−)=f(5+)=f(5−)amp;f(6−)=3amp;f(6+)=4amp;f(7−)=f(7+)=4amp;f(8−)=4amp;f(8)=5\begin{aligned}& f(0)=3 \\& f\left(0^{+}\right)=3 \\& f\left(1^{-}\right)=3 \\& f\left(1^{+}\right)=2 \\& f\left(2^{-}\right)=2 \\& f\left(2^{+}\right)=3 \\& f\left(3^{+}\right)=f\left(3^{-}\right)=3=f\left(4^{+}\right)=f\left(4^{-}\right)=f\left(5^{+}\right)=f\left(5^{-}\right) \\& f\left(6^{-}\right)=3 \\& f\left(6^{+}\right)=4 \\& f\left(7^{-}\right)=f\left(7^{+}\right)=4 \\& f\left(8^{-}\right)=4 \\& f(8)=5\end{aligned}

amp;⇒f(x) is not continuous at amp;x=1,2,6,8amp;⇒∑a∈sa=1+2+6+8amp;=17\begin{aligned}& \Rightarrow f(x) \text { is not continuous at } \\& x=1,2,6,8 \\& \Rightarrow \sum_{a \in s} a=1+2+6+8 \\& \quad=17\end{aligned}

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