JEE MAIN - Mathematics (2024 - 6th April Evening Shift - No. 21)

Let $$[t]$$ denote the greatest integer less than or equal to $$t$$. Let $$f:[0, \infty) \rightarrow \mathbf{R}$$ be a function defined by $$f(x)=\left[\frac{x}{2}+3\right]-[\sqrt{x}]$$. Let $$\mathrm{S}$$ be the set of all points in the interval $$[0,8]$$ at which $$f$$ is not continuous. Then $$\sum_\limits{\text {aes }} a$$ is equal to __________.
Answer
17

Explanation

$$\begin{aligned} f(x) & =\left[\frac{x}{3}+3\right]-[\sqrt{x}] \\ & =\left[\frac{x}{2}\right]-[\sqrt{x}]+3 \end{aligned}$$

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

$$\Rightarrow$$ Critical points,

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

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

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