JEE MAIN - Mathematics (2025 - 28th January Morning Shift - No. 22)
Let $\mathrm{f}(x)=\left\{\begin{array}{lc}3 x, & x<0 \\ \min \{1+x+[x], x+2[x]\}, & 0 \leq x \leq 2 \\ 5, & x>2\end{array}\right.$
where [.] denotes greatest integer function. If $\alpha$ and $\beta$ are the number of points, where $f$ is not continuous and is not differentiable, respectively, then $\alpha+\beta$ equals _______ .
where [.] denotes greatest integer function. If $\alpha$ and $\beta$ are the number of points, where $f$ is not continuous and is not differentiable, respectively, then $\alpha+\beta$ equals _______ .
Answer
5
Explanation
$$\begin{aligned} & f(x)=\left\{\begin{array}{ccc} 3 x & ; & x<0 \\ \min \{1+x, x\} & ; & 0 \leq x<1 \\ \min \{2+x, x+2\} & ; & 1 \leq x<2 \\ 5 & ; & x>2 \end{array}\right. \\ & f(x)=\left\{\begin{array}{ccl} 3 \mathrm{x} & ; & x<0 \\ \mathrm{x} & ; & 0 \leq x<1 \\ \mathrm{x}+2 & ; & 1 \leq x<2 \\ 5 & ; & x>2 \end{array}\right. \end{aligned}$$
_28th_January_Morning_Shift_en_22_1.png)
Not continuous at $\mathrm{x} \in\{1,2\} \Rightarrow \alpha=2$
Not diff. at $\mathrm{x} \in\{0,1,2\} \Rightarrow \beta=3$
$$\alpha+\beta=5$$
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