JEE MAIN - Mathematics (2024 - 31st January Evening Shift - No. 17)
Consider the function $$f:(0, \infty) \rightarrow \mathbb{R}$$ defined by $$f(x)=e^{-\left|\log _e x\right|}$$. If $$m$$ and $$n$$ be respectively the number of points at which $$f$$ is not continuous and $$f$$ is not differentiable, then $$m+n$$ is
0
1
2
3
Explanation
$$\begin{aligned} & f:(0, \infty) \rightarrow R \\ & f(x)=e^{-\left|\log _e x\right|} \end{aligned}$$
$$\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{e}^{|\ln \mathrm{x}|}}=\left\{\begin{array}{l} \frac{1}{\mathrm{e}^{-\ln \mathrm{x}}} ; 0<\mathrm{x}<1 \\ \frac{1}{\mathrm{e}^{\ln \mathrm{x}}} ; \mathrm{x} \geq 1 \end{array}\right.$$
$$\left\{\begin{array}{l} \frac{1}{\frac{1}{x}}=x ; 0< x<1 \\ \frac{1}{x}, x \geq 1 \end{array}\right.$$
_31st_January_Evening_Shift_en_17_1.png)
$$\mathrm{m}=0$$ (No point at which function is not continuous)
$$\mathrm{n}=1$$ (Not differentiable)
$$\therefore \mathrm{m}+\mathrm{n}=1$$
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