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QJEEM24Ma9b9c: JEE Main 2024 Mathematics - Class 12 JEEM24Ma9b9c Solution: Consider the function defined by . If and be respectively the…

Consider the function f:(0,∞)→Rf:(0, \infty) \rightarrow \mathbb{R} defined by f(x)=e−∣log⁡ex∣f(x)=e^{-\left|\log _e x\right|} . If mm and nn be respectively the number of points at which ff is not continuous and ff is not differentiable, then m+nm+n is

Multiple Choice Options

  • A 0
  • B 1
  • C 2
  • D 3

Step-by-Step Verified Solution

amp;f:(0,∞)→Ramp;f(x)=e−∣log⁡ex∣\begin{aligned}& f:(0, \infty) \rightarrow R \\& f(x)=e^{-\left|\log _e x\right|}\end{aligned}

f(x)=1e∣ln⁡x∣={1e−ln⁡x;0lt;xlt;11eln⁡x;x≥1\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.

{11x=x;0lt;xlt;11x,x≥1\left\{\begin{array}{l}\frac{1}{\frac{1}{x}}=x ; 0< x<1 \\\frac{1}{x}, x \geq 1\end{array}\right.

JEE Main 2024 (Online) 31st January Evening Shift Mathematics - Limits, Continuity and Differentiability Question 33 English Explanation

m=0\mathrm{m}=0 (No point at which function is not continuous)

n=1\mathrm{n}=1 (Not differentiable)

∴m+n=1\therefore \mathrm{m}+\mathrm{n}=1

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