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MCA NIMCET Differentibility PYQ


MCA NIMCET PYQ







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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2019 PYQ

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MCA NIMCET PYQ
Let $f(x)=\begin{cases}{{x}^2\sin \frac{1}{x}} & {,\, x\ne0} \\ {0} & {,x=0}\end{cases}$
Then which of the follwoing is true





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2024 PYQ

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MCA NIMCET PYQ
Let $f\colon\mathbb{R}\rightarrow\mathbb{R}$ be a function such that $f(0)=\frac{1}{\pi}$ and $f(x)=\frac{x}{e^{\pi x}-1}$ for $x\ne0$, then





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2024 PYQ

Solution

Analysis of Continuity and Differentiability

Function:

\[ f(x) = \begin{cases} \dfrac{x}{e^{\pi x} - 1}, & x \neq 0 \\ \dfrac{1}{\pi}, & x = 0 \end{cases} \]

✅ Continuity at \( x = 0 \):

\[ \lim_{x \to 0} f(x) = \frac{1}{\pi} = f(0) \quad \Rightarrow \quad \text{Function is continuous at } x = 0 \]

✏️ Differentiability at \( x = 0 \):

\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = -\frac{1}{2} \]

✅ Final Result:

  • Function is continuous at \( x = 0 \)
  • Function is differentiable at \( x = 0 \)
  • \( f'(0) = \boxed{-\frac{1}{2}} \)

MCA NIMCET PYQ
The set of points, where  is differential in 





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2018 PYQ

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MCA NIMCET PYQ
The slope of the function 





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2017 PYQ

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