MCQ
$x = 1$ બિંદુ આગળ વિધેય$f(x)=\begin{cases}x^3-1;&1< x<\infty\\x-1;&-\infty< x\leq1\end{cases}$ એ ..........
- Aસતત અને વિકલનીય
- ✓સતત અને અવિકલનીય
- Cઅસતત અને વિકલનીય
- Dઅસતત અને અવિકલનીય
$f(x)=\begin{cases}x^3-1;&1< x<\infty\\x-1;&-\infty< x\leq1\end{cases}$
$\lim_{x \rightarrow 1^+}x^3-1=\lim_{x \rightarrow 1^-}(x-1)=f(1)\\ 0 \ \ \ \ = \ \ \ 0 \ \ \ = \ \ \ \ \ 0$
$x=1$ આગળ $f$ એ સતત છે
વિક્લનીય
$RHD$
$\lim_{x \rightarrow 1^+}\frac{f(x)-f(1)}{x-1}=\lim_{x \rightarrow 1^+}\frac{x^3-1-0}{x-1}\Rightarrow\frac{(x-1)(x^2+x+1)}{x-1}$
$\lim_{x \rightarrow 1^+}(x^2+x+1)\Rightarrow3=f'(1)$
$LHD$
$\lim_{x \rightarrow 1^-}\frac{x-1-0}{x-1}\Rightarrow\lim_{x \rightarrow 1^-}\frac{x-1}{x-1}=0=f'(1)$
$f$ એ $x=1$ આગળ વિકલનીય છે
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