MCQ
Let $f(x) = \frac{{x\,\, - \,\,1}}{{2\,{x^2}\,\, - \,\,7x\,\, + \,\,5}}$ . Then :
- A$x\overset{limit}{\rightarrow}1 \,\, f(x) = - \frac{1}{3}$
- B$x\overset{limit}{\rightarrow}0 \,\, f(x) = - \frac{1}{5}$
- C$f(x) \neq 0$
- ✓All of the above
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$STATEMENT -1$ : For each real $\mathrm{t}$, there exists a point $\mathrm{c}$ in $[\mathrm{t}, \mathrm{t}+\pi]$ such that $\mathrm{f}^{\prime}(\mathrm{c})=0$. because
$STATEMENT -2$: $f(t)=f(t+2 \pi)$ for each real $t$.