- A$f^{\prime \prime}(0)=1$
- ✓$\mathrm{f}^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{24-\pi^2}{2 \pi}$
- C$f^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{12-\pi^2}{2 \pi}$
- D$f^{\prime \prime}(0)=0$
$ f^{\prime \prime}(x)=6 x \sin \left(\frac{1}{x}\right)-3 \cos \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right)-\frac{\sin \left(\frac{1}{x}\right)}{x} $
$ f^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{12}{\pi}-\frac{\pi}{2}=\frac{24-\pi^2}{2 \pi}$
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$1.$ Which of the following is true for $0 < x < 1$ ?
$(A)$ $0 < $ f(x) $ < \infty$
$(B)$ $-\frac{1}{2} < f(x) < \frac{1}{2}$
$(C)$ $-\frac{1}{4} < f(x) < 1$
$(D)$ $-\infty < $ f $($ x $) < 0$
$2.$ If the function $e^{-x} f(x)$ assumes its minimum in the interval $[0,1]$ at $x=\frac{1}{4}$, which of the following is true?
$(A)$ $f^{\prime}(x)$
$(B)$ $f^{\prime}(x)>f(x), 0$
$(C)$ f $^{\prime}(x)$
$(D)$ $f^{\prime}(x)$
Give the answer question $1$ and $2.$
$STATEMENT -1$ : 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 $t$.