- A$(0 , \pi )$
- ✓$\left[ {\frac{\pi }{4}\,\,,\,\,\pi } \right)$
- C$\left( {0\,\,,\,\,\frac{\pi }{4}} \right]$
- D$\left( {0\,\,,\,\,\frac{\pi }{2}} \right)$
$=5\left(x^{2}-\frac{8}{5} x+\frac{4}{5}\right)$
$=\left(5\left(x-\frac{4}{5}\right)^{2}+\frac{4}{5}\right)$
Thus, $R_{g}=\left[\frac{4}{5}, \infty\right)$
$\frac{4}{5} \leq g(x)<\infty$
$\log _{5 / 4}\left(\frac{4}{5}\right) \leq \log _{5 / 4}(g(x))<\log _{5 / 4}(\infty)$
$-1 \leq \log _{5 / 4}(g(x))<\infty$
$\tan ^{-1}(-1) \leq \tan ^{-1}\left(\log _{5 / 4}(g(x))\right)<\tan ^{-1}(\infty)$
$-\frac{\pi}{4} \leq \tan ^{-1}\left(\log _{5 / 4}((g(x)))<\frac{\pi}{2}\right.$
$-\frac{\pi}{4} \leq f(x)<\frac{\pi}{2}$
Thus, $\quad R_{f}=\left[-\frac{\pi}{4}, \frac{\pi}{2}\right)$
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$(A)$ $I_n=I_{n+2}$
$(B)$ $\sum_{m=1}^{10} I_{2 m+1}=10 \pi$
$(C)$ $\sum_{m=1}^{10} I_{2 m}=0$
$(D)$ $ I_n=I_{n+1}$