Which of the following statements is incorrec | vedclass

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(A) Let $I = \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$. Here,$a = \frac{\pi}{6}$ and $b = \frac{\pi}{3}$,so $a+b = \frac{\pi}{2}$. $g(\alpha) =

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$g(\alpha)$ is a strictly increasing function

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(A) Let $I = \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$. Here,$a = \frac{\pi}{6}$ and $b = \frac{\pi}{3}$,so $a+b = \frac{\pi}{2}$. $g(\alpha) = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin^{\alpha} x}{\sin^{\alpha} x + \cos^{\alpha} x} dx$ ... $(i)$ Using the property,$g(\alpha) = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin^{\alpha}(\frac{\pi}{2}-x)}{\sin^{\alpha}(\frac{\pi}{2}-x) + \cos^{\alpha}(\frac{\pi}{2}-x)} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos^{\alpha} x}{\cos^{\alpha} x + \sin^{\alpha} x} dx$ ... $(ii)$ Adding $(i)$ and $(ii)$: $2g(\alpha) = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin^{\alpha} x + \cos^{\alpha} x}{\sin^{\alpha} x + \cos^{\alpha} x} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 dx = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$. Thus,$g(\alpha) = \frac{\pi}{12}$,which is a constant function. $A$ constant function is neither strictly increasing nor strictly decreasing. Therefore,statements $A$,$B$,and $C$ are all incorrect.

Which of the following statements is incorrect for the function $g(\alpha)$ for $\alpha \in R$ such that $g(\alpha)=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin^{\alpha} x}{\cos^{\alpha} x+\sin^{\alpha} x} dx$?

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