Let f : R rightarrow R be a twice differentia | vedclass

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(A) Given $g(x) = f((	an x - 1)^{2} + a - 1)$.Taking the derivative, $g^{\prime}(x) = f^{\prime}((	an x - 1)^{2} + a - 1) \cdot 2(	an x - 1) \cdot

Vedclass · Accepted Answer

Neither $(I)$ nor $(II)$ is True

Vedclass · Answer

(A) Given $g(x) = f((	an x - 1)^{2} + a - 1)$.Taking the derivative, $g^{\prime}(x) = f^{\prime}((	an x - 1)^{2} + a - 1) \cdot 2(	an x - 1) \cdot \sec^{2}x$.Since $f^{\prime\prime}(x) > 0$, $f^{\prime}(x)$ is a strictly increasing function.We are given $f^{\prime}(a-1) = 0$.If $(	an x - 1)^{2} > 0$, then $(	an x - 1)^{2} + a - 1 > a - 1$, which implies $f^{\prime}((	an x - 1)^{2} + a - 1) > f^{\prime}(a - 1) = 0$.For $x \in (0, \frac{\pi}{4})$, $	an x For $x \in (\frac{\pi}{4}, \frac{\pi}{2})$, $	an x > 1$, so $(	an x - 1) > 0$. Thus $g^{\prime}(x) = (	ext{positive}) \cdot (	ext{positive}) \cdot (	ext{positive}) > 0$. So $g$ is increasing in $(\frac{\pi}{4}, \frac{\pi}{2})$.Therefore, neither statement $(I)$ nor $(II)$ is true.

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