Let f:R R be a twice differentiable function such that f^''(x)0 for all x R and f^'(a-1)=…

Question

Let $f:R	o R$ be a twice differentiable function such that ${f}^{''}(x)>0$ for all $x\in R$ and ${f}^{'}(a-1)=0$, where $a$ is a real number.

Let $g(x)=f({	an }^{2}x-2	an x+a),0Consider the following two statements:

(I) g is increasing in $(0,\frac{\pi }{4})$

(II) g is decreasing in $

Accepted Answer

$$ \begin{aligned} g(x) & =f\left((\tan x-1)^2+a-1\right) \ g^{\prime}(x) & =f^{\prime}\left((\tan x-1)^2+a-1\right) \cdot 2(\tan x-1) \cdot \sec ^2 x . \,\,\,\,\,\,\,\,...(i)\end{aligned} $$

Given $f^{\prime \prime}(x)>0 \quad \forall x \in R$

$\therefore f^{\prime}(x)$ is increasing function

Case (i) $0$$ \begin{aligned} & 0 1 \ & -1f^{\prime}(a-1)=0 \end{aligned} $$

From (i) $g\^{\prime}(x)=(+)(-)(+)=-v e$

$g(x)$ is decreasing in $\left(0, \frac{\pi}{4}\right)$

Similarly we can prove that $g(x)$ is increasing in $

Solutions · 1