Let \(g(x)=3 f\left(\frac{x}{3}\right)+f(3-x) \forall x \in(0,3)\) and \(f^{\prime\prime}(x)>0 \forall x \in(0,3)\) then…
Let \(g(x)=3 f\left(\frac{x}{3}\right)+f(3-x) \forall x \in(0,3)\) and \(f^{\prime\prime}(x)>0 \forall x \in(0,3)\) then \(g(x)\) decreases in interval \((0, \alpha)\), then \(\alpha\) is (22 Jan, Shift I, Memory Based)
\(\frac{9}{4}\)
\(g(x)=3 f\left(\frac{x}{3}\right)+f(3-x) \text { and } f^{\prime \prime}(x)>0 \forall x \in(0,3)\)
\(\Rightarrow f^{\prime}(x)\) is increasing function
\(\begin{aligned}& g^{\prime}(x)=3 \times \frac{1}{3} \cdot f^{\prime}\left(\frac{x}{3}\right)-f^{\prime}(3-x) \\& =\mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)-\mathrm{f}^{\prime}(3-\mathrm{x})\end{aligned}\)
If g is decreasing in \((0, \alpha)\)
\(\begin{aligned}& \mathrm{g}^{\prime}(\mathrm{x})<0 \\& \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)-\mathrm{f}^{\prime}(3-\mathrm{x})<0 \\& \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)<\mathrm{f}^{\prime}(3-\mathrm{x}) \\& \Rightarrow \frac{\mathrm{x}}{3}<3-\mathrm{x} \\& \Rightarrow \mathrm{x}<\frac{9}{4}\end{aligned}\)
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