Let g(x) = 3f(x/3) + f(3 − x) and f′′(x) > 0 for all x ∈ (0, 3).

Question

Let \(g(x)=3 f\left(\frac{x}{3}\right)+f(3-x)\) and \(f^{\prime \prime}(x)>0\) for all \(\mathrm{x} \in(0,3)\). If g is decreasing in \((0, \alpha)\) and increasing in \((\alpha, 3)\), then \(8 \alpha\) is

(1) 24

(2) 0

(3) 18

(4) 20

Answer 1

Correct option is (3) 18

\(g(x)=3 f\left(\frac{x}{3}\right)+f(3-x)\) and \(f^{\prime \prime}(x)>0 \forall x \in(0,3)\)

\(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})\) is increasing function

\(g^{\prime}(x)=3 \times \frac{1}{3} \cdot f^{\prime}\left(\frac{x}{3}\right)-f^{\prime}(3-x)\)

\(=\mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)-\mathrm{f}^{\prime}(3-\mathrm{x})\)

If g is decreasing in \((0, \alpha)\)

\(\mathrm{g}^{\prime}(\mathrm{x})<0\)

\(\mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)-\mathrm{f}^{\prime}(3-\mathrm{x})<0\)

\(\mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)<\mathrm{f}^{\prime}(3-\mathrm{x})\)

\(\Rightarrow \frac{\mathrm{x}}{3}<3-\mathrm{x}\)

\(\Rightarrow \mathrm{x}<\frac{9}{4}\)

Therefore \(\alpha=\frac{9}{4}\)

Then \(8 \alpha=8 \times \frac{9}{4}=18\)