Let g(x) = 3f(x/3) + f(3-x)∀x∈(0, 3) and f''(x) > 0∀x ∈ (0, 3) then g(x) decreases in interval (0, α), then α is

Question

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

(1) \(\frac{7}{4}\)

(2) \(\frac{2}{3}\)

(3) \(\frac{9}{4}\)

(4) \(\frac{7}{3}\)

Answer 1

Correct option is (3) \(\frac{9}{4}\)

\(g(x)=3 f\left(\frac{x}{3}\right)+f(3-x)\)  

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

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

\(g^{\prime \prime}=\frac{f^{\prime \prime}(x)}{3}+f^{\prime \prime}(3-x)\) 

\(\Rightarrow g^{\prime}(x)>0\) 

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

\(f^{\prime}(x)>0 \Rightarrow f^{\prime}(x)\) is increasing