Given,

F(1) = 0, F(3) = 4

F'(x) < 0 for all x $$\in$$(1, 3)

and f(x) = xF(x)

Now, f'(x) = F(x) + xF'(x)

$$\Rightarrow$$ f'(1) = F(1) + 1 . F'(1)

$$\Rightarrow$$ f'(1) = 0 + F'(1) ....... (1)

As F'(x) < 0 for all x $$\in$$ (1, 3)

$$\therefore$$ F'(1) < 0

From (1), we get

f'(1) = F'(1) < 0

From Lagrange theorem

$$F'(2) = {{F(3) - F(1)} \over {3 - 1}}$$

$$ = {{ - 4 - 0} \over 2}$$

= $$-$$2

As f(x) = xF(x)

$$\therefore$$ f(3) = 3 . F(3) = 3 . ($$-$$4) = $$-$$12

and f(1) = 1 . F(1) = 0

$$\therefore$$ $$f'(2) = {{f(3) - f(1)} \over {3 - 1}}$$

$$ = {{ - 12} \over 2} = - 6$$

As, f'(x) = F(x) + x . F'(x)

$$\therefore$$ f'(2) = F(2) + 2 . F'(2)

$$ \Rightarrow - 6 = {{f(2)} \over 2} + 2.\,( - 2)$$

$$ \Rightarrow {{f(2)} \over 2} = - 2$$

$$\Rightarrow$$ f(2) = $$-$$4

$$\therefore$$ f(2) < 0

$$f'(x) = {{f(3) - f(1)} \over {3 - 1}}$$ when x $$\in$$ (1, 3)

f(3) = $$-$$12

and f(1) = 0

$$\therefore$$ $$f'(x) = {{ - 12} \over 2} = - 6$$

$$\therefore$$ f'(x) $$\ne$$ 0 for any x $$\in$$(1, 3)