Let $F: R \rightarrow R$ be a thrice differentiable function. Suppose that $F(1)=0, F(3)=-4$ and $F'(x)<0$ for all $x \in(1 / 2,3)$. Let $f(x)=x F(x)$ for all $x \in R$. Then which of the following statements is (are) correct?
(a) $f'(1)<0$
(b) $f(2)<0$
(c) $f'(x) \neq 0$ for all $x \in(1,3)$
(d) $f'(x)=0$ for some $x \in(1,3)$
Answer & explanation
Correct answer: option 2
We have, $f(x)=x F(x)$ for all $x \in R$
$\Rightarrow f'(x)=F(x)+x F'(x)$ for all $x \in R$
$\Rightarrow f'(1)=F(1)+F'(1)$
$\Rightarrow f'(1)=0+F'(1)=F'(1)<0\left[∵ F'(x)<0\right.$ for all $x \in\left(\frac{1}{2}, 3\right)\left.\right]$
So, statement (a) is correct.
It is given that $F'(x)<0$ for all $x \in(1 / 2,3)$. So, F(x) is decreasing on the interval (1/2, 3).
∴ $F(2)<F(1)$
$\Rightarrow F(2)<0$ [∵ F(1) = 0]
$\Rightarrow 2 F(2)<0$
$\Rightarrow f(2)<0$ [∵ f(x) = x F(x)]
So, statement (b) is correct.
Again $f(x)=x F(x)$ for all $x \in R$.
$\Rightarrow f'(x)=F(x)+x F'(x)$ for all $x \in R$
Now, $F'(x)<0$ for all $x \in(1 / 2,3)$
$\Rightarrow F'(x)<0$ for all $x \in(1,3)$ and $F(x)$ is decreasing on $(1,3)$
$\Rightarrow F'(x)<0$ for all $x \in(1,3)$ and $F(x)<F(1)$ for all $x \in(1,3)$
$\Rightarrow F'(x)<0$ and $F(x)<0$ for all $x \in(1,3)$
$\Rightarrow F'(x)+x F(x)<0$ for all $x \in(1,3)$
$\Rightarrow f'(x)<0$ for all $x \in(1,3)$
$\Rightarrow f'(x) \neq 0$ for all $x \in(1,3)$
So, statement (c) is correct and statement (d) is incorrect.