Question

Let $F:R\to R$ be a thrice differentiable function. Supose that $F\left(1\right)=0,F\left(3\right)=–4$ and $F^{\prime }\left(x\right)<0$ for all $x\in (1/2,3)$. Let $f\left(x\right)=xF\left(x\right)$ for all $x\in R$.

The correct statement(s) is(are)

• A. $f^{\prime }\left(1\right)<0$
• B. $f\left(2\right)<0$
• C. $f^{\prime }\left(x\right)\ne 0$ for any $x\in (1,3)$
• D. $f^{\prime }\left(x\right)=0$ for some $x\in (1,3)$

Answer 1

Answer: (A), (B), (C)

$f^{\prime }\left(x\right)\ne 0$ for any $x\in (1,3)$

$f\left(x\right)=xF\left(x\right)\Rightarrow f^{\prime }\left(x\right)=x{F}^{\prime }\left(x\right)+F\left(x\right)\Rightarrow f^{\prime }\left(1\right)=F^{\prime }\left(1\right)+F\left(1\right)=F^{\prime }\left(1\right)\therefore f^{\prime }\left(1\right)<0$

$f\left(2\right)=2F\left(2\right)$
As its is given $F^{\prime }\left(x\right)<0\forall x\in R$
So $F\left(x\right)$ is decreasing in nature.
$F\left(2\right)<F\left(1\right)\Rightarrow F\left(2\right)<0$
$\therefore f\left(2\right)<0$

$f^{\prime }\left(x\right)=x{F}^{\prime }\left(x\right)+F\left(x\right)$
We know that $F\left(x\right)\le 0$ and $x{F}^{\prime }\left(x\right)<0\forall x\in (1,3)$

So $f^{\prime }\left(x\right)<0\forall x\in (1,3)$