Question

Let $f:R\to (0,\infty )$ and $g:R\to R$ be twice differentiable functions such that $f^{\prime \prime }$ and $g^{\prime \prime }$ are continuous functions on $R$. Suppose $f^{\prime }\left(2\right)=g\left(2\right)=0,f^{\prime \prime }\left(2\right)\ne 0$ and $g^{\prime }\left(2\right)\ne 0$. If $\underset{x\to 2}{lim}\frac{f\left(x\right)g\left(x\right)}{f^{\prime }\left(x\right)g^{\prime }\left(x\right)}=1,$ then

• A. $f$ has a local minimum at $x=2$
• B. $f$ has a local maximum at $x=2$
• C. $f^{\prime \prime }\left(2\right)>f\left(2\right)$
• D. $f\left(x\right)-f^{\prime \prime }\left(x\right)=0$ for at least one $x\in R$

Answer 1

Answer: (A), (D)

The correct options are
A $f$ has a local minimum at $x=2$
D $f\left(x\right)-f^{\prime \prime }\left(x\right)=0$ for at least one $x\in R$

$f^{\prime }\left(2\right)=g\left(2\right)=0,f^{\prime \prime }\left(2\right)\ne 0$ and $g^{\prime }\left(2\right)\ne 0$
$\underset{x\to 2}{lim}\frac{f\left(x\right)g\left(x\right)}{f^{\prime }\left(x\right)g^{\prime }\left(x\right)}=1$

Applying L'Hospital Rule, we get
$\underset{x\to 2}{lim}\frac{f^{\prime }\left(x\right)g\left(x\right)+f\left(x\right)g^{\prime }\left(x\right)}{f^{\prime \prime }\left(x\right)g^{\prime }\left(x\right)+f^{\prime }\left(x\right)g^{\prime \prime }\left(x\right)}=1\Rightarrow \frac{f\left(2\right)}{f^{\prime \prime }\left(2\right)}=1\Rightarrow f^{\prime \prime }\left(2\right)=f\left(2\right)$
$\because f^{\prime }\left(2\right)=0$ and the range of $f$ is $(0,\infty )$
$\therefore f^{\prime \prime }\left(2\right)>0$
Hence, $f\left(2\right)$ will be local minima.