Question

Let $f:R\to (0,\infty )$ and $g:R\to R$ be twice differential such that $f^{\prime \prime }$ and $g^{\prime \prime }$ are continuous 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)

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

$\underset{x\to 2}{lim}\frac{f\left(x\right)g\left(x\right)}{f^{\prime }\left(x\right)g^{\prime }\left(x\right)}=1$
$\because \underset{x\to 2}{lim}\frac{f\left(x\right)g\left(x\right)}{f^{\prime }\left(x\right)g^{\prime }\left(x\right)}$ $\left(\frac{0}{0}\right)$ Indeterminant form as $f^{\prime }\left(2\right)=0,g\left(2\right)=0$ $\therefore$ using L.H.
$\underset{x\to 2}{lim}\frac{f^{\prime }\left(x\right)g\left(x\right)+g^{\prime }\left(x\right)f\left(x\right)}{f^{\prime \prime }\left(x\right)g^{\prime }\left(x\right)+g^{\prime }\left(x\right)f^{\prime }\left(x\right)}=\frac{f^{\prime }\left(2\right)g\left(2\right)+g^{\prime }\left(2\right)f\left(2\right)}{f^{\prime \prime }\left(2\right)g^{\prime }\left(2\right)+g^{\prime }\left(2\right)f^{\prime }\left(2\right)}=\frac{g^{\prime }\left(2\right)f\left(2\right)}{f^{\prime \prime }\left(2\right)g^{\prime }\left(2\right)}=1\Rightarrow f^{\prime \prime }\left(2\right)=f\left(2\right)$
and $f^{\prime }\left(2\right)=0$ & range of $f\left(x\right)\in (0,\infty )$
So $f^{\prime \prime }\left(2\right)=f\left(2\right)=+ve$ so $f\left(x\right)$ has point of minima at $x=2$
and $f\left(2\right)=f^{\prime \prime }\left(2\right)$ so $f\left(x\right)=f^{\prime \prime }\left(x\right)$ have at least one solution in $x\in R$.