Let $f: R \rightarrow(0, \infty)$ and $g: R \rightarrow R$ be twice differentiable functions such that $f''$ and $g''$ are continous functions on $R$. Suppose $f'(2)=g(2)=0, f''(2) \neq 0$ and $g''(2) \neq 0$. If $\lim\limits_{x \rightarrow 2} \frac{f(x) g(x)}{f'(x) g'(x)}=1$, then

(a) f has a local maximum at $x=2$

(b) f has a local minimum at $x=2$

(c) $f''(2)>f(2)$

(d) $f(x)-f''(x)=0$ for at least one $x \in R$.

(b), (d)

We have

$\lim\limits_{x \rightarrow 2} \frac{f(x) g(x)}{f'(x) g'(x)}=1$

$\Rightarrow \lim\limits_{x \rightarrow 2} \frac{f'(x) g(x)+f(x) g'(x)}{f''(x) g'(x)+f'(x) g''(x)}=1$      [Using L' Hospital's rule]

$\Rightarrow \frac{f'(2) g(2)+f(2) g'(2)}{f''(2) g'(2)+f'(2) g''(2)}=1$      [∵ f', f'', g', g'' are continuous]

$\Rightarrow \frac{f(2) g'(2)}{f''(2) g'(2)}=1$               [∵ f'(2) = g(2) = 0]

$\Rightarrow f''(2)=f(2)$

$\Rightarrow f''(2)>0$                     [∵ f : R → (0, ∞)   ∴ f(2) > 0]

Thus, we have f'(2) = 0 and f''(2) > 0. So, f has a local minimum at x = 2.

Again, f''(2) = f(2)

⇒ f(2) - f''(2) = 0

⇒ f(x) - f''(x) = 0 at x = 2

⇒ f(x) - f''(x) = 0 for at least one x ∈ R.