Discussion Forum

Let f:R→ (0,∞) and g: R→ R be twice differentiable functions such that f ’’ and g’’ are continuous functions on R

 Suppose  $f'(2)=g(2)=0, f''(2)\neq 0$ and $g'(2)\neq 0$, If  $\lim_{x \rightarrow 2}\frac{f(x)g(x)}{f'(x)g'(x)}=1$,  then

Answer:

Explanation:

Here,  $\lim_{x \rightarrow 2}\frac{f(x).g(x)}{f'(x).g'(x)}=1$

 $\Rightarrow \lim_{x \rightarrow 2}\frac{f(x)g'(x)+f'(x)g(x)}{f''(x)g'(x)+f'(x)g''(x)}=1$

  [using L' hospital's rule]

$\Rightarrow   \frac{f(2)g'(2)+f'(2)g(2)}{f''(2)g'(2)+f'(2)g''(2)}=1$

$\Rightarrow       \frac{f(2)g'(2)}{f''(2)g'(2)}=1$

 $[\therefore f'(2)=g(2)=0]$

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

$\therefore$            $f(x)-f"(x)=0$ , for atleast one x ε R

 $\Rightarrow$ Option (d) is correct

   Also, f:R → (0,∞ )

$\Rightarrow$       f(2)>0

$\therefore$   f''(2)=f(2)>0  [ from Eq .(i)]

Since f '(2)=0 and f''(2) >0

$\therefore$ f(x) attains local minimum at x=2

$\Rightarrow$  Option (a) is correct.