Let $f: \mathbb{R} \rightarrow(0, \infty)$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ be twice differentiable functions such that $f^{\prime \prime}$ and $g^{\prime \prime}$ are continuous functions on $\mathbb{R}$. Suppose $f^{\prime}(2)=g(2)=0, \quad f^{\prime \prime}(2) \neq 0$ and $g^{\prime}(2) \neq 0$. If $\lim _{x \rightarrow 2} \frac{f(x) g(x)}{f^{\prime}(x) g^{\prime}(x)}=1$, then

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

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

(C) $\quad f^{\prime \prime}(2)>f(2)$

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Application of Derivatives

Let

f : R \to (0, \infty)

and

g : R \to 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 \to 2} \frac{f ( x ) g ( x )}{f ^{'} ( x ) g ^{'} ( x )} = 1

, then

Answer: (A), (D)

Apply L'Hôpital's rule to the limit:

lim_{x \to 2} \frac{f ( x ) g ( x )}{f ^{'} ( x ) g ^{'} ( x )} = lim_{x \to 2} \frac{f ^{'} ( x ) g ( x ) + f ( x ) g ^{'} ( x )}{f ^{''} ( x ) g ^{'} ( x ) + f ^{'} ( x ) g ^{''} ( x )} = \frac{g ^{'} ( 2 ) f ( 2 )}{f ^{''} ( 2 ) g ^{'} ( 2 )} = \frac{f ( 2 )}{f ^{''} ( 2 )} = 1

. Thus

f (2) = f^{''} (2)

. Since

f (2) > 0

(as

f : R \to (0, \infty)

), we have

f^{''} (2) > 0

, so

f

has a local minimum at

x = 2

, confirming (A). For (D), consider

h (x) = f (x) - f^{''} (x)

. We have

h (2) = 0

. Since

f^{'} (2) = 0

and

f

has a local minimum at

x = 2

, by Rolle's theorem applied appropriately, there exists at least one point where

f (x) = f^{''} (x)

, confirming (D).

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · One or more correct options