Application of Derivatives 4 Question 29
####31. Let $f: R \rightarrow(0, \infty)$ and $g: R \rightarrow R$ be twice differentiable functions such that $f^{\prime \prime}$ and $g^{\prime \prime}$ are continuous functions on $R$. Suppose $f^{\prime}(2)=g(2)=0, f^{\prime \prime}(2) \neq 0$ and $g^{\prime}(2) \neq 0$.
If $\lim _{x \rightarrow 2} \frac{f(x) g(x)}{f^{\prime}(x) g^{\prime}(x)}=1$, then
(2016 Adv.)
(a) $f$ has a local minimum at $x=2$
(b) $f$ has a local maximum at $x=2$
(c) $f^{\prime \prime}(2)>f(2)$
(d) $f(x)-f^{\prime \prime}(x)=0$, for atleast one $x \in R$
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Answer:
Correct Answer: 31. (b)
Solution:
- Here,
$ \lim _{x \rightarrow 2} \frac{f(x) \cdot g(x)}{f^{\prime}(x) \cdot g^{\prime}(x)}=1 $
$\Rightarrow \lim _{x \rightarrow 2} \frac{f(x) g^{\prime}(x)+f^{\prime}(x) g(x)}{f^{\prime \prime}(x) g^{\prime}(x)+f^{\prime}(x) g^{\prime \prime}(x)}=1$
$ \Rightarrow \quad \frac{f(2) g^{\prime}(2)+f^{\prime}(2) g(2)}{f^{\prime \prime}(2) g^{\prime}(2)+f^{\prime}(2) g^{\prime \prime}(2)}=1 $
[using L’ Hospital’s rule]
$ \Rightarrow \quad \frac{f(2) g^{\prime}(2)}{f^{\prime \prime}(2) g^{\prime}(2)}=1 \quad\left[\because f^{\prime}(2)=g(2)=0\right] $
$\Rightarrow \quad f(2)=f^{\prime \prime}(2)$
$\therefore f(x)-f^{\prime \prime}(x)=0$, for atleast one $x \in R$.
$\Rightarrow$ Option (d) is correct.
Also, $f: R \rightarrow(0, \infty)$
$ \begin{array}{lrl} \Rightarrow & f(2)>0 \\ \therefore & f^{\prime \prime}(2)=f(2)>0 \end{array} $
[from Eq. (i)]
Since, $f^{\prime}(2)=0$ and $f^{\prime \prime}(2)>0$
$\therefore f(x)$ attains local minimum at $x=2$.
$\Rightarrow$ Option (a) is correct.
