SATHEE: Application of Derivatives 4 Question 29

Application of Derivatives 4 Question 29

####31. 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

(2016 Adv.)

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

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

(d) $f (x) - f^{''} (x) = 0$ , for atleast one $x \in R$

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Answer:

Correct Answer: 31. (b)

Solution:

Here,

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

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

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

[using L’ Hospital’s rule]

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

$\Rightarrow f (2) = f^{''} (2)$

$∴ f (x) - f^{''} (x) = 0$ , for atleast one $x \in R$ .

$\Rightarrow$ Option (d) is correct.

Also, $f : R \to (0, \infty)$

$\begin{array}{lrl} \Rightarrow & f (2) > 0 \\ ∴ & f^{''} (2) = f (2) > 0 \end{array}$

[from Eq. (i)]

Since, $f^{'} (2) = 0$ and $f^{''} (2) > 0$

$∴ f (x)$ attains local minimum at $x = 2$ .

$\Rightarrow$ Option (a) is correct.

Application of Derivatives 4 Question 29

Answer:

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Application of Derivatives 4 Question 29

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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