SATHEE: Application of Derivatives 2 Question 31

Application of Derivatives 2 Question 31

####31. Show that $1 + x \log (x + \sqrt{x^{2} + 1}) \geq \sqrt{1 + x^{2}} \forall x \geq 0$ .

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Answer:

Correct Answer: 31. $\frac{1}{2} - \frac{π}{3}$ ( $1 + \frac{π}{3}$ ), $\frac{\sqrt{3}}{2} - \frac{π}{6}$ ( $1 + \frac{π}{6}$ )

Solution:

Let $f (x) = 1 + x \log (x + \sqrt{x^{2} + 1}) - \sqrt{1 + x^{2}}$

$∴ f^{'} (x) = x \cdot \frac{1 + \frac{x}{\sqrt{x^{2} + 1}}}{x + \sqrt{x^{2} + 1}} + \log (x + \sqrt{x^{2} + 1})$

$\begin{aligned} - \frac{x}{\sqrt{x^{2} + 1}} = & \frac{x}{\sqrt{x^{2} + 1}} + \log (x + \sqrt{x^{2} + 1}) - \frac{x}{\sqrt{x^{2} + 1}} \\ \Rightarrow & f^{'} (x) = & \log (x + \sqrt{x^{2} + 1}) \\ \Rightarrow & f^{'} (x) \geq 0 [∵ \log (x + \sqrt{x^{2} + 1}) \geq 0] \end{aligned}$

$∴ f (x)$ is increasing for $x \geq 0$ .

$\Rightarrow f (x) \geq f (0)$

$\Rightarrow 1 + x \log (x + \sqrt{1 + x^{2}}) - \sqrt{1 + x^{2}} \geq 1 + 0 - 1$

$\Rightarrow 1 + x \log (x + \sqrt{1 + x^{2}}) \geq \sqrt{1 + x^{2}}, \forall x \geq 0$

Application of Derivatives 2 Question 31

Answer:

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Application of Derivatives 2 Question 31

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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