Application of Derivatives 2 Question 31
31. Show that $1+x \log \left(x+\sqrt{x^{2}+1}\right) \geq \sqrt{1+x^{2}} \quad \forall x \geq 0$.
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Answer:
Correct Answer: 31. (d)
Solution:
- Let $f(x)=1+x \log \left(x+\sqrt{x^{2}+1}\right)-\sqrt{1+x^{2}}$
$$ \therefore \quad f^{\prime}(x)=x \cdot \frac{1+\frac{x}{\sqrt{x^{2}+1}}}{x+\sqrt{x^{2}+1}}+\log \left(x+\sqrt{x^{2}+1}\right) $$
$$ \begin{aligned} & -\frac{x}{\sqrt{x^{2}+1}}= & \frac{x}{\sqrt{x^{2}+1}}+\log \left(x+\sqrt{x^{2}+1}\right)-\frac{x}{\sqrt{x^{2}+1}} \\ \Rightarrow & f^{\prime}(x)= & \log \left(x+\sqrt{x^{2}+1}\right) \\ \Rightarrow & & \quad f^{\prime}(x) \geq 0 \quad\left[\because \log \left(x+\sqrt{x^{2}+1}\right) \geq 0\right] \end{aligned} $$
$\therefore f(x)$ is increasing for $x \geq 0$.
$\Rightarrow \quad f(x) \geq f(0)$
$\Rightarrow 1+x \log \left(x+\sqrt{1+x^{2}}\right)-\sqrt{1+x^{2}} \geq 1+0-1$
$\Rightarrow 1+x \log \left(x+\sqrt{1+x^{2}}\right) \geq \sqrt{1+x^{2}}, \forall x \geq 0$
