Applications Of Derivatives Question 181
Question: If $ f(x)=2x+{{\cot }^{-1}}x+\log (\sqrt{1+x^{2}}-x) $ , then $ f(x) $
Options:
A) Increases in [0 , $ \infty $ )
B) Decreases in [0 , $ \infty $ )
C) Neither increases nor decreases in (0, $ \infty $ )
D) Increases in (- $ \infty $ , $ \infty $ )
Show Answer
Answer:
Correct Answer: A
Solution:
We have $ f(x)=2x+{{\cot }^{-1}}x+\log (\sqrt{1+x^{2}}-x) $
$ \therefore f’(x)=2-\frac{1}{1-x^{2}}+\frac{1}{\sqrt{1+x^{2}}-x}( \frac{x}{\sqrt{1-x^{2}}}-1 ) $
$ =\frac{1+2x^{2}}{1+x^{2}}-\frac{1}{\sqrt{1+x^{2}}}=\frac{1+2x^{2}}{1+x^{2}}-\frac{\sqrt{(1+x^{2})}}{1+x^{2}} $
$ =\frac{x^{2}+\sqrt{1+x^{2}}(\sqrt{1+x^{2}}-1)}{1+x^{2}}\ge 0 $ for all x
Hence f(x) is an increasing function on $ (-\infty ,,\infty ) $ and in particular on $ [0,\ \infty ) $ .
BETA
