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f(x)=2x -(tan^-1x)- ln{x+(1+x^2)^1/2} whre x belongs to R, then f(x)---------- in (-infinity,+infinity)
  1. increasing b)decreasing
  2. nonicreasing d non decresing

moidin afsan , 10 Years ago
Grade 11
anser 1 Answers
Jitender Singh

Last Activity: 10 Years ago

Ans:
f(x) = 2x-tan^{-1}(x) - ln(x+\sqrt{1+x^{2}})
f'(x) = 2-\frac{1}{1+x^{2}} - \frac{1 + \frac{1.2x}{2\sqrt{1+x^{2}}}}{x+\sqrt{1+x^{2}}}
f'(x) = 2-\frac{1}{1+x^{2}} - \frac{1}{\sqrt{1+x^{2}}}
f'(x) = \frac{2x^{2}+1-\sqrt{1+x^{2}}}{1+x^{2}}
f'(x) = \frac{(2x^{2}+1)-(\sqrt{1+x^{2}})}{1+x^{2}}
2x^{2}+1\geq \sqrt{1+x^{2}}
\Rightarrow f'(x)\geq 0
=>f(x) is non decreasing in the given interval.
Thanks & Regards
Jitender Singh
IIT Delhi
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