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Question Number 39695 by maxmathsup by imad last updated on 09/Jul/18

l e t f (x) = l n (2 x a r c t a n \sqrt{2 x^{2} - 1)}

1) f i n d D_{f}

2) c a l c u l a t e f^{^{'}} (x) a n d d e t e r m i n e i t s s i g n .

3) d e t e r m i n e t h e e q u a t i o n o f a s s y m p t o t e a t p o n t A (1, f (1))

3) f i n d a a n d b f r o m R / f (x) \sim a (x - 1) + b (x \to 1)

4) c a l c u l a t e \int_{0}^{1} f (x) d x

5) c a l c u l a t e f^{^{″}} (x)

Commented by maxmathsup by imad last updated on 14/Jul/18

1) w e h a v e f (x) = l n (2 x) + l n (a r c t a n (\sqrt{2 x^{2} - 1}))

x \in D_{f} \Leftrightarrow x > 0 a n d 2 x^{2} - 1 > 0 a n d a r c t a n (\sqrt{2 x^{2} - 1}) > 0 b u t

2 x^{2} - 1 > 0 \Leftrightarrow x^{2} > \frac{1}{2} \Rightarrow∣ x ∣> \frac{1}{\sqrt{2}} \Rightarrow x \in] - \infty, - \frac{1}{\sqrt{2}} [\cup] \frac{1}{\sqrt{2}}, + \infty [\Rightarrow

D_{f} =] \frac{1}{\sqrt{2}}, + \infty [

2) f^{,} (x) = \frac{1}{x} + \frac{(a r c t a n {(\sqrt{2 x^{2} - 1)})}^{^{'}}}{a r c t a n (\sqrt{2 x^{2} - 1})} = \frac{1}{x} + \frac{1}{a r c t a n (\sqrt{2 x^{2} - 1})} {\frac{\frac{4 x}{2 \sqrt{2 x^{2} - 1}}}{1 + 2 x^{2} - 1}}

= \frac{1}{x} + \frac{1}{x a r c t a n (\sqrt{2 x^{2} - 1}) \sqrt{2 x^{2} - 1}} .

Commented by maxmathsup by imad last updated on 14/Jul/18

w e h a v e x > \frac{1}{\sqrt{2}} \Rightarrow f^{^{'}} (x) > 0

2) e q u a t i o n o f t a n g e n t t o C_{f} a t p o i n t A (1, f (1)) i s y = f^{^{'}} (1) (x - 1) + f (1)

b u t f (1) = l n (2) + l n (\frac{π}{4}) a n d f^{^{'}} (1) = 1 + \frac{4}{π} \Rightarrow

y = (1 + \frac{4}{π}) (x - 1) + l n (2) + l n (\frac{π}{4}) .

Commented by maxmathsup by imad last updated on 14/Jul/18

3) i n t h i s Q c h a n g e a s s y m p t o t e b y t a n g e n t .

Commented by maxmathsup by imad last updated on 14/Jul/18

4) a = (1 + \frac{4}{π}) a n d b = l n (2) + l n (\frac{π}{4})