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Question Number 35873 by Tinkutara last updated on 25/May/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 28/May/18
	$let sin^(−1) x=θ sinθ=x tanθ=(x/(√(1−x^2 ))) so θ=tan^(−1) ((x/(√(1−x^2 )))) f(x)=(1/Π){tan^(−1) ((x/(√(1−x^2 ))))+tan^(−1) x}+((x+1)/((x+1)^2 +4)) f(x)=(1/Π){tan^(−1) ((((x/(√(1−x^(2 ) )))+(x/1))/(1−(x^2 /(√(1−x^2 ))))))}+((x+1)/((x+1)^2 +4)) f(x)=(1/Π){tan^(−1) (((x+x(√(1−x^2 )) )/((√(1−x^2 )) −x^2 )))}+((x+1)/((x+1)^2 +4)) (1−x^2 )=(1+x)(1−x) 1−x^2 =0 at x=±1 1−x^2 <0 when x>1 1−x^2 <0 when x<−1 1−x^2 >0 when x [−1,1] ∣f(x)∣_(x=−1) =(1/Π){tan^− (1)}+0 =(1/4) f(x) at x=1 f(x)=(1/Π){tan^(−1) (−1)}+(2/8) =(1/Π)×−(Π/4)+(1/4) =0 range of f(x) [0,(1/4)] pls check pls$
	$l e t {s i n}^{- 1} x = θ$ $s i n θ = x$ $t a n θ = \frac{x}{\sqrt{1 - x^{2}}} s o θ = {t a n}^{- 1} (\frac{x}{\sqrt{1 - x^{2}}})$ $f (x) = \frac{1}{Π} {{t a n}^{- 1} (\frac{x}{\sqrt{1 - x^{2}}}) + {t a n}^{- 1} x} + \frac{x + 1}{{(x + 1)}^{2} + 4}$ $f (x) = \frac{1}{Π} {{t a n}^{- 1} (\frac{\frac{x}{\sqrt{1 - x^{2}}} + \frac{x}{1}}{1 - \frac{x^{2}}{\sqrt{1 - x^{2}}}})} + \frac{x + 1}{{(x + 1)}^{2} + 4}$ $f (x) = \frac{1}{Π} {{t a n}^{- 1} (\frac{x + x \sqrt{1 - x^{2}}}{\sqrt{1 - x^{2}} - x^{2}})} + \frac{x + 1}{{(x + 1)}^{2} + 4}$ $(1 - x^{2}) = (1 + x) (1 - x)$ $1 - x^{2} = 0 a t x = \pm 1$ $1 - x^{2} < 0 w h e n x > 1$ $1 - x^{2} < 0 w h e n x < - 1$ $1 - x^{2} > 0 w h e n x [- 1, 1]$ $∣ f (x) ∣_{x = - 1}$ $= \frac{1}{Π} {{t a n}^{-} (1)} + 0$ $= \frac{1}{4}$ $f (x) a t x = 1$ $f (x) = \frac{1}{Π} {{t a n}^{- 1} (- 1)} + \frac{2}{8}$ $= \frac{1}{Π} \times - \frac{Π}{4} + \frac{1}{4}$ $= 0$ $r a n g e o f f (x) [0, \frac{1}{4}]$ $p l s c h e c k$ $p l s$
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