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If 0$<$x$<$1 then

(a) $\sqrt{\frac{1-x}{1+x}} < \frac{log(1+x)}{sin^{-1}x} < 1$

(b) $\sqrt{\frac{1-x}{1+x}} > \frac{log(1+x)}{sin^{-1}x} > 1$

(c) $\sqrt{\frac{1-x}{1+x}} > \frac{log(1+x)}{sin^{-1}x} < 1$

(d) $\sqrt{\frac{1-x}{1+x}} < \frac{log(1+x)}{sin^{-1}x} > 1$
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Let we have a function g(x)  =  √ (1-x) / (1+x)  -  ( log(1+x) / sin-1x )

Now  g(1)  =  0 - log(2) / (π/2)    =   -2 * 0.301 / π    =   -0.602 / π

        g(0)  =  1 -  Limx --> 0  [(1/1+x)] / [1 / sqrt(1 - x2)]

                =  1 - 1  =  0

Now as it is clear that :

        g(0) > g(1) hence it is decreasing function and max value is at x = 0..

Thus g(x) <  0 always for 0 < x < 1    [ Equality holds at x = 0 ]

Thus ,

              √ (1-x) / (1+x)  -  ( log(1+x) / sin-1x )   <  0 

==>        √ (1-x) / (1+x)       <    log(1+x) / sin-1x

And on verification at points 0 and 1 , we will find each of them cant exceed the value of 1..

Hence A) should be the correct option..

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$\sqrt{\frac{1-x}{1+x}} $...........i

take $x=cos\theta$

putting in eqn i

$\sqrt{\frac{1-x}{1+x}} =tan\frac{\theta }{2}$.....................ii

$\frac{log\left ( 1+x \right )}{\sin^{-1}x}$

$=\frac{log\left ( 2cos^{2}\frac{\theta }{2} \right )}{\sin^{-1}cos\theta }$....................iii

Now, put any value for $\theta$

Say $\theta =\frac{\pi }{3}$

And now, get values where option a) and d) satisfies

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