Inequality

Question

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

Comments

ur a) and d) option same

rt?

a) or d) will be ans

Answer 1

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 -  Lim_{x --> 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..

Comments

Here $\log(1+x)$ is $ \mathrm{e}$ based NOT $10$ based.

$\frac{\mathrm{d}}{\mathrm{d}x}(\log(1+x))=\frac{1}{1+x}$ when $\log$ is $\mathrm{e}$ based.

If so, then $\log(1+1)=\log(2)\approx 0.693 ~$ not that $\log_{10}{2}\approx 0.301$.

But here you used $\log(2) \approx 0.301 ~$ which is wrong by the way.

Answer 2

$\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

Comments

This time, I really got confused as to mark which one as the best answer, both are such outstanding answers. For such students, who have just started preparing, Habib's answer would provide them with concepts of increasing and decreasing functions, so marked it as best. Though for gate aspirants, Srestha Di's answer is quite intuitive and time-saving. Thanx, @Srestha Di.

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wc :)

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Only the option (a) is correct. Because here $\log$ is e based. Even if you think here $\log$ is 10 based, you find option (c) as correct as well (!!!). To prove this take $x=\frac{1}{2}$ where $x \in (0,1)$.

Then $\sqrt{\frac{1-x}{1+x}}=\frac{1}{\sqrt{3}}\approx 0.577$

and $\frac{\log(1+x)}{\sin^{-1}x}=\frac{\log(3/2)}{\pi/6} \approx 0.336 ~$ [here taking 10 based $\log$]

Which is really surprising! That's why $\log$ must be e based here. If you take $\log$ e based, then $\frac{\log(3/2)}{\pi/6} \approx 0.774 ~$ which definitely suits the whole question.