# If $f(x) = \sin \log_e (\frac{\sqrt{4-x^2}}{1-x})$ then find the range of this function.

Problem :

If $f(x) = \sin \log_e (\frac{\sqrt{4-x^2}}{1-x})$ then find the range of this function.

My approach :

$\frac{\sqrt{4-x^2}}{1-x} >0 \Rightarrow 1-x >0$ also $4-x^2 >0$

$\Rightarrow x \in (-2,1)$ Domain of f(x) is (-2,1)

Now how to find the range of this function please suggest on this .. thanks..

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Note that

$$\lim_{x\to -2^+}\frac{\sqrt{4-x^2}}{1-x}=0\;\;,\;\;\lim_{x\to 1^-}\frac{\sqrt{4-x^2}}{1-x}=\infty$$

and thus

$$\left\{\alpha\in\Bbb R\;;\;\alpha=\log\frac{\sqrt{4-x^2}}{1-x}\;,\;\;x\in(-2,1)\right\}=\Bbb R$$

and from here

$$\text{Im}\left(\sin\log\frac{\sqrt{4-x^2}}{1-x}\right)=[-1,1]$$

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but answer is (-1, sin1) – sultan Dec 29 '13 at 3:59
Then either "the answer" is wrong or I am. Unless shown some mistake in the above I'm going to go for "they are wrong and I am right". – DonAntonio Dec 29 '13 at 4:30

Let $\displaystyle \frac{4-x^2}{(1-x)^2}=y$

$\displaystyle \implies 4-x^2=y(1+x^2-2x)\iff x^2(y+1)-(2y)x+y-4=0$

As $x$ is real, the discriminant $\displaystyle (2y)^2-4(y+1)(y-4)=4+3y\ge0\iff y\ge-\frac43$

$\displaystyle\implies \frac{4-x^2}{(1-x)^2}\ge-\frac43$

As we need $\displaystyle \frac{\sqrt{4-x^2}}{1-x}>0,$ from the above argument it can assume any positive real value

So, the range will be the range $\sin y$ for real $y$

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(1-x) and not $(1-x)^2$ ... – sultan Dec 28 '13 at 16:50
@sultan, I have found the range of the square of $$\frac{\sqrt{4-x^2}}{1-x}$$ – lab bhattacharjee Dec 28 '13 at 16:51
but answer is (-1,sin1)... – sultan Dec 29 '13 at 3:59