Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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..

share|improve this question

2 Answers 2

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

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

share|improve this answer
    
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 $y$

share|improve this answer
1  
(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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.