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.

The no. of real values of $x$ satisfying $\displaystyle \tan^{-1}\left(\frac{x}{1-x^2}\right)+\tan^{-1}\left(\frac{1}{x^3}\right) = \frac{3\pi}{4}$

options ::

(a) $0$

(b) $1$

(c) $2$

(d) Infinitely many

My Try:: Using The formula $\tan^{-1}A+\tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$

So $\displaystyle \tan^{-1}\left(\frac{\frac{x}{1-x^2}+\frac{1}{x^3}}{1-\frac{x}{x^3.(1-x^2)}}\right) = \frac{3\pi}{4}$

So $\displaystyle \frac{x^4+1-x^2}{x^3-x^5-x}= -1$

So $(x-1)(x^4-x+1) = 0$

$x = 1$ and $x^4 -x+1 = 0$

So Iam getting only one value of $x$ which is $x=1$ and How can i calculate any real value of $x$ exists from $x^4-x+1 = 0$

OR any other method by using we can solve this Question

Thanks

share|improve this question
    
The equation $x^4-x+1=0$ has no real solution. –  tetori Jan 26 '13 at 10:28
    
@juantheron I noticed you've asked several question on MSE, but you've never accepted an answer. You should accept your favorite answer if it is satisfactory. –  Git Gud Jan 26 '13 at 10:46
    
O sorry Git Gud .... –  juantheron Jan 26 '13 at 11:15

1 Answer 1

up vote 1 down vote accepted
  1. If $|x|<1$, then $x^4+1>1>x$. If $|x|\geq 1$, then $x^4+1>x^4\geq x$. In both cases $$ x^4-x+1>0 $$ so the equation $x^4-x+1=0$ have no real solutions.

  2. Note that $x=1$ is not a solution of original equation.

share|improve this answer
    
Thanks Norbert for solving $x^4-x+1 = 0$ but i did not understand why $x= 1$ is not a solution of original equation because at $x = 1$ the equation is $\displaystyle \tan^{-1}(\infty)+\tan^{-1}(1) = \frac{\pi}{2}+\frac{\pi}{4} = \frac{3\pi}{4}$ which is exactly equaal to $R.H.S$ –  juantheron Jan 26 '13 at 10:46
2  
Expression $\tan^{-1}(\infty)$ makes no sense. Even if you are talking about limits it is not ok because $$\lim\limits_{x\to 1+}\tan^{-1}\frac{x}{1-x^2}=\frac{\pi}{2}$$$$\lim\limits_{x\to 1-}\tan^{-1}\frac{x}{1-x^2}=-\frac{\pi}{2}.$$ Not to mention that you can't substitute $x=1$ into expression $\frac{x}{1-x^2}$ –  no identity Jan 26 '13 at 10:51
    
Oh Thanks Norbert i did not consider it means $L.H.S$ and $R.H.S$ at $x=1$ So our final answer of the question is no value of $x$ means option (a) –  juantheron Jan 26 '13 at 11:16

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.