logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
0 votes

The number of real solutions of $\tan^{-1}\sqrt{x(x+1)}+\sin^{-1}\sqrt{x^2+x+1}=\large\frac{\pi}{2}$ is

$(a)\;0\qquad(b)\;1\qquad(c)\;2\qquad(d)\;infinite$

Can you answer this question?
 
 

1 Answer

0 votes
$\tan^{-1}\sqrt{x(x+1)}=\large\frac{\pi}{2}$$-\sin^{-1}\sqrt{(x^2+x+1)}$
$\Rightarrow \tan^{-1}\sqrt{x(x+1)}=\cos^{-1}\sqrt{x^2+x+1}$
$\Rightarrow \cos^{-1}\large\frac{1}{\sqrt{x^2+x+1}}$$=\cos^{-1}\sqrt{x^2+x+1}$
$\Rightarrow x^2+x+1=1$
$\Rightarrow x(x+1)=0$
$x=0,-1$
Hence (c) is the correct answer.
answered Oct 15, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...