Question

The sum of the solutions of the equation
$2{\sin }^{-1}\left(\sqrt{x^2+x+1}\right)+{\cos }^{-1}\left(\sqrt{x^2+x}\right)=\frac{3\pi }{2}\mathrm{is}$

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

The correct option is B $-1$

$0\le \sqrt{x^2+x+1}\le 1\mathrm{and}0\le \sqrt{x^2+x}\le 1\therefore x=-1,0$

$\mathrm{For}x=-1,L.H.S.=2{\sin }^{-1}1+{\cos }^{-1}0=2\times \frac{\pi }{2}+\frac{\pi }{2}=\frac{3\pi }{2}.\therefore x=-1\mathrm{is}a\mathrm{solution}.$

$\mathrm{For}x=0,L.H.S.=2{\sin }^{-1}1+{\cos }^{-1}0=\frac{3\pi }{2}.\mathrm{Therefore},x=0\mathrm{is}\mathrm{also}a\mathrm{solution}.\mathrm{Thus},\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{solutions}=0+\left(-1\right)=-1.$