Question

Sum of the squares of all the solution(s) of the equation, $2{\sin }^{-1}(x+2)={\cos }^{-1}(x+3)$ is

• A. 4
• B. 6.25
• C. 10.25
• D. none

Answer 1

Answer: (A)

$\text{ Let }{\sin }^{-1}(x+2)=\alpha x+2=\sin \alpha$
$\therefore 2\alpha ={\cos }^{-1}(x+3)$
$\cos 2\alpha =x+3=(x+2)+1=1+\sin \alpha$
$1-2{\sin }^2\alpha =1+\sin \alpha$
$\sin \alpha (1+2\sin \alpha )=0$
$\Rightarrow \sin \alpha =0\text{ or }\sin \alpha =-1/2$
$\therefore x=-2\text{ or }x=-2.5\text{ (rejected) as it does not satisfy the original equation }$
$\therefore \therefore x^2=6.25$