Question

If the equation ${\sin }^{-1}(x^2+x+1)+{\cos }^{-1}(\lambda x+1)=\frac{\pi }{2}$ has exactly two solutions for $\lambda ϵ[a,b)$ then the value of $(a+b)$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

${\sin }^{-1}(x^2+x+1)+{\cos }^{-1}(\lambda x+1)=\frac{\pi }{2}$

The above equations has exactly the solutions For $\lambda ϵ(a,b)$

${\sin }^{-1}(x^2+x+1)=\frac{\pi }{2}-{\cos }^{-1}(\lambda x+1)$

Apply $\sin$ on both sides

$\therefore x^2+x+1=2x+1$

$x(x+(1-\lambda ))=0$

$x=0$ or $x=\lambda -1$

$0\le x^2+x+1\le 1$

$0\le {(\lambda -1)}^2+(\lambda -1)+1\le 1$

$0\le {\lambda }^2-\lambda +1\le 1$

$\therefore {\lambda }^2-\lambda +1\le 1$

${\lambda }^2-\lambda \le 0$

$\therefore \lambda ϵ(0,1),\lambda \ne 1$ ($\therefore x$ must have exactly two solutions)

$\therefore a=0,b=1$

$a+b=1$