Question

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

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

$\therefore {\sin }^{-1}\left(x^2+x+1\right)+{\cos }^{-1}(\lambda x+1)=\frac{\pi }{2}$
If $x^2+x+1=\lambda x+1$
$\Rightarrow x^2+(1-\lambda )x=0$
$\Rightarrow x=0$ or $x=\lambda -1$
${\sin }^{-1}\left(x^2+x+1\right)$ will exists if $-1\le x^2+x+1\le 1$.
$\Rightarrow x^2+x\le 0\Rightarrow -1\le x<0$
For exactly two solution
$-1\le \lambda -1<0\Rightarrow 0\le \lambda <1$
$\therefore a=0\text{ and }b=1$