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 $-1 \leq x^2+x+1 \leq 1$.
$\Rightarrow x^2+x \leq 0 \Rightarrow-1 \leq x<0$
For exactly two solution
$-1 \leq \lambda-1<0 \Rightarrow 0 \leq \lambda<1 $
$\therefore a =0 \text { and } b =1 $