Question

lf the equation ${\sin }^{-1}(x^2+x+1)+{\cos }^{-1}(\lambda x+1)=\frac{\pi }{2}$ has exactly two solutions, then $\lambda$ can not have the integral value(s)

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

Answer: (A), (C), (D)

$1$
$-1$
$2$

${\sin }^{-1}(x^2+x+1)+{\cos }^{-1}(\lambda x+1)=\frac{\pi }{2}$
${\cos }^{-1}(\lambda x+1)=\frac{\pi }{2}-{\sin }^{-1}(x^2+x+1)$
Taking $\cos$ on both sides,
$\lambda x+1=x^2+x+1$
$x^2+(1-\lambda )x=0$
$x(x+1-\lambda )=0$
$x=0$ or $\lambda =x+1$

${\sin }^{-1}(x^2+x+1)$ is defined for $-1\le x^2+x+1\le 1$
$x^2+x+2\ge 0$ is always true.
$x(x+1)\le 0\Rightarrow x\in [-1,0]$
$\Rightarrow x+1\in [0,1]$
$\lambda \in [0,1]$
$\lambda \ne -1,2$

Also when $\lambda =1$, there is only one solution to the given equation i.e., $x=0$
So, $\lambda \ne 1$.