Question

If the equation ${\sin }^{-1}(x^2+x+1)+{\cos }^{-1}(ax+1)=\frac{\pi }{2}$ has exactly two distinct solutions then value of $a$ could not be

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

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

-1
1
2

$si{n}^{-1}(x^2+x+1)+co{s}^{-1}(ax+1)=\frac{\pi }{2}$
$co{s}^{-1}(ax+1)=\frac{\pi }{2}-si{n}^{-1}(x^2+x+1)$
$co{s}^{-1}(ax+1)=co{s}^{-1}(x^2+x+1)$
$x^2+x+1-ax-1=0$
$x^2+x(1-a)=0$
$x=\frac{a-1\pm (1-a)}{2}$
Therefore
$x=0$ and $x=(a-1)$
Also $-1\le ax+1\le 1$
$\frac{-2}{a}\le x\le 0$
Clearly $a$ cannot be 1 as we would get coincident values of x.
If a=-1, x=-2 then
$co{s}^{-1}(2+1)$
$co{s}^{-1}\left(3\right)$ does not exists.
If a=2, x=1
$si{n}^{-1}(x^2+x+1)$
$=si{n}^{-1}\left(3\right)$
Which does not exists.
Hence only $a=0$ and $x=-1$ satisfies the above equation.