Mathematics - Inverse Trigonometric Function Question with Solution | TestHub

The given equation holds if $x^2+x+1=ax+1$

and $-1\le x^2+x+1\le 1$ $\Rightarrow x(x+1+a)=0$

and $-1\le x\le 0$ $\Rightarrow x=0$ or $a-1$

and $-1\le x\le 0$

$\therefore x=0$ is one solution

and for another different solution $-1\le a-1<0$ $\Rightarrow 0\le a\le 1$.

So only integral value a can have is 0 .