JEE Main 2026 Apr 5 Shift 1, Mathematics Q17: Methods to Evaluate Limits

For small $x$, ${\sin }^2((\alpha +1)x)\sim (\alpha +1{)}^2{x}^2$, and using $\cos \theta \approx 1-{\theta }^2/2$ on each factor:

$1-\cos (\alpha x)\cos ((\alpha +1)x)\cos ((\alpha +2)x)\approx \frac{x^2}{2}({\alpha }^2+(\alpha +1{)}^2+(\alpha +2{)}^2)$.

Setting the limit equal to $2$:

$\frac{{\alpha }^2+(\alpha +1{)}^2+(\alpha +2{)}^2}{2(\alpha +1{)}^2}=2\Rightarrow {\alpha }^2+(\alpha +2{)}^2=3(\alpha +1{)}^2$.

Expanding: $2{\alpha }^2+4\alpha +4=3{\alpha }^2+6\alpha +3$, i.e. ${\alpha }^2+2\alpha -1=0$.

The product of the roots is $-1$.