MHTCET 25 Apr 2025 Shift 2 Paper

Concept:Use the inverse tangent addition formula to combine terms, then solve the resulting cubic equation for $x<0$.Explanation:Start with $\tan^{-1}(x+1)+\tan^{-1}(x-1)+\tan^{-1}x=\tan^{-1}3$.Apply $\tan^{-1}a+\tan^{-1}b=\tan^{-1}\left(\frac{a+b}{1-ab}\right)$ to the first two terms.$\tan^{-1}\left(\frac{2x}{2-x^2}\right)+\tan^{-1}x=\tan^{-1}3$Again apply the same formula.$\tan^{-1}\left(\frac{\frac{2x}{2-x^2}+x}{1-\frac{2x^2}{2-x^2}}\right)=\tan^{-1}3$Simplify the fraction.$\tan^{-1}\left(\frac{4x-x^3}{2-3x^2}\right)=\tan^{-1}3$Since the tangents are equal, the arguments are equal.$\frac{4x-x^3}{2-3x^2}=3$This gives the cubic equation $x^3-9x^2-4x+6=0$.Checking the given condition $x<0$, the valid root is $x=-1$.Substitute into $500x^4+270x^2+997$.$500(-1)^4+270(-1)^2+997=500+270+997=1767$Answer:The value is $1767$, which corresponds to option B.