MHTCET 18 Apr 2026 Shift 2 Paper

Concept:Simplify the argument of $\cot^{-1}$ using logarithm rules, then apply the inverse tangent addition identity and differentiate.Explanation:Start with the given expression:$$y=\cot^{-1}\left(\frac{x-\log x^{x^{2}}}{\log e^{x^{2}}+\log x^{x}}\right)$$Use logarithm properties:$$\log x^{x^{2}}=x^{2}\log x,\quad \log e^{x^{2}}=x^{2},\quad \log x^{x}=x\log x$$Substitute these into the expression:$$y=\cot^{-1}\left(\frac{x-x^{2}\log x}{x^{2}+x\log x}\right)$$Factor $x$ from numerator and denominator. Since $x>0$, cancel $x$:$$y=\cot^{-1}\left(\frac{1-x\log x}{x+\log x}\right)$$Let $a=x$ and $b=\log x$. Given $x\log x<1$, we have $ab<1$. Therefore,$$\tan^{-1}x+\tan^{-1}(\log x)=\tan^{-1}\left(\frac{x+\log x}{1-x\log x}\right)$$Thus,$$\cot\left[\tan^{-1}x+\tan^{-1}(\log x)\right]=\frac{1-x\log x}{x+\log x}$$Hence,$$y=\tan^{-1}x+\tan^{-1}(\log x)$$Differentiate with respect to $x$:$$\frac{dy}{dx}=\frac{1}{1+x^{2}}+\frac{1}{1+(\log x)^{2}}\cdot\frac{1}{x}$$$$=\frac{1}{1+x^{2}}+\frac{1}{x\left[1+(\log x)^{2}\right]}$$Answer:$$\frac{dy}{dx}=\frac{1}{1+x^{2}}+\frac{1}{x\left[1+(\log x)^{2}\right]}$$This matches Option B.