MHTCET 18 Apr 2026 Shift 1 Paper

Concept:Use $\log_e x = t$ and combine the two $\tan^{-1}$ terms using their addition formula. If the combined expression is constant, its derivative is zero.Explanation:Let $t = \log_e x = \ln x$. Since $x \in \left(e^{-1/3}, e^{1/12}\right)$, we get $-\frac13 < t < \frac1{12}$.Using logarithm rules:$$y = \tan^{-1}\left(\frac{1-3t}{1+3t}\right) + \tan^{-1}\left(\frac{4+3t}{1-12t}\right).$$Let $u = \frac{1-3t}{1+3t}$ and $v = \frac{4+3t}{1-12t}$.Both $u$ and $v$ are positive in the given interval.Now,$$\tan y = \frac{u+v}{1-uv}.$$Compute:$$u+v = \frac{5(1+9t^2)}{(1+3t)(1-12t)},$$and$$1-uv = \frac{-3(1+9t^2)}{(1+3t)(1-12t)}.$$Therefore,$$\tan y = -\frac{5}{3},$$which is a constant.Hence $y$ is constant throughout the interval, so$$\frac{dy}{dx} = 0.$$Answer:$0$   i.e. Option B.