MHTCET 10 May 2024 Shift 1 Paper

Concept:For a curve $y=f(x)$, the slope of the tangent is $\frac{dy}{dx}$, and the slope of the normal is $-\frac{1}{\frac{dy}{dx}}$.Explanation:Given curve: $y=x\log x$.Differentiate with respect to $x$:$$\frac{dy}{dx}=1+\log x$$This is the slope of the tangent.The given line is $2x-2y+3=0$, whose slope is:$$2x-2y+3=0 \Rightarrow y=x+\frac{3}{2}$$So the slope of the line is $1$.The normal is parallel to this line, hence the slope of the normal is $1$.Therefore, the slope of the tangent must be $-1$, since tangent and normal are perpendicular.$$1+\log x=-1$$$$\log x=-2$$$$x=e^{-2}$$Substitute $x=e^{-2}$ in $y=x\log x$:$$y=e^{-2}\log(e^{-2})=-2e^{-2}$$So the point of contact is $\left(e^{-2},-2e^{-2}\right)$.Equation of the normal passing through this point with slope $1$ is:$$y+2e^{-2}=1\left(x-e^{-2}\right)$$$$x-y=3e^{-2}$$Answer:Option B: $x-y=3e^{-2}$