JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Application of Derivatives

For tangents parallel to the x-axis, we need $\frac{dy}{dx}=0$.

Differentiating the curve implicitly:



For horizontal tangent: $9y-2=0\Rightarrow y=2/9$.

Substituting into the original equation: $(2/9{)}^5-9x(2/9)+2x=0$, which gives $x=(2/9{)}^5/(-16)<0$. However, checking shows this doesn't satisfy the original equation properly, so $M=0$.

For tangents parallel to the y-axis, we need $5y^4-9x=0$.

Substituting $x=5y^4/9$ into the original equation:




This gives $y=0$ or $y=5/18$.

For $y=0$: $x=0$, giving point $(0,0)$.
For $y=5/18$: $x=5(5/18{)}^4/9$, giving another valid point.

Therefore, $N=2$ and $M+N=0+2=2$.