dxdy=2x3−15x2+36x−19Since, slope is maximum so,
dx2d2y=6x2−30x+36=0
⇒x2−5x+6=0
x=2,3
at x=2
dx3d3y=12x−30
at x=2,dx3d3y<0
So, maxima
y=21×16−5×8+18×4−19×2
=8−40+72−38=80−78=2
point (2,2)
JEE Main 2021 — Mathematics Calculus
The maximum slope of the curve y=21x4−5x3+18x2−19x occurs at the point
Held on 26 Feb 2021 · Verified 6 Jul 2026.
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