Verified Solution • 2025–26 Session moderate

QJEEM21Ma1229: JEE Main 2021 Mathematics - Class 12 JEEM21Ma1229 Solution: The maximum slope of the curve occurs at the point :

The maximum slope of the curve y=12x4−5x3+18x2−19xy = {1 \over 2}{x^4} - 5{x^3} + 18{x^2} - 19x occurs at the point :

Multiple Choice Options

  • A (3,212)\left( {3,{{21} \over 2}} \right)
  • B (0, 0)
  • C (2, 9)
  • D (2, 2)

Step-by-Step Verified Solution

Given, y=12x4−5x3+18x2−19xy = {1 \over 2}{x^4} - 5{x^3} + 18{x^2} - 19x

dydx=12×4x3−15x2+36x−19{{dy} \over {dx}} = {1 \over 2} \times 4{x^3} - 15{x^2} + 36x - 19

⇒\Rightarrow Slope M = 2x3−15x2+36x−192{x^3} - 15{x^2} + 36x - 19

At max of slope dMdx=0{{dM} \over {dx}} = 0

∴\therefore dMdx=6x2−30x+36=0{{dM} \over {dx}} = 6{x^2} - 30x + 36 = 0

⇒6(x2−5x+6)=0\Rightarrow 6({x^2} - 5x + 6) = 0

⇒6(x−2)(x−3)=0\Rightarrow 6(x - 2)(x - 3) = 0

∴\therefore x=2,3x = 2,3

Now, d2Mdx2=6(2x−5){{{d^2}M} \over {d{x^2}}} = 6(2x - 5)

at x = 2,{{{d^2}M} \over {d{x^2}}} = 6(4 - 5) = - 6 < 0

∴\therefore at x = 2 slope is maximum.

At x = 2,

y = 8 - 40 + 72 - 38 = 2

∴\therefore Required point = (2, 2)

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