JEE MAIN - Mathematics (2021 - 26th February Morning Shift - No. 12)
The maximum slope of the curve $$y = {1 \over 2}{x^4} - 5{x^3} + 18{x^2} - 19x$$ occurs at the point :
$$\left( {3,{{21} \over 2}} \right)$$
(0, 0)
(2, 9)
(2, 2)
Explanation
Given, $$y = {1 \over 2}{x^4} - 5{x^3} + 18{x^2} - 19x$$
$${{dy} \over {dx}} = {1 \over 2} \times 4{x^3} - 15{x^2} + 36x - 19$$
$$ \Rightarrow $$ Slope M = $$2{x^3} - 15{x^2} + 36x - 19$$
At max of slope $${{dM} \over {dx}} = 0$$
$$ \therefore $$ $${{dM} \over {dx}} = 6{x^2} - 30x + 36 = 0$$
$$ \Rightarrow 6({x^2} - 5x + 6) = 0$$
$$ \Rightarrow 6(x - 2)(x - 3) = 0$$
$$ \therefore $$ $$x = 2,3$$
Now, $${{{d^2}M} \over {d{x^2}}} = 6(2x - 5)$$
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)
$${{dy} \over {dx}} = {1 \over 2} \times 4{x^3} - 15{x^2} + 36x - 19$$
$$ \Rightarrow $$ Slope M = $$2{x^3} - 15{x^2} + 36x - 19$$
At max of slope $${{dM} \over {dx}} = 0$$
$$ \therefore $$ $${{dM} \over {dx}} = 6{x^2} - 30x + 36 = 0$$
$$ \Rightarrow 6({x^2} - 5x + 6) = 0$$
$$ \Rightarrow 6(x - 2)(x - 3) = 0$$
$$ \therefore $$ $$x = 2,3$$
Now, $${{{d^2}M} \over {d{x^2}}} = 6(2x - 5)$$
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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