The maximum slope of the curve y=1/2x^4-5x^3+18x^2-19x occurs at the point

Correct answer: (B) 2,2. dy/dx=2x^3-15x^2+36x-19 Since, slope is maximum so, (d^2y)/(dx^2)=6x^2-30x+36=0 ⇒x^2-5x+6=0 x=2,3 at x=2 (d^3y)/(dx^3)=12x-30 at x=2,(d^3y)/(dx^3)<0 So, maxima y=1/2×16-5×8+18×4-19×2 =8-40+72-38=80-78=2 point 2,2

Q44 / 90 · JEE Main 2021 (26 Feb Shift 1) · Mathematics

The maximum slope of the curve $y = \frac{1}{2} x^{4} - 5 x^{3} + 18 x^{2} - 19 x$ occurs at the point

A
$(3, \frac{21}{2})$
Correct answer
B
$(2, 2)$
C
$(2, 9)$
D
$(0, 0)$

Solution

\frac{d y}{d x} = 2 x^{3} - 15 x^{2} + 36 x - 19

Since, slope is maximum so,

$\frac{d^{2} y}{d x^{2}} = 6 x^{2} - 30 x + 36 = 0$

$\Rightarrow x^{2} - 5 x + 6 = 0$

$x = 2, 3$

at $x = 2$

$\frac{d^{3} y}{d x^{3}} = 12 x - 30$

at $x = 2, \frac{d^{3} y}{d x^{3}} < 0$

So, maxima

$y = \frac{1}{2} \times 16 - 5 \times 8 + 18 \times 4 - 19 \times 2$

$= 8 - 40 + 72 - 38 = 80 - 78 = 2$

point $(2, 2)$

Hence, the correct answer is option: 'B'.

The maximum slope of the curve y=12x4-5x3+18x2-19x occurs at the point

Solution

Relevant Questions

The maximum slope of the curve $y = \frac{1}{2} x^{4} - 5 x^{3} + 18 x^{2} - 19 x$ occurs at the point