If the Rolle's theorem holds for the func | vedclass

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Question

$f\left( x \right) = 2{x^3} + a{x^2} + bx$

let, $a=-1,b=1$

Given that $f\left( x \right)$ satisfy Roll theorem in interval $[-1,1]$

$f\l

Vedclass · Accepted Answer

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Vedclass · Answer

$f\left( x ight) = 2{x^3} + a{x^2} + bx$

let, $a=-1,b=1$

Given that $f\left( x ight)$ satisfy Roll theorem in interval $[-1,1]$

$f\left( x ight)$ must satify two conditions.

$(1)f(a)=f(b)$

$(2)f'(c)=0$

($c$ should be between $a$ and $b$ )

$f\left( a ight) = f\left( 1 ight) - 2{\left( 1 ight)^3} + a{\left( 1 ight)^2} + b\left( 1 ight) = 2 + a + b$

$f\left( b ight) = f\left( { - 1} ight) - 2{\left( { - 1} ight)^3} + a{\left( { - 1} ight)^2} + b\left( { - 1} ight) =  - 2 + a - b$

$f\left( a ight) = f\left( b ight)$

$2 + a + b =  - 2 + a - b$

$2b =  - 4$

$b =  - 2$

( given that $c = \frac{1}{2}$)

$f'\left( x ight) = 6{x^2} + 2ax + b$ 

at $x = \frac{1}{2},f'\left( x ight) = 0$

$0 = 6{\left( {\frac{1}{2}} ight)^2} + 2a\left( {\frac{1}{2}} ight) + b$

$\frac{3}{2} + a + b = 0$

$\frac{3}{2} + a - 2 = 0$

$a = 2 - \frac{3}{2} = \frac{1}{2}$

$2a + b = 2 	imes \frac{1}{2} - 2 = 1 -  - 1$

If the Rolle's theorem holds for the function $f(x) = 2x^3 + ax^2 + bx$ in the interval $[-1, 1 ]$ for the point $c = \frac{1}{2}$ , then the value of $2a + b$ is

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