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