If the function f(x) = x^3 - 6 x^2 + ax + b satisfies Rolle’s the… - Vidyadip

MCQ

If the function $f(x) = {x^3} - 6{x^2} + ax + b$ satisfies Rolle’s theorem in the interval $[1,\,3]$ and $f'\left( {{{2\sqrt 3 + 1} \over {\sqrt 3 }}} \right) = 0$, then $a =$ ..............

A
$- 11$
B
$- 6$
C
$6$
✓
$11$

✓

Answer

Correct option: D.

$11$

d
(d) $f(x) = {x^3} - 6{x^2} + ax + b$

==> $f'(x) = 3{x^2} - 12x + a$

==> $f'(c) = 0$ ==> $f'\left( {2 + \frac{1}{{\sqrt 3 }}} \right) = 0$

==> $3{\left( {2 + \frac{1}{{\sqrt 3 }}} \right)^2} - 12\left( {2 + \frac{1}{{\sqrt 3 }}} \right) + a = 0$

==> $3\left( {4 + \frac{1}{3} + \frac{4}{{\sqrt 3 }}} \right) - 12\left( {2 + \frac{1}{{\sqrt 3 }}} \right) + a = 0$

==> $12 + 1 + 4\sqrt 3 - 24 - 4\sqrt 3 + a = 0$ ==> $a = 11$.

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