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5. Continuity and Differentiation
medium
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
$6$
D
$11$
Solution
(d) $f(x) = {x^3} – 6{x^2} + ax + b$
==> $f'(x) = 3{x^2} – 12x + a$
==> $f'(c) = 0$ ==> $f'\left( {2 + \frac{1}{{\sqrt 3 }}} \right) = 0$
==> $3{\left( {2 + \frac{1}{{\sqrt 3 }}} \right)^2} – 12\left( {2 + \frac{1}{{\sqrt 3 }}} \right) + a = 0$
==> $3\left( {4 + \frac{1}{3} + \frac{4}{{\sqrt 3 }}} \right) – 12\left( {2 + \frac{1}{{\sqrt 3 }}} \right) + a = 0$
==> $12 + 1 + 4\sqrt 3 – 24 – 4\sqrt 3 + a = 0$ ==> $a = 11$.
Standard 12
Mathematics
