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Expansion of (x^2 - y^2)

If f is def...

Question

If $f$ is defined in $[1, 3]$ by $f (x) = x^{3} + {b x}^{2} + a x,$ such that $f (1) - f (3) = 0$ and $f^{'} (c) = 0$ where $c = 2 + \frac{1}{\sqrt{3}},$ then $(a, b) =$

(- 6, 11)

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(2 - \frac{1}{\sqrt{3}}, 2 \frac{1}{\sqrt{3}})

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(11, - 6)

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(6, 11)

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Solution

The correct option is C $(11, - 6)$
$f^{1} (x) = 3 x^{2} + 2 b x + a$
$f^{1} (c) = 0$
${3 c}^{2} + 2 b c + a = 0$
$\Rightarrow a = 11, b = - 6$

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