It is given that for the function $f (x) = x^{3} - 6 x^{2} + a x + b$ Rolle's theorem holds in $[1, 3]$ with $c = 2 + \frac{1}{\sqrt{3}}$ . Find the value of $^{'} a^{'}$ and $^{'} b^{'}$ .

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Solution

Given $f (1) = f (3) = 0$

$1^{3} - 6 \times 1 + a + b = 3^{3} - 6 \times 3^{2} + 3 a + b = 0$

$\Rightarrow a + b - 5 = 0$ and $3 a + b - 27 = 0$

on solving

$a + b = 5$
$3 a + b = 27$
___________
$- 2 a = - 22$
$∴ a = 11$ $b = - 6$

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Similar questions

f (x) = x^{3} + b x^{2} + a x + 5 o n [1, 3]

c = 2 + \frac{1}{\sqrt{3}}

f (x) = x^{3} + b x^{2} + a x, x \in [1, 3]

c = 2 + \frac{1}{\sqrt{3}}

2 + \frac{1}{\sqrt{3}} .

f (x) = x^{3} - 6 x^{2} + a x + b .

[1, 3]

c = 2 + \frac{1}{\sqrt{3}}

(a, b)

f (x) = x^{3} - 6 x^{2} + a x + b

[1, 3]

c = \frac{2 \sqrt{3} + 1}{\sqrt{3}}

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