Question
It is given that the Rolle's theorem holds for the function $f(x) = x^3 + bx^2 + cx, \text{x}\in[1,2]$ at the point $\text{x}=\frac{4}{3},$ the values of b and c.

Answer

So, $f(1) = f(2)$
$\Rightarrow (1)^3 + b(1)^2 + c(1) = (2)^3 + b(2)^2 + c(2)$
$\Rightarrow 1 + b + c = 8 + 4b + 2c$
$\Rightarrow 3b + c + 7 = 0$ ....(i)
And $\text{f}'\Big(\frac{4}{3}\Big)=0$
$\Rightarrow3\Big(\frac{4}{3}\Big)^2+2\text{b}\Big(\frac{4}{3}\Big)+\text{c}=0$ [As, $f'(x) = 3x^2 + 2bx + c$]
$\Rightarrow\frac{16}{3}+\frac{8\text{b}}{3}+\text{c}=0$
$\Rightarrow8\text{b}+3\text{c}+16=0\ ....(\text{ii})$
(ii) - (i) × 3, we get
$8b - 9b + 16 - 21 = 0$
$\Rightarrow -b - 5 = 0$
$\Rightarrow b = -5$
Substituting $b = -5$ in (i), we get
$3(-5) + c + 7 = 0$
$\Rightarrow -15 + c + 7 = 0$
$\Rightarrow c = 8$

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