First slide

Mean value theorems

Question

Rolle’s theorem holds for the function $f (x) = x^{3} + b x^{2} + c x, 1 \leq x \leq 2$ at the point $\frac{4}{3}$ , then the value of $100 c - 500 b$ must be

Difficult

A

3300
B

4400
C

3400
D

2400

Solution

Now, $f (1) = f (2)$
$\begin{matrix} \Rightarrow 1 + b + c = 8 + 4 b + 2 c \\ \Rightarrow 3 b + c = - 7 ..... (i) \end{matrix}$

And $\begin{matrix} f' (\frac{4}{3}) = 0 \Rightarrow 3 {(\frac{4}{3})}^{2} + 2 b (\frac{4}{3}) + c = 0 \\ \Rightarrow 16 + 8 b + 3 c = 0 ..... (i i) \end{matrix}$

From Eqs. (i) and (ii), we get

$\begin{matrix} b = - 5, c - 8 \\ \Rightarrow 100 c - 500 b = 800 + 2500 = 3300 \end{matrix}$

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