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The value of C in Lagrange’s mean value theorem for $f (x) = l x^{2} + m x + n, (l \neq .0) [a, b] is$

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$\frac{a + b}{2}$

$\frac{a}{2}$

$\frac{a - b}{2}$

$\frac{b}{2}$

answer is D.

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Detailed Solution

$\begin{aligned} f^{1} (c) = \frac{f (b) - f (a)}{b - a} \\ 2 l c + m = \frac{l (b^{2} - a^{2}) + m (b - a)}{b - a} \\ c = \frac{a + b}{2} \end{aligned}$

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