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The value of 'C' in Lagrange's mean value theorem for $f (x) = {(x - a)}^{m} {(x - b)}^{n} in [a, b] is$

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$\frac{ma + nb}{m + n}$

$\frac{a + b}{m + n}$

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

$\frac{mb + na}{m + n}$

answer is A.

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Detailed Solution

$f (a) = 0 = f (b) f^{1} (x) = {(x - a)}^{m} n {(x - b)}^{n - 1} + m {(x - a)}^{m - 1} {(x - b)}^{n} f^{1} (r) = 0 \Rightarrow \frac{n}{x - b} + \frac{m}{x - a} = 0 \Rightarrow nx - na + mx - mb = 0 \Rightarrow x = \frac{mb + na}{m + n}$

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