$f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} 3 & x = 0 - x^{2} + 3 x + a & 0 < x < 1 m x + b & 1 \leq x \leq 2 \end{matrix}$ satisfies the hypothesis of the Langrange's theorem then the value of $(a + b) m$ is

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Solution

$A c c o r d i n g t o l a g r a n g e^{'} s t h e o r e m a c o n t i n u o s f u n c t i o n w i l l h a v e i t^{'} s d e r i v a t e f u n c t i o n w i l l a l s o b e c o n t i n u o u s . s o f o r c o n t i n u i t y o f f (x) f (0) = 3 = a f (1) = - 1 + 3 + 3 = m + b m + b = 5 a n d i t^{'} s d e r i v a t i v e f u n c t i o n s h o u l d a l s o b e c o n t i n u o u s . f (x) = - 2 x + 30 < x < 1 = m 1 < x < 2 f^{'} (1) = - 2 + 3 = m m = 1 s o b = 4 m (a + b) = 1. (3 + 4) = 7$

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