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# Solution for If from Lagrange'S Mean Value Theorem, We Have F ′ ( X 1 ) = F ′ ( B ) − F ( a ) B − a , Then (A) a < X1 ≤ B (B) a ≤ X1 < B (C) a < X1 < B (D) a ≤ X1 ≤ B - CBSE (Commerce) Class 12 - Mathematics

ConceptMaximum and Minimum Values of a Function in a Closed Interval

#### Question

If from Lagrange's mean value theorem, we have $f' \left( x_1 \right) = \frac{f' \left( b \right) - f \left( a \right)}{b - a}, \text { then }$

(a) a < x1 ≤ b
(b) a ≤ x1 < b
(c) a < x1 < b
(d) a ≤ x1 ≤ b

#### Solution

(c) a < x1 < b

In the Lagrange's mean value theorem,$c \in \left( a, b \right)$ such that  $f'\left( c \right) = \frac{f\left( b \right) - f\left( a \right)}{b - a}$.

So, if there is $x_1$ such that

$f'\left( x_1 \right) = \frac{f\left( b \right) - f\left( a \right)}{b - a}$ then $x_1 \in \left( a, b \right)$.

$\Rightarrow a < x_1 < b$

Is there an error in this question or solution?

#### APPEARS IN

Solution for question: If from Lagrange'S Mean Value Theorem, We Have F ′ ( X 1 ) = F ′ ( B ) − F ( a ) B − a , Then (A) a < X1 ≤ B (B) a ≤ X1 < B (C) a < X1 < B (D) a ≤ X1 ≤ B concept: Maximum and Minimum Values of a Function in a Closed Interval. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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