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Question 110 of 148

Q.If f(x)f(x) and g(x)g(x) are two functions as defined in Generalized law of mean then Lagrange's law of mean is a particular case of Generalised law of mean for :

(a) f′(x)=0f'(x) = 0
(b) g′(x)=0g'(x) = 0
(c) g(x)g(x) is an identity function
(d) f(x)f(x) is an identity function
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Lagrange's Mean Value Theorem is the special case of Cauchy's Generalised Mean Value Theorem when g(x)=xg(x)=x (the identity function).

  1. Cauchy's Generalised Mean Value Theorem states: if f,gf,g are continuous on [a,b][a,b], differentiable on (a,b)(a,b), and g′(x)≠0g'(x)\neq0, then there exists c∈(a,b)c\in(a,b) with f(b)−f(a)g(b)−g(a)=f′(c)g′(c)\dfrac{f(b)-f(a)}{g(b)-g(a)}=\dfrac{f'(c)}{g'(c)}.
  2. Lagrange's Mean Value Theorem states: there exists c∈(a,b)c\in(a,b) with f′(c)=f(b)−f(a)b−af'(c)=\dfrac{f(b)-f(a)}{b-a}.
  3. To recover Lagrange's form from Cauchy's form, we need g(b)−g(a)=b−ag(b)-g(a)=b-a and g′(c)=1g'(c)=1. …

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