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The value of 'c' in Rolle's theorem forin (a, b) where a>0 isx²+abf(x) = logx(a + b)1) A.M. of a, b2) G.M. of a, b1 14)a b3) H.M. of a, b |
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Answer» Step-by-step explanation: We have, The function f(x)=log (a+b)x x +ab
in[a,b] Now, x∈(a,b) ∴ (a+b)x x 2 +ab
∈R +ve
We know that, (1). log with the positive quantities as it domain is CONTINUOUS. ∴f(x) is continuous on [a,b]. (2).f(x)=log(x 2 +ab)−log(a+b)x =log(x 2 +ab)−log(a+b)x−logx On differentiation and we get, f ′ (x)= x 2 +ab 2x
−0− x 1
∵x∈(a,b) So, f ′ (x) is EXISTS on (a,b). (3).f(a)=log( (a+b)a a 2 +ab
)=log(1)=0 f(b)=log( (a+b)b b 2 +ab
)=log(1)=0 ∴f(a)=f(b) Then, There exists at LEAST one real. c∈(a,b) such that f ′ (c)=0. Now, f ′ (c)=0 ⇒ c 2 +ab 2c
− c 1
=0 ⇒ c 2 +ab 2c
= c 1
⇒2c 2 =c 2 +ab ⇒c 2 =ab ⇒c= ab
∈R +ve
Hence, this is the answer. |
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