The value of c of Lagrange's mean value theorem for f (x )= 25-x^2 on [1 ,5 ] is

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Correct answer: A. (a) : According to lagrange's mean value theorem for any function f (x ) ,a<x<b, there exist c ∈ (a ,b ) such that f^ ' (c )= f (b )-f (a )b-a Here, a=1 ,b=5 and f (x )= 25-x^2 ∴f (5 )= 25-25=0 And f (1 )= 25-1= 24=2 6 ∴f^ ' (c )= 0-2 65-1 ,c ∈ (1 ,5 )= -2 64 ⇒f^ ' (c )- 62 Now f (x )= 25-x^2 ⇒f^ ' (x )= 12 25-x^2 (-2x )= -x 25-x^2 ⇒f^ ' (c )= -c 25-c^2 ...(ii) = log_10 ⁡e log_e ⁡x= (0.4343 ) log ⁡x ⇒f^ ' (x )= 0.4343x On equating (i) and (ii), we get -c 25-c^2= - 62 ⇒2c= 150-6c^2 ⇒4c^2=150-6c^2 ⇒10c^2=150 ⇒c^2=15 ⇒c= 15 Since, c ∈ (1 ,5 ), So, c= 15

$The value of c of Lagrange's mean value theorem for f (x) = 25 - x^{2} on [1, 5] is$

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