If f (x) is differentiable in the interval [2, 5], where f(2)=15 and f(5)=12, then there exists a number c, 2 < c < 5 for which f ' (c) is equal to
Options:
A .  
12
B .  
15
C .  
110
D .  
7
Answer: Option C : C As f (x) is differentiable in [2 , 5], therefore, it is also continuos in [2, 5]. Hence, by mean value theorem, there exists a real number c in (2, 5) such that f′(c)=f(5)−f(2)5−2⇒f′(c)=12−153=110. Hence (c) is the correct answer.
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