Question

Let f(x) = √(x - 1) = √(x + 24 - 10√(x - 1)); 1 < x < 26 be real valued function. Then f'(x) for 1 < x < 26 is

(A) 0

(B) 1/√(x - 1)

(C) 2√(x - 1) - 5

(D) None of these 

Answer 1

Answer is (A) 0

From Rolle’s theorem in (1, 26), f(1) = f(26) = 5. In the given interval, the function satisfies all conditions of Rolle’s theorem. Therefore, in [1, 26], at least, there is a point for which f'(x) = 0.