JEE MAIN 2023 LMVT

Question

If a function f(x) is continuous on [1, 3] and differentiable in (1, 3), then according to LMVT, there exists a point c∈ (1,3) such that f'(c) = (3)-(1) What is the value 3-1 of c for the function f(x) = x(x - 2)?

Answer 1

f(x) = x( x-2) in interval [1, 3]

f'(x) = 2x-2

a = 1 and b = 3

f(a) = -1 and f(b) = 3

(f(b) - f(a) }/{ b- a }  = {3 + 1 }/{3-1}  = 4/2 = 2

As per the LMVT statement, there is a point c ∈ (1, 3) such that f'(c) = [f(b) – f(a)]/ (b – a), i.e. f'(c) = 2

2c-2 = 2

c = 2

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