Question

If f(x) and g(x), 0 ≤ x ≤ 1 be differentiable functions such that f(0) = 2 g(0) = 0, f(1) = 6, g(1) = 2 then in (0,1)

• A. f '(x) = 0 ∀x ∈ R
• B. f ' (x) = 2g'(x), for at least one value of x in (0,1)
• C. 0
• D. None of these

Answer 1

Answer: (B)

f ' (x) = 2g'(x), for at least one value of x in (0,1)

Answer 2

By LMVT; f '(x) = $\frac{f(1)–f(0)}{1 - 0}$= 6 – 2 = 4

g'(x) $\frac{g(1)–g(0)}{1 - 0}$= $\frac{2–0}{1}$= 2

⇒ f ' (x) = 2g' (x) for at least one value of x in (0,1)