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Question:

Let f(x) = xsin(πx), x > 0. Then for all natural numbers n, f'(x) vanishes at:

a unique point in the interval (n, n+1)

a unique point in the interval (n+1/2, n+1)

two points in the interval (n, n+1)

a unique point in the interval (n, n+1/2)

Solution:

The correct options are
B a unique point in the interval (n+1/2, n+1)
C a unique point in the interval (n, n+1)
Given that f(x) = xsin(πx) ⇒ f'(x) = sin(πx) + πxcos(πx)
f'(x) = 0 ⇒ sin(πx) + πxcos(πx) = 0
⇒ tan(πx) = -πx
⇒ πx ∈ ((2n+1)π/2, (n+1)π) ⇒ x ∈ (n+1/2, n+1) and also x ∈ (n, n+1).