Question

P(x) is a polynomial function with real coefficients. Let a, b ∈ R with a < b, are two consecutive roots of the equation P(x) = 0, then show that there exists atleast one 'c' such that a ≤ c ≤ b and P'(c) + 100 P(c) = 0. 

Answer 1

Let ekuk f(x) = e^100x P(x), x∈ [a, b] 

f'(x) = e^100x p(x) + e^100x p(x) (100) 

(i) f(x) is continuous & differentiable 

(ii) f(a) = e^100a p(a) f(b) = e^100b p(b) 

⇒ By rolles theorem

 f'(x) = 0 

⇒ e^100c p'(c) + e^100c P(c) (100) = 0 

⇒ p'(c) + p(c) (100) = 0