Let f(x) and g(x) be differentiable functions having no common zeros so that f(x) g ′ (x) ≠ f ′ (x) g(x). Prove that between any two zeros of f(x), there exist atleast one zero of g(x).
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Let two consecutive zero of f(x) be a and b
f(a) = 0 = f(b). If possible, suppose g(x) has no zero. Define φ (x) = 
φ (x) satisfies conditions in Rolle’s theorem,
⇒ φ′ = 0 for at least one c ∈ (a, b)
⇒ f ′ g(c) – f(c) g ′ = 0
Which is a contradiction to given condition f(x) g ′ (x) ≠ f ′ (x) g(x)
Hence our supposition that g(x) has no zero is wrong
⇒ g(x) has at least one zero.
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems