Question

Let $f: (a,b) \to \mathbb{R}$ be twice differentiable function such that $f(x) = \int_a^x g(t)dt$ for a differentiable function $g(x)$. If $f(x)=0$ has exactly five distinct roots in $(a,b)$, then $g(x)g'(x)=0$ has at least :

• A. three roots in (a, b)
• B. five roots in (a, b)
• C. seven roots in (a, b)
• D. twelve roots in (a, b)

Hint:

Rolle's Theorem is a powerful tool for finding the number of roots of derivatives. If a function $h(x)$ has $n$ roots, its derivative $h'(x)$ must have at least $n-1$ roots located between the roots of $h(x)$.

Answer 1

The Correct Option is C

Given: \[ f(x) = \int_a^x g(t)\,dt \Rightarrow f'(x)=g(x), \quad f''(x)=g'(x) \] It is given that \( f(x)=0 \) has exactly five distinct roots in \( (a,b) \). Let them be: \[ a < c_1 < c_2 < c_3 < c_4 < c_5 < b \] Step 1: Zeros of \( g(x) \) Between each pair of consecutive roots of \( f(x) \), Rolle’s theorem guarantees at least one root of \( f'(x)=g(x) \). Hence, \( g(x)=0 \) has at least: \[ 5 - 1 = 4 \text{ distinct roots} \] Step 2: Zeros of \( g'(x) \) Between consecutive roots of \( g(x) \), Rolle’s theorem again guarantees roots of \( g'(x) \). Thus, \( g'(x)=0 \) has at least: \[ 4 - 1 = 3 \text{ distinct roots} \] Step 3: Zeros of \( g(x)g'(x)=0 \) The equation \( g(x)g'(x)=0 \) is satisfied when either: \[ g(x)=0 \quad \text{or} \quad g'(x)=0 \] Total minimum number of distinct roots: \[ 4 + 3 = 7 \] \[ \boxed{\text{At least } 7 \text{ roots}} \]