Question

Let \(f\) and \(g\) be differentiable on the interval \(I\) and let \(a, b \in I, a<b\). Then
(A) If \(f(a)=0=f(b)\), the equation \(f^{\prime}(x)+f(x) g^{\prime}(x)=0\) is solvable in \((a, b)\).
(B) If \(f(a)=0=f(b)\), the equation \(f^{\prime}(x)+f(x) g^{\prime}(x)=0\) may not be solvable in \((a, b)\).
(C) If \(g(a)=0=g(b)\), the equation \(g^{\prime}(x)+k g(x)=0\) is solvable in \((a, b), k \in \mathbb{R}\)
(D) If \(g(a)=0=g(b)\), the equation \(g^{\prime}(x)+k g(x)=0\) may not be solvable in \((a, b), k \in \mathbb{R}\)

Answer 1

Ans : \((A, C)\)
Hint \(: f(a)=0=f(b) \quad \Rightarrow f^{\prime}(a) \cdot f^{\prime}(b)<0\)
let \(h(x)=f^{\prime}(x)+f(x) g^{\prime}(x) \quad h(a)=f^{\prime}(a), h(b)=f^{\prime}(b)\)
\(\therefore h(a) \cdot h(b)<0 \quad \Rightarrow h(x)=0\) has root(s) between \((a, b)\)
Similarly, \(g^{\prime}(x)+k g(x)=0\) has root(s) between \((a, b)\) as \(g(a)=0=g(b)\)