WBJEE · Maths · Continuity and Differentiability
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 & Solution
Correct Answer
(C) If \(g(a)=0=g(b)\), the equation \(g^{\prime}(x)+k g(x)=0\) is solvable in \((a, b), k \in \mathbb{R}\)
Step-by-step Solution
Detailed explanation
From option (a) We have, \(f(a)=f(b)=0\) \(\Rightarrow f^{\prime}(a) \cdot f^{\prime}(b) < 0\) Again, let \(h(x)=f^{\prime}(x)+f(x) g^{\prime}(x)\) \(\Rightarrow h(a)=f^{\prime}(a)+fg^{\prime}(a)=f^{\prime}(a)\) and \(h(b)=f^{\prime}(b)+f(b) g^{\prime}(b)=f^{\prime}(b)\)…
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