JEE Mains · Maths · STD 12 - 6. Application of derivatives
Let a function \(f:[0,5] \rightarrow \mathrm{R}\) be continuous. \(f(1)=3\) and \(\mathrm{F}\) be defined as \(\mathrm{F}(\mathrm{x})=\int\limits_{1}^{\mathrm{x}} \mathrm{t}^{2} \mathrm{g}(\mathrm{t}) \mathrm{dt},\) where \(\mathrm{g}(\mathrm{t})=\int\limits_{1}^{\mathrm{t}} \mathrm{f}(\mathrm{u}) \mathrm{du}\) Then for the function \(\mathrm{F}\), the point \(\mathrm{x}=1\) is
- A a point of local minima
- B not a critical point
- C a point of inflection
- D a point of local maxima
Answer & Solution
Correct Answer
(A) a point of local minima
Step-by-step Solution
Detailed explanation
\(\mathrm{F}^{\prime}(\mathrm{x})=\mathrm{x}^{2} \mathrm{g}(\mathrm{x})=\mathrm{x}^{2} \int_{1}^{\mathrm{x}} \mathrm{f}(\mathrm{u}) \mathrm{du} \Rightarrow \mathrm{F}^{\prime}(1)=0\)…
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