Mathematics - Area Under Curves Question with Solution | TestHub

MathematicsArea Under CurvesArea bounded by two curvesEasy2 minPYQ_2007

MathematicsEasysingle choice

The function $f (x) = x^{2} e^{- 2 x}, x > 0$ . Then the maximum value of $f (x)$ is

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$Given : f (x) = x^{2} e^{- 2 x}, x > 0 f^{'} (x) = x^{2} \cdot e^{- 2 x} (- 2) + e^{- 2 x} \cdot 2 x put f^{'} (x) = 0 \Rightarrow 2 e^{- 2 x} \cdot x (- x + 1) = 0 \Rightarrow x = 1 or x = 0 f^{''} (x) = (- 4 x^{2} - 6 x + 1) e^{- 2 x} f^{''} (1) = - 9 e^{- 2 x} < 0 f^{''} (0) = e^{- 2 x} > 0 ∴ value of f (x) is maximum at x = 1 ∵ f (x) = x^{2} \cdot e^{- 2 x} \Rightarrow f (1) = e^{- 2} = \frac{1}{e ^{2}}$

Stream:VITEEESubject:MathematicsTopic:Area Under CurvesSubtopic:Area bounded by two curves

⏱ 2mℹ️ Source: PYQ_2007

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