Let $$f(x)=$$∫$$1x$$ e−$$t2/21$$−$$t2dt$$ then

Correct option is 2f has minimum at x = – 1

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Let $f (x) = \int_{1}^{x} e^{- t^{2} / 2} (1 - t^{2}) d t$ then

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f has maximum at x = 0

f has minimum at x = – 1

f has maximum at x = – 1

f has no critical point

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detailed solution

Correct option is B

f′(x)=e−x2/21−x2>0 for −1

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Let f(x)=∫1x e−t2/21−t2dt then

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Ready to Test Your Skills?

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Let $f (x) = \int_{1}^{x} e^{- t^{2} / 2} (1 - t^{2}) d t$ then