The inflection points on the graph of function y=∫0x (t−1)(t−2)2dt are

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x=–1

x=3/2

x=4/3

x=1

answer is C.

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Detailed Solution

dydx=(x−1)(x−2)2 so d2ydx2=(x−2)(3x−4).The points of inflection are given by d2ydx2=0 so x=2, x=4/3 are points of inflection.

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