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The inflection points on the graph of function y=0x(t1)(t2)2dt are

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a
x=–1
b
x=3/2
c
x=4/3
d
x=1

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detailed solution

Correct option is C

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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