The point of extremum of f(x)=∫0x(t−2)2(t−1)dt is a

f(x)=∫0x (t-2)2 (t-1) dt f1(x)=(x-2)2 (x-1) (by Leibritz Rule) If f1(x)=0 ⇒ (x-2)2 (x-1)=0 ∴ At x=1, f(x) has minimum and at x=2, f(x) has neither maximum nor minimum exists

The point of extremum of f(x)=∫0x(t−2)2(t−1)dt is a

f(x)=∫0x (t-2)2 (t-1) dt f1(x)=(x-2)2 (x-1) (by Leibritz Rule) If f1(x)=0 ⇒ (x-2)2 (x-1)=0 ∴ At x=1, f(x) has minimum and at x=2, f(x) has neither maximum nor minimum exists

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The point of extremum of $f (x) = \int_{0}^{x} {(t - 2)}^{2} (t - 1) d t$ is a

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max at x=1

max at x=2

min at x=1

min at x=2

answer is C.

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Detailed Solution

$f (x) = \int_{0}^{x} {(t - 2)}^{2} (t - 1) dt f^{1} (x) = {(x - 2)}^{2} (x - 1) (by Leibritz Rule) If f^{1} (x) = 0 \Rightarrow {(x - 2)}^{2} (x - 1) = 0 ∴ At x = 1, f (x) has minimum and at x = 2, f (x) has neither maximum nor minimum exists$