Definite Integration Question 345
Question: Let $ g(x)=\int_0^{x}{f(t)dt} $ where $ \frac{1}{2}\le f(t)\le 1,t\in [0,1] $ and $ 0\le f(t)\le \frac{1}{2} $ for $ t\in (1,2] $ , then
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Options:
A) $ -\frac{3}{2}\le g(2)<\frac{1}{2} $
B) $ 0\le g(2)<2 $
C) $ \frac{3}{2}<g(2)\le \frac{5}{2} $
D) $ 2<g(2)<4 $
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Answer:
Correct Answer: B
Solution:
$ g(2)=\int_0^{2}{f(t)dt=\int_0^{1}{f(t)dt+\int_1^{2}{f(t)dt}}} $
$ As\frac{1}{2}\le f(t)\le 1for0\le t\le 1, $
$ \int_0^{1}{\frac{1}{2}dt\le \int_0^{1}{f(t)dt\le \int_0^{1}{tdt}}} $ or $ \frac{1}{2}\le \int_0^{1}{f(t)dt\le 1} $ …..(i)
$ As0\le f(t)\le \frac{1}{2}for1<t\le 2, $
$ \int_1^{2}{0dt\le \int_1^{2}{f(t)dt\le \int_1^{2}{\frac{1}{2}dt}}} $
$ \therefore 0\le \int_1^{2}{f(t)dt\le \frac{1}{2}} $ …..(ii)
Adding (i) and (ii), $ 1/2\le g(2)\le 3/2 $
$ \therefore g(2) $ satisfies the inequality $ 0\le g(2)<2 $ .
BETA
