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Question

Let g(x)=x0f(t)dt, where 12f(t)1 for t[0,1] and 0f(t)12 for t(1,2]. Then

A
32g(2)<12
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B
0g(2)<2
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C
32<g(2)52
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D
2<g(2)<4
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Solution

The correct option is B 0g(2)<2
Given g(x)=x0f(t)dt
g(2)=20f(t)dt=10f(t)dt+21f(t)dt
Now 12f(t)1 for t[0,1]
We get 1012dt10f(t)dt101dt1210f(t)dt1 ...(1)
Again 0f(t)12 for t[1,2]
110dt21f(t)dt2112dt021f(t)dt12 ...(2)
From (1) and (2)
1210f(t)dt+21f(t)dt3212g(2)320g(2)<2

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