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Question

Let g(x)=x0f(t)dt, where f is such that 12f(x)1 for t[0,1] and 0f(t)12 for t[1,2] Then, g(2) satisfies the inequality

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

The correct option is C 12g(2)32
g(x)=x0f(t)dt=1/20f(t)dt+11/2f(t)dt
As, t[0,1]12f(t)1
1012dt10f(t)dt101dt
1210f(t)dt1...(1)
As, t[1,2]0f(t)12
210dt21f(t)dt2112dt
021f(t)dt12...(2)
Add equation (1) and (2)
1210f(t)dt+21f(t)dt32
12g(2)32



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