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Question

Let g(x)=x0f(t)dt, where f is such that 12f(t)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
32<(2)52
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D
2<g(2)<4
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Solution

The correct option is C 0g(2)<2
Given, g(x)=x0f(t)dt

g(2)=20f(t)dt =10f(t)dt+21f(t)dt

Now, 12f(t) for t[0,1]

We get 1012dt10f(t)dt101dt1210f(t)dt1 ...(i)

Again, 0f(t)12 for t[1,2] ...(ii)

210dt21f(t)dt2112dt 021f(t)dt12

From Eqs.(i) and (ii), we get

1210f(t)dt+21f(t)dt32 12g(2)32 0g(2)<2.

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