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Question

Let g(x)=x0f(t) dt where 12f(t)1,tϵ[0,1] and 0f(t)12 for tϵ(1,2), then [IIT Screening 2000]


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

g(2)=20f(t)dt=10f(t)dt+21f(t)dt
As 12f(t)1 for 0t1,
1012dt10f(t)dt10t dt or 1210f(t) dt1(i)
As 0f(t)12 for 1<t2,210 dt21f(t) dt2112 dt
021f(t) dt12(ii)
Adding (i) and (ii), 12g(2)32
g(2) satisfies the inequality 0g(2)<2


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