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Question

Let g(x)=x0f(t)dt, where f is such that 12f(x)1 for tϵ[0,1] and 0f(t)12 for tϵ[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)<2
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Solution

The correct option is C 12g(2)32
If g(x)=x0f(t)dt

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

given, if t[0,1]then0.5f(t)10.5l11eqn(1)

and if t[1,2]then0f(t)0.50l20.5eqn(2)

adding eqn(1) and eqn(2), we get
0.5l1+l21.50.5g(2)1.5

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