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Question

If I=x02t2[t]dt,xR+, where [] denotes the greatest integer function, then the value of I is

A
1ln2([x]+2{x}1)
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B
1ln2([x]+2{x})
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C
1ln2([x]2{x})
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D
1ln2([x]+2{x}+1)
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Solution

The correct option is A 1ln2([x]+2{x}1)
Let x[n,n+1), nI
Now,
I=x02t2[t]dtI=x02{t}dtI=n02{t}dt+xn2{t}dtI=n102{t}dt+xn2{t}dtI=n102tdt+xn2tndtI=n[2tln2]10+[12n2tln2]xnI=nln2+12nln2(2x2n)I=nln2+1ln2(2xn1)I=[x]+2{x}1ln2

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