If J=1∫0x1+x8dx
Consider the following statements I)J>14II)J>π4
Then
A
Only I is true
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B
Only II is true
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C
Both I and II are true
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D
Neither I nor II is true
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Solution
The correct option is A Only I is true J=1∫0x1+x8dx∵0<x8<1
For x∈(0,1), we have 1>x8>0⇒2>1+x8>1⇒12<11+x8<1⇒x2<x1+x8<x
Now, 1∫0x2dx<1∫0x1+x8dx<1∫0xdx⇒14<J<12<π4