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Question

lf 0<a<c, 0<b<c then 0axbxcxdx=

A
logbclogac
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B
logalogblogc
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C
1logb/c1loga/c
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D
logaclogbc
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Solution

The correct option is C 1logb/c1loga/c

0(axcxbxcx)dx
=0(ac)xdx0(bc)xdx
=(ac)xlog(ac)0(bc)xlog(bc)0
=⎢ ⎢11log(ac)⎥ ⎥⎢ ⎢ ⎢11log(bc)⎥ ⎥ ⎥
=1log(bc)1log(ac)


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