Question
$\int_{1/e}^e {|\log x|\,dx = } $
$ = [x - x\log x]_{1/e}^1 + [x\log x - x]_1^e$
$ = (1 - 0) - \left\{ {\frac{1}{e} - \frac{1}{e}( - 1)} \right\} + e - e + 1$
$ = 2 - \frac{2}{e} = 2\left( {1 - \frac{1}{e}} \right)$.
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$I.$ $P\,\left( {\frac{{{A^c}}}{{{B^c}}}} \right) = \frac{3}{4}$
$II.$ घटनायें $A$ तथा $B$ परस्पर अपवर्जी है
$III.$ $P\left( {\frac{A}{B}} \right) + P\left( {\frac{A}{{{B^c}}}} \right) = 1$