Correct Answer - Option 1 : A

_{15} + A

_{14} + (A

_{13 }. A

_{12} + A̅

_{13 }. A̅

_{12})

Given address range: 1000 H to 2FFF

∴ Total No. of address lines = 13 i.e. \(\begin{array}{*{20}{c}} {2{\rm{FFF}}}\\ {\underline {1000} }\\ {1{\rm{FFF}}} \end{array}\)

(1FFF)_{H} = (0001 1111 1111 1111)_{2} = 13 address lines

i.e. \(\begin{array}{*{20}{c}} {{{\rm{A}}_{15}}}&{{{\rm{A}}_{14}}}&{{{\rm{A}}_{13}}}&{{{\rm{A}}_{12}}}&{{{\rm{A}}_{11}}}&{{{\rm{A}}_{10}}}& \cdots &{{{\rm{A}}_2}}&{{{\rm{A}}_1}}&{{{\rm{A}}_0}}&{}\\ 0&0&0&1&0&0&{}&0&0&0&{ = {{\left( {1000} \right)}_{\rm{H}}}}\\ {}&{}& \vdots &{}&{}& \vdots &{}&{}&{}&{}&{}\\ 0&0&1&0&1&1&{}&1&1&1&{ = {{\left( {2{\rm{FFF}}} \right)}_{\rm{H}}}} \end{array}\)

i.e. \(\begin{array}{*{20}{c}} {{{\rm{A}}_{15}}}&{{{\rm{A}}_{14}}}&{{{\rm{A}}_{13}}}&{{{\rm{A}}_{12}}}\\ 0&0&0&1\\ 0&0&1&0 \end{array}\)

To provide \(\overline {{\rm{CS}}}\) as low, the condition is

A_{15 }= A_{14} = 0 and A_{13} = A_{12} = (01) or (10)

i.e. A15 = A14 = 0 and A13 , A12 shouldn't be (11) , (00)

Thus it is A15 + A14 + (A 13 . A12 + A̅13 . A̅ 12)

Hence **option (1) is correct**