VIN is active and VREF = 0, the circuit is drawn as:\({V_{OUT1}} = - \frac{{{R_F}}}{{{R_{IN}}}}{V_{IN}}\) When VIN = 0 and VREF is active, the circuit is drawn as:\({V_{OU{T_2}}} = \left( {\frac{{{V_{REF}} - {R_2}}}{{{R_1} + {R_2}}}} \right)\left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right)\)Total Vout = Vout1 + Vout2\({V_{out}} = \left( {\frac{{{V_{REF}} - {R_2}}}{{{R_1} + {R_2}}}} \right)\left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right) - \frac{{{R_E}}}{{{R_{IN}}}}\;{V_{IN}}\)Given VREF is fixed then V+ is fixed, i.e.\({V^ + } = \frac{{{V_{REF}} - {R_2}}}{{{R_1} + {R_2}}}\)\({V_{OUT}} = \left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right)\left( {\frac{{{V_{REF}}\;{R_2}}}{{{R_1} + {R_2}}}} \right) - \frac{{{R_F}}}{{{R_{IN}}}}{V_{IN}}\)For VIN = 0.1 V, we have VOUT = 1 V\(1 = \left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right)\left( {\frac{{{V_{REF}}\;{R_2}}}{{{R_1} + {R_2}}}} \right) - \frac{{{R_F}}}{{{R_{IN}}}}\left( {0.1} \right)\) ---(i)For VIN = 10 V, we have VOUT = 6 V\(6 = \left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right)\left( {\frac{{{V_{REF}}\;{R_2}}}{{{R_1} + {R_2}}}} \right) - \frac{{{R_F}}}{{{R_{IN}}}}\left( 1 \right)\) ---(ii)From (i)\(\left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right)\left( {\frac{{{V_{REF}} \cdot {R_2}}}{{{R_1} + {R_2}}}} \right) = 1 + \frac{{{R_F}}}{{{R_{IN}}}}\left( {0.1} \right)\)Putting above value in eq (ii)\(6 = 1 + \frac{{{R_F}}}{{{R_{IN}}}}\left( {0.1} \right) - \frac{{{R_F}}}{{{R_{IN}}}}\left( 1 \right)\)\(5 = \frac{{{R_F}}}{{{R_{IN}}}}\left( { - 0.9} \right)\)\(\frac{{{R_F}}}{{{R_{IN}}}} = - \frac{5}{{0.9}} = - 5.55\)Since only positive values are given in the OPTIONS:\(\frac{{{R_F}}}{{{R_{IN}}}} = 5.55\)