A) 32
B) 16
C) 8
D) 10
Correct Answer: B
Solution :
Given \[(x-2a)\,(x-4a)\,(x-6a)\,(x-8a)+k{{a}^{4}}\] is a perfect square. \[=\,\left( x-2a \right)\,\left( x-8a \right)\,\left( x-4a \right)\,\left( x-6a \right)+k{{a}^{4}}\] \[=\,\,\left( {{x}^{2}}-2ax-8ax+16{{a}^{2}} \right)\] \[\left( {{x}^{2}}-6ax-4ax+24{{a}^{2}} \right)+k{{a}^{4}}\] \[=\left( {{x}^{2}}-10ax+16{{a}^{2}} \right)\,\left( {{x}^{2}}-10ax+24{{x}^{2}} \right)\] Let \[{{x}^{2}}-10\,ax=y\] \[\left( y+16{{a}^{2}} \right)\,\left( y+24{{a}^{2}} \right)+k{{a}^{4}}\] \[=\,\,\,{{y}^{2}}+24{{a}^{2}}y+16{{a}^{2}}y+384{{a}^{4}}+k{{a}^{4}}\] \[=\,\,\,{{y}^{2}}+40{{a}^{2}}y+384{{a}^{2}}+k{{a}^{4}}\] \[\left| {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \right|\] \[=\,\,\,{{y}^{2}}+2.20{{a}^{2}}.y+384{{a}^{4}}+k{{a}^{4}}\] Compare \[=a=y\Rightarrow b=20{{a}^{2}}\] \[{{b}^{2}}={{\left( 20{{a}^{2}} \right)}^{2}}=400{{a}^{4}}\] So, \[400{{a}^{4}}-384{{a}^{4}}=16{{a}^{4}}\Rightarrow \,\,\,\underline{\mathbf{k=16}}\]
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