A) 4
B) 2
C) 3
D) 1
Correct Answer: D
Solution :
[d] \[\because \,\,\,\,\,\,\,\,\,\,\,({{e}^{4x}}-2{{e}^{2x}}+1)+({{e}^{3x}}-2{{e}^{2x}}+{{e}^{x}})=0\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,{{({{e}^{2x}}-1)}^{2}}+{{e}^{x}}{{({{e}^{x}}-1)}^{2}}=0\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,{{({{e}^{x}}-1)}^{2}}\underbrace{[{{({{e}^{x}}+1)}^{2}}+{{e}^{x}}]}_{Always\text{ }positive\text{ }terms}=0\] Hence \[{{e}^{x}}-1=0\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=0\] is the only solution
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