A) \[y-1=0\]
B) \[x-1=0\]
C) \[x+y-1=0\]
D) \[x-y+1=0\]
Correct Answer: A
Solution :
\[y=(2x-1){{e}^{2(1-x)}}\] \[\Rightarrow \] \[\frac{dy}{dx}=2{{e}^{2(1-x)}}-2(2x-1){{e}^{2(1-x)}}\] \[=2{{e}^{2(1-x)}}(2-2x)\] \[=4{{e}^{2(1-x)}}(1-x)\] Put, \[\frac{dy}{dx}=0\Rightarrow x=1\] Now, \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=-8{{e}^{2(1-x)}}(1-x)-4{{e}^{2(1-x)}}\] \[\Rightarrow \] \[{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{x=1}}=-4<0\] So, y is maximum at\[x=1,\]when\[x=1,\text{ }y=1\] Thus, the point of maximum is (1,1). The equation of the tangent at (1,1) is \[y-1=0(x-1)\]or \[y=1\]
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