Question

The equation of the tangent to the curve $ y=(2x-1){{e}^{2(1-x)}} $ at the points its maximum, is

• A. $ y-1=0 $
• B. $ x-1=0 $
• C. $ x+y-1=0 $
• D. $ x-y+1=0 $

Answer 1

Answer: (A)

$ 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 $