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)

$y=(2x-1)e^{2(1-x)}$ $\Rightarrow$ $\frac{dy}{dx}=2e^{2(1-x)}-2(2x-1)e^{2(1-x)}$ $=2e^{2(1-x)}(2-2x)$ $=4e^{2(1-x)}(1-x)$ Put, $\frac{dy}{dx}=0\Rightarrow x=1$ Now, $\frac{d^{2}y}{d{x}^2}=-8e^{2(1-x)}(1-x)-4e^{2(1-x)}$ $\Rightarrow$ ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=1}=-4<0$ So, y is maximum at $x=1,$ when $x=1,y=1$ Thus, the point of maximum is (1,1). The equation of the tangent at (1,1) is $y-1=0(x-1)$ or $y=1$