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

y = 1

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x = 1

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x + y = 1

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x - y = - 1

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Solution

The correct option is A $y = 1$
Given, $y = (2 x - 1) e^{2 (1 - x)}$
$∴ y^{'} (x) = - 2 (2 x - 1) e^{2 (1 - x)} + 2 e^{2 (1 - x)} = 2 e^{2 (1 - x)} (- 2 x + 2)$
Thus $y^{'} (x) = 0 \Rightarrow x = 1$
Since $y^{''} (1) < 0$ so $(1, 1)$ is the point of maximum and the equation of tangent is
$y - 1 = 0 (x - 1)$ ,i.e. $y = 1$

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List-I	List-II
a) Equation of tangent to the curve $y = b e^{- x / a}$ at $x = 0$	1) $x - 2 y = 2$
b) Equation of tangent to the curve $y = x^{2} + 1$ at $(1, 2)$	2) $y = 2 x$
c) Equation of normal to the curve $y = 2 x - x^{2}$ at $(2, 0)$	3) $x - y = π$
d) Equation of normal to the curve $y = sin x$ at $x = π$	4) $\frac{x}{a} + \frac{y}{b} = 1$

y = e^{- | x |}

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