First slide

Tangents and normals

Question

$The curve x + y - \log_{e} (x + y) = 2 x + 5 has a vertical tangent at the point (α, β) . Then α + β =$

Moderate

A

-1
B

1
C

2
D

-2

Solution

$\begin{array}{l} x + y - \log_{e} (x + y) = 2 x + 5 \\ Differentiate both sides with respect to x \\ \Rightarrow 1 + y' - \frac{1}{x + y} \cdot (1 + y') = 2 \\ \Rightarrow (1 + y') (x + y) - (1 + y') = 2 x + 2 y \\ \Rightarrow x + y + y' (x + y - 1) - 1 = 2 x + 2 y \\ y^{'} = \frac{x + y + 1}{x + y - 1} \\ \sin c e s l o p e o f v e r t i c a l \tan g e n t i s u n d e f i n e d, d e n o m i n a t o r = 0 \\ x + y - 1 = 0 \Rightarrow α + β - 1 = 0 \\ \Rightarrow α + β = 1 \end{array}$

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