The curve $y - e^{x y} + x = 0$ has a vertical tangent at

(1, 1)

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(0, 1)

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(1, 0)

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(0, 0)

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Solution

The correct option is D $(1, 0)$
Given, $y - e^{x y} + x = 0$
Differentiating w.r.t $x$
$\frac{d y}{d x} - e^{x y} (y + x \frac{d y}{d x}) + 1 = 0$
$\Rightarrow \frac{d y}{d x} = \frac{1 - y e^{x y}}{x e^{x y} - 1}$
Thus for vertical tangent, $\frac{d y}{d x} = \infty$
$\Rightarrow x e^{x y} - 1 = 0$
Hence required point is $(1, 0)$

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