The tangent to the curve, $y = x e^{x^{2}}$ passing through the point $(1, e)$ also passes through the point:

(\frac{4}{3}, 2 e)

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(2, 3 e)

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(\frac{5}{3}, 2 e)

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(3, 6 e)

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Solution

The correct option is B $(\frac{4}{3}, 2 e)$
$y = x e^{x^{2}}$

$\frac{d y}{d x} {∣ ∣_{(1, e)} = (x \cdot e^{x^{2}} \cdot 2 x + e^{x^{2}}) ∣ ∣}_{1, e} = 2 \cdot e + e = 3 e$

$T : y - e = 3 e (x - 1)$

$y = 3 e x - 3 e + e$

$y = (3 e) x - 2 e$
Out of the options only option:A

$(\frac{4}{3}, 2 e)$ lies on it as $2 e = 3 e \times \frac{4}{3} - 2 e \Rightarrow 2 e = 2 e$

Suggest Corrections

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