Question:
The tangent to the curve, $y=x^{x^{2}}$ passing through the point (1,e) also passes through the point :
Correct Option: 1
Solution:
$y=x e^{x^{2}}$
$\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{(1, e)}=\left.\left(\mathrm{e} \cdot \mathrm{e}^{\mathrm{x}^{2}} \cdot 2 \mathrm{x}+\mathrm{e}^{\mathrm{x}^{2}}\right)\right|_{(1, \mathrm{e})}=2 \cdot \mathrm{e}+\mathrm{e}=3 \mathrm{e}$
$\mathrm{T}: \mathrm{y}-\mathrm{e}=3 \mathrm{e}(\mathrm{x}-1)$
$y=3 e x-3 e+e$
$y=(3 e) x-2 e$
$\left(\frac{4}{3}, 2 e\right)$ lies on it
Option (1)
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