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# The tangent to the curve $y=e^{2x}$ at the point $(0,1)$ meets x-axis at:

$\begin{array}{1 1}(A)\;(0,1) \\ (B)\;\bigg(\frac{-1}{2}\bigg) \\ (C)\;(2,0) \\ (D)\;(0,2) \end{array}$

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## 1 Answer

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Toolbox:
• Slope of a line is $ax+by+c=0$ is $- \bigg( \large\frac{ Coefficient\: of \: x}{ Coefficient\: of \: y} \bigg)$
• If $y = f(x),\: then \: \bigg( \large\frac{dy}{dx} \bigg)_p =$ slope of the normal to $y = f(x)$ at point $p$
Step 1
$y=e^{2x}$ is the equation of the curve.
On differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}=2e^{2x}$
The slope at the point $(0,1)$ is
$\large\frac{dy}{dx_{(0,1)}}=2.e^{\circ} \: \: \: ( \because e^{\circ}=1)$
$= 2$
The $y$ coordinate when the slope is 2 and when it meets the $x$ - axis.
$y=0$
Hence the required point is (2,0)
Hence the correct option is C
answered Aug 11, 2013

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