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Let $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function. If $\mathrm{px}+\mathrm{my}+\mathrm{n}=0$ is a tangent drawn to the curve $y=f(x)$ at $x=\alpha$, then at $x$ $=0, \frac{d}{d x}\left(f\left(\alpha e^{2 x}\right)\right)=$
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Verified Answer
The correct answer is:
$\frac{-2 p \alpha}{m}$
Since $p x+m y+n=0$ is tangent drawn to the curve $y=f(x)$ at $x=\alpha$.
Hence $\frac{d y}{d x}=f^{\prime}(x)=($ Slope of $p x+m y+n=0)$
$$
\Rightarrow \quad \frac{d f(\mathrm{x})}{d x}=\left(\frac{-p}{m}\right) \Rightarrow f(x)=\frac{-p}{m} x+c
$$
when $x=\alpha e^2 x$ then (where $\mathrm{c}=$ arbitrary constant)
$$
f\left(\alpha e^{2 x}\right)=\frac{-p}{m}\left(\alpha e^{2 x}\right)+c
$$
Hence $\frac{d}{d x}\left[f\left(\alpha e^{2 x}\right)\right]=-\frac{p \alpha}{m} \frac{d}{d x}\left(e^{2 x}\right)+\frac{d c}{d x}$
Hence $\frac{d}{d x}\left[f\left(\alpha e^{2 x}\right)\right]=\frac{-2 p \alpha}{m} e^{2 x}$
when $x=0$, then
$$
\frac{d}{d x}\left[f\left(\alpha e^{2 x}\right)\right]=\frac{-2 p \alpha}{m} e^0=\frac{-2 p \alpha}{m}
$$
Hence $\frac{d y}{d x}=f^{\prime}(x)=($ Slope of $p x+m y+n=0)$
$$
\Rightarrow \quad \frac{d f(\mathrm{x})}{d x}=\left(\frac{-p}{m}\right) \Rightarrow f(x)=\frac{-p}{m} x+c
$$
when $x=\alpha e^2 x$ then (where $\mathrm{c}=$ arbitrary constant)
$$
f\left(\alpha e^{2 x}\right)=\frac{-p}{m}\left(\alpha e^{2 x}\right)+c
$$
Hence $\frac{d}{d x}\left[f\left(\alpha e^{2 x}\right)\right]=-\frac{p \alpha}{m} \frac{d}{d x}\left(e^{2 x}\right)+\frac{d c}{d x}$
Hence $\frac{d}{d x}\left[f\left(\alpha e^{2 x}\right)\right]=\frac{-2 p \alpha}{m} e^{2 x}$
when $x=0$, then
$$
\frac{d}{d x}\left[f\left(\alpha e^{2 x}\right)\right]=\frac{-2 p \alpha}{m} e^0=\frac{-2 p \alpha}{m}
$$
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