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If the normal to the curve $y=f(x)$ at $(1,2)$ make an angle $\frac{3 \pi}{4}$ with positive $X$-axis, then $f^{\prime}(1)=$
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We have,
$$
\begin{aligned}
y & =f(x) \\
\frac{d y}{d x} & =f^{\prime}(x) \\
\left(\frac{d y}{d x}\right)_{1,2} & =f^{\prime}(1)
\end{aligned}
$$
Slope of normal $=\frac{-1}{f^{\prime}(1)}$
$\therefore$ Normal of curve makes an angle $\frac{3 \pi}{4}$ with positive direction of $X$-axis
$$
\therefore \quad \tan \frac{3 \pi}{4}=\frac{-1}{f^{\prime}(1)} \Rightarrow-1=\frac{-1}{f^{\prime}(1)} \Rightarrow f^{\prime}(1)=1
$$
$$
\begin{aligned}
y & =f(x) \\
\frac{d y}{d x} & =f^{\prime}(x) \\
\left(\frac{d y}{d x}\right)_{1,2} & =f^{\prime}(1)
\end{aligned}
$$
Slope of normal $=\frac{-1}{f^{\prime}(1)}$
$\therefore$ Normal of curve makes an angle $\frac{3 \pi}{4}$ with positive direction of $X$-axis
$$
\therefore \quad \tan \frac{3 \pi}{4}=\frac{-1}{f^{\prime}(1)} \Rightarrow-1=\frac{-1}{f^{\prime}(1)} \Rightarrow f^{\prime}(1)=1
$$
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