If the normal to the curve $y=\mathrm{f}(x)$ at the point (3,4) makes an angle of $\left(\frac{3…
If the normal to the curve $y=\mathrm{f}(x)$ at the point (3,4) makes an angle of $\left(\frac{3 \pi}{4}\right)$ with the positive X -axis, then the value of $\mathrm{f}^{\prime}(3)$ is
-1
$-\frac{3}{4}$
$\frac{4}{3}$
1
Solution
$\begin{aligned} & \text { Slope of the normal }=\frac{-1}{\frac{\mathrm{~d} y}{\mathrm{~d} x}} \\ & \Rightarrow \tan \frac{3 \pi}{4}=\frac{-1}{\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)_{(3,4)}} \\ & \Rightarrow\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{(3,4)}=\mathrm{i} \\ & \Rightarrow \mathrm{f}^{\prime}(3)=1\end{aligned}$