Question

If the normal to the curve $y=f\left(x\right)$ of the point $(3,4)$ makes an angle $\raisebox{1ex}{$3\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ with the positive axis, then $f\text{'}\left(3\right)=$

• A. $-1$
• B. $\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$
• C. $\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$
• D. $1$

Answer 1

The correct option is D

$1$

The explanation for the correct answer.

Given: $y=f\left(x\right)$

The slope of the tangent at $(3,4)$ is $\frac{d}{dx}\left(f\right(x\left)\right)$

$\Rightarrow f\text{'}{\left(x\right)}_{(3,4)}=f\text{'}\left(3\right)$

The slope of the normal is $\frac{-1}{f\text{'}{\left(x\right)}_{(3,4)}}=\frac{-1}{f\text{'}\left(3\right)}$

Normal makes an angle $\frac{3\mathrm{\pi }}{4}$with the x-axis

Therefore slope is $\tan \left(\frac{3\mathrm{\pi }}{4}\right)$

$$\begin{gathered} \Rightarrow \frac{-1}{f\text{'}\left(3\right)}=\tan \left(\frac{3\mathrm{\pi }}{4}\right) \\ \Rightarrow \frac{-1}{f\text{'}\left(3\right)}=\tan \left(\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow =-\tan \frac{\mathrm{\pi }}{4} \\ \Rightarrow \frac{-1}{f\text{'}\left(3\right)}=-1 \\ \Rightarrow f\text{'}\left(3\right)=1 \end{gathered}$$

Hence option (D) is correct.