Question

If the normal to the curve $y=f\left(x\right)$ at $(3,4)$ makes angle $\frac{3\pi }{4}$ with $\overline{OX}$ then $f^1\left(3\right)=$

• A. $-1$
• B. $1$
• C. $(-3/4)$
• D. $\left(4/3\right)$

Answer 1

The correct option is B $1$

$\because y=f\left(x\right)$

$y^{‘}=f^{‘}\left(x\right)$

Slope of tangent $=f^{‘}\left(x\right)$

Slope of normal $=-\frac{1}{f^{‘}\left(x\right)}$

$\Rightarrow \tan \frac{3\pi }{4}=-\frac{1}{f^{‘}\left(3\right)}$

$\Rightarrow -1=-\frac{1}{f^{‘}\left(3\right)}$

$\Rightarrow f^{‘}\left(3\right)=1$