If the normal to the curve y = f(x) at the point (3, 4) makes an angle $\frac{3 \pi}{4}$ with the positive x-axis, then f'(3) =
Answer & explanation
Correct answer: option 4
We have,
$y=f(x)$
$\Rightarrow \frac{d y}{d x}=f'(x) \Rightarrow\left(\frac{d y}{d x}\right)_{(3,4)}=f'(3) \Rightarrow-\frac{1}{\left(\frac{d y}{d x}\right)_{(3,4)}}=-\frac{1}{f'(3)}$
It is given that the slope of the normal $=\tan \frac{3 \pi}{4}$
∴ $\tan \frac{3 \pi}{4}=-\frac{1}{f'(3)} \Rightarrow f'(3)=1$