Question

If the normal to the curve $y=f\left(x\right)$ at the point $(3,4)$ make an angle $\frac{3\mathrm{\pi }}{4}$​ with the positive $x-$axis, then $f\prime \left(3\right)$ is

• A. $-1$
• B. $\frac{-3}{4}$
• C. $\frac{4}{3}$
• D. $1$

Answer 1

The correct option is D

$1$

Explanation for the correct answer:

The first order derivative of the curve $y=f\left(x\right)$ at a point gives the slope of the tangent to the curve at that point.

$\Rightarrow$$\frac{dy}{dx}=f\text{'}\left(x\right)$ is the slope of the tangent at a point on the curve.

$f\text{'}\left(3\right)$ is the slope of the tangent at point $(3,4)$

The normal makes $\frac{3\mathrm{\pi }}{4}$ angle with positive $x-$axis.

Hence, the slope of the normal$=\tan \left(\frac{3\mathrm{\pi }}{4}\right)=-1$

A normal is perpendicular to the tangent

Hence, the product of the slopes of the tangent and normal are negative reciprocals of each other

$\Rightarrow$slope of tangent $\times$slope of normal$=-1$

$\Rightarrow$ $f\text{'}\left(3\right)\times -1=-1$

$\Rightarrow$ $f\text{'}\left(3\right)=1$

Hence, option (D) is the correct answer.