If the normal to the curve `y=f(x)` at the point `(3,4)` makes an angle `(3pi)/4` with the positive x-axis, then `f'(3)=` (a) `-1` (b) `-3/4` (c) `4/3` (d) `1`

Options

A. `-1`
B. `-3//4`
C. `4//3`
D. `1`

Answer

D

Explanation

Slope of tangent of `y=f(x)` at `(3,4)` is
`((dy)/(dx))_((3,4))=f'(x)_((3,4))`
Therefore, slope of normal `=-1/(f'(x)_((3,4)))=-1/(f'(3))`
But `-1/(f'(3))=tan((3pi)/4)` [Given]
`implies(-1)/(f'(3))=tan((pi)/2+(pi)/4)=-"tan"(pi)/4=-1`
`impliesf'(3)=1`