If the normal to the curve y=f (x) at the point (3, 4) makes an angle 3 p/4 with the positive x-axis then f' (3) = (A) –1(B) –3/4 (C) 4/3

y =f(x)

Slope of the normal at (x,y) to the curve = -1/f'(x)

Given, -1/f'(x) = tan (3pi/4) =-1,

So, f'(3)=1.

Slope of normal to y = f(x) at (3,4) is -1/f'(3). Thus

-1 / f'(3) = tan (3∏ / 4 ) = tan(∏/2 + ∏/4) = -cot(∏/4) = -1

=> f'(3) = 1

Hope that helps :)