The value of f’(x) is -1 at the point P on a continuous curve y = f(x). What is the angle which the tangent to the curve at P makes with the positive direction of x axis?

The correct option is 3π/4.

Let's understand why this is the correct answer step-by-step.

First, recall the geometric interpretation of the derivative of a function at a point. The derivative of a function $y=f\left(x\right)$ with respect to $x$ at a point $P$, denoted as $f^{\prime }\left(x\right)$ or $\frac{dy}{dx}$, represents the slope ($m$) of the tangent line to the curve at that point $P$.

Thus, we have:
$m=f^{\prime }\left(x\right)$

In this problem, we are given that the value of the derivative at point $P$ is $-1$:
$m=-1$

Next, let $\theta$ be the angle that the tangent line makes with the positive direction of the x-axis (measured counter-clockwise). By definition, the slope of a line is the tangent of this angle:
$m=\tan \left(\theta \right)$

Equating the two expressions for the slope $m$, we get:
$\tan \left(\theta \right)=-1$

We need to find the angle $\theta$ in the standard range $[0,\pi )$ for the inclination of a line. We know that:
$\tan \left(\frac{\pi }{4}\right)=1$

Since the tangent value is negative, the angle $\theta$ lies in the second quadrant. Using the trigonometric identity $\tan (\pi -x)=-\tan \left(x\right)$, we find:
$\tan \left(\pi -\frac{\pi }{4}\right)=-\tan \left(\frac{\pi }{4}\right)=-1$

Simplifying the angle, we get:
$\theta =\pi -\frac{\pi }{4}=\frac{3\pi }{4}$

Therefore, the angle which the tangent to the curve makes with the positive direction of the x-axis is $\frac{3\pi }{4}$.