The value of b for which the function f (x) = sin x – bx + c is decreasing for x ∈ R is given by

The correct option is b > 1.

To determine the range of values for $b$ such that the function $f\left(x\right)=sinx-bx+c$ is decreasing for all real numbers $x\in R$, we need to analyze its first derivative.

A differentiable function is strictly decreasing on an interval if its first derivative is less than zero for all points in that interval. Thus, we require:
$$f^{\prime }\left(x\right)<0\text{for all}x\in R$$

First, let's find the derivative of the given function $f\left(x\right)$ with respect to $x$:
$$f^{\prime }\left(x\right)=\frac{d}{dx}(sinx-bx+c)$$
$$f^{\prime }\left(x\right)=cosx-b$$

For the function to be decreasing for all $x\in R$, we must satisfy the inequality:
$$cosx-b<0$$
which simplifies to:
$$cosx<b$$

We know that the range of the cosine function, $cosx$, for any real number $x$ is bounded by:
$$-1\le cosx\le 1$$

For the condition $b>cosx$ to hold true for every possible value of $x$, the value of $b$ must be strictly greater than the maximum possible value of $cosx$.

Since the maximum value of $cosx$ is $1$, we conclude that:
$$b>1$$