The function \[f(x)=\sin x-kx-c\] where k and c are constants, decreases always when:
a) \[k>1\]
b) \[k\ge 1\]
c) \[k<1\]
d) \[k\le 1\]
e) \[k<-1\]

Hint: A function f(x) is strictly decreasing if f’(x) < 0 for all x in its domain. So we will find the 1st derivative of f(x) w.r.t x. Now just apply the condition for decreasing function and find the value of k from there.

Complete step-by-step answer:
We are given a real valued function \[f(x)=\sin x-kx-c\] where k and c are constants. We have to find the value of k for which f(x) decreases always.
Now from the theory of calculus we know that a function will be decreasing in its domain if the first derivative of the function is strictly less than zero.
Now for our given function,\[f(x)=\sin x-kx-c\]. So, its first derivative or \[f'(x)=\cos x-k\].
As we know,\[\dfrac{d}{dx}(\sin x)=\cos x\] and \[\dfrac{d}{dx}(kx)=k\] [k being a constant] and \[\dfrac{d}{dx}(c)=0\].
Now in order to be f(x) a decreasing function we get \[f'(x)=\cos x-k<0\].
Or,\[\cos xNow we know \[\cos x\] is a real valued bounded function. It has minimum value -1 and maximum value as 1.
Now we get the maximum value of \[\cos x\] is 1. So, in order to solve the above equation the maximum value of \[\cos x\] will be there.
So, we get \[1\le k\]
Or, \[k\ge 1\]
Hence, the required value of k is \[k\ge 1\] for all real k.
Hence, the correct option to the question is (b) \[k\ge 1\].

Note: Since it has been told that the function decreases always so the equality case in the derivative has also to be considered. Don’t take the value of k to be greater than 1 because the function cos has a range of $\left[ -1,1 \right]$ i.e. closed brackets are there, so you can use the equality case also.