Question

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) $k<-1$

Answer 1

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.