Question

The range of \[f(x)=\cos x-\sin x\]is
A.\[\left[ -1,1 \right]\]
B.\[\left( -1,2 \right)\]
C.\[\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]\]
D.\[\left( -\sqrt{2},\sqrt{2} \right)\]

Answer 1

Hint: The domain of any given function is the set of input values for which the function is defined. When determining the domain of a function, two situations are considered. Firstly, division by zero is not allowed. So, one must eliminate the values of \[x\] which makes the denominator zero.
Secondly, for even roots such as square roots, the radicant cannot be negative. We must make sure that the radicand is nonnegative. The range can be defined as the set of all possible \[y\] -values the relation can produce from the \[x\]-values. There is no direct way to find the range algebraically. However, one strategy that works most of the time is to find the domain of the inverse function (if it exists). First, swap the \[x\] and \[y\] variables everywhere they appear in the equation and then solve for y. Find the domain of this new equation and it will be the range of the original.

Complete step-by-step answer:
\[f(x)=\cos x-\sin x\]
Rewriting the above equation by dividing and multiplying by \[\sqrt{2}\]
\[f(x)=\sqrt{2}\left[ \dfrac{\cos x}{\sqrt{2}}-\dfrac{\sin x}{\sqrt{2}} \right]\]
As we know that \[\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\] and \[\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\]
Applying these we can write the equation as
\[f(x)=\sqrt{2}\left[ \sin \dfrac{\pi }{4}\cos x-\cos \dfrac{\pi }{4}\sin x \right]\]
Further solving using the trigonometric formulas we get
\[f(x)=\sqrt{2}\sin \left( \dfrac{\pi }{4}-x \right)\]
We know that the domain of \[\sin x\] is \[\left[ -1,1 \right]\], that means
\[-1\le \sin x\le 1\]
So,
\[-1\le \sin \left( \dfrac{\pi }{4}-x \right)\le 1\]
Multiplying by \[\sqrt{2}\] we get
\[-\sqrt{2}\le \sqrt{2}\sin \left( \dfrac{\pi }{4}-x \right)\le \sqrt{2}\]
We can write the above equation as
\[-\sqrt{2}\le f(x)\le \sqrt{2}\]
Therefore, the range of \[f(x)=\cos x-\sin x\] is \[\left[ -\sqrt{2},\sqrt{2} \right]\].
Hence, option \[(D)\] is the correct answer.
So, the correct answer is “Option D”.

Note: The range of the real function is the set of all real values taken at points in its domain. In order to find the range of real functions, first solve the equation and find the values. The set of values obtained is the range of the function.