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Question: The range of \(f(x) = \cos x - \sin x,\) is...

The range of f(x)=cos⁡x−sin⁡x,f(x) = \cos x - \sin x, is

A

(−1,1)( - 1,1)

B

[−1,1)\lbrack - 1,1)

C

[−π2,π2]\left\lbrack - \frac{\pi}{2},\frac{\pi}{2} \right\rbrack

D

[−2,2]\lbrack - \sqrt{2},\sqrt{2}\rbrack

Answer

[−2,2]\lbrack - \sqrt{2},\sqrt{2}\rbrack

Explanation

Solution

Let,f(x)=cos⁡x−sin⁡x⇒f(x)=2(12cos⁡x−12sin⁡x)f(x) = \cos x - \sin x \Rightarrow f(x) = \sqrt{2}\left( \frac{1}{\sqrt{2}}\cos x - \frac{1}{\sqrt{2}}\sin x \right)

⇒f(x)=2cos⁡(x+π4)\Rightarrow f(x) = \sqrt{2}\cos\left( x + \frac{\pi}{4} \right)

Now since, −1≤cos⁡(x+π4)≤1- 1 \leq \cos\left( x + \frac{\pi}{4} \right) \leq 1 ⇒−2≤f(x)≤2\Rightarrow - \sqrt { 2 } \leq f ( x ) \leq \sqrt { 2 }

⇒f(x)∈[−2,2]\Rightarrow f(x) \in \lbrack - \sqrt{2},\sqrt{2}\rbrack

Trick : ∵\because Maximum value of cos⁡x−sin⁡x\cos x - \sin x is 2\sqrt{2} and

minimum value of cos⁡x−sin⁡x\cos x - \sin x is −2- \sqrt{2}.

Hence, range of f(x)=[−2,2]f(x) = \lbrack - \sqrt{2},\sqrt{2}\rbrack.