The range of $f (x) = cos x - sin x$ is

(- 1, 1)

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[- 1, 1]

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[- \frac{π}{2}, \frac{π}{2}]

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[- \sqrt{2}, \sqrt{2}]

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Solution

The correct option is B $[- \sqrt{2}, \sqrt{2}]$
range of $f (x) = cos x - sin x$

$= \sqrt{2} [\frac{cos x}{\sqrt{2}} - \frac{sin x}{\sqrt{2}}]$

$= \sqrt{2} [sin \frac{π}{4} cos x - cos \frac{π}{4} sin x]$

$f (x) = \sqrt{2} [sin (\frac{π}{4} - x)]$

we know that $sin θ$ lies in $[- 1, 1]$

$\Rightarrow - 1 \leq sin θ \leq 1$

So,

$\Rightarrow - 1 \leq sin (\frac{π}{4} - x) \leq 1$

$\Rightarrow - \sqrt{2} \leq \sqrt{2} sin (\frac{π}{4} - x) \leq \sqrt{2}$

$\Rightarrow - \sqrt{2} \leq f (x) \leq \sqrt{2}$

$∴ f (x) \in [- \sqrt{2}, \sqrt{2}]$ .

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