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Q.

The range of y such that the evaluation in $x, y + \cos x = \sin x$ has a real solution

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a

$[- 1, 1]$

b

$[- \frac{1}{2}, \frac{1}{2}]$

c

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

d

$[- 2, 2]$

answer is B.

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Detailed Solution

$\begin{array}{l} y = \sin x - \cos x \\ = \sqrt{2} (\sin x \cos \frac{π}{4} - \cos x \sin \frac{π}{4}) \\ = \sqrt{2} \sin (x - \frac{π}{4}) \\ Since - 1 \leq \sin (x - \frac{π}{4}) \leq 1 \\ ⇒ - \sqrt{2} \leq \sqrt{2} \sin (x - \frac{π}{4}) \leq \sqrt{2} \\ \Rightarrow - \sqrt{2} \leq y \leq \sqrt{2} \end{array}$

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