Question:medium

The range of \( 2 \left| \sin x + \cos x \right| - \sqrt{2} \) is:

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To determine the range of functions involving trigonometric identities, transform the expressions into standard forms and consider the range of trigonometric functions such as sine and cosine.
Updated On: Apr 17, 2026
  • \( \left[ -\sqrt{2}, \sqrt{2} \right] \)
  • \( \left[ -3\sqrt{2}, \sqrt{2} \right] \)
  • \( \left( -3\sqrt{2}, \sqrt{2} \right) \)
  • None of these
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The Correct Option is A

Solution and Explanation

The range of \( \sin x + \cos x \) is determined. Step 1: Transform to an equivalent expression \[ \sin x + \cos x = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right) \] Given that the sine function is bounded by \( -1 \leq \sin \theta \leq 1 \), we have: \[ -\sqrt{2} \leq \sin x + \cos x \leq \sqrt{2} \] Applying the modulus operation yields: \[ 0 \leq \left| \sin x + \cos x \right| \leq \sqrt{2} \] Multiplying by 2: \[ 0 \leq 2 \left| \sin x + \cos x \right| \leq 2\sqrt{2} \]
Step 2: Modify for the specified expression \[ -\sqrt{2} \leq 2 \left| \sin x + \cos x \right| - \sqrt{2} \leq \sqrt{2} \] Conclusion: \[ \boxed{\left[ -\sqrt{2}, \sqrt{2} \right]} \]
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