CBSE Class 12 Mathematics Relations and Functions Q #2134
COMPETENCY BASED
APPLY
2 Marks Generic-Curriculum VSA
Determine the range of $f(x) = \cos x + \sin x + \sqrt{2}$.
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Detailed Solution

Step 1: Simplify the trigonometric expression

We use the harmonic addition formula. The expression $\cos x + \sin x$ can be written as $\sqrt{2}(\frac{1}{\sqrt{2}}\cos x + \frac{1}{\sqrt{2}}\sin x)$. This simplifies to: $$ \cos x + \sin x = \sqrt{2} \sin(x + \frac{\pi}{4}) $$

Step 2: Determine the range of the trigonometric part

We know that the range of the sine function $\sin(\theta)$ is $[-1, 1]$. Therefore, the range of $\sqrt{2} \sin(x + \frac{\pi}{4})$ is: $$ [-\sqrt{2}, \sqrt{2}] $$

Step 3: Apply the constant shift

The function is $f(x) = (\cos x + \sin x) + \sqrt{2}$. Adding $\sqrt{2}$ to the range found in Step 2: $$ [-\sqrt{2} + \sqrt{2}, \sqrt{2} + \sqrt{2}] = [0, 2\sqrt{2}] $$

Final Answer: [0, 2\sqrt{2}]

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must apply the trigonometric identity for linear combinations of sine and cosine to determine the range of a composite function.
Knowledge Dimension: PROCEDURAL
Justification: The student follows a specific algorithmic process (harmonic addition formula) to transform the function into a standard form to identify the range.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. This question tests the student's ability to manipulate trigonometric expressions beyond basic textbook definitions, aligning with NEP 2020's focus on conceptual understanding.

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