Class CBSE Class 12 Mathematics Applications of Derivatives Q #931
KNOWLEDGE BASED
UNDERSTAND
2 Marks 2023 VSA
Find the maximum and minimum values of the function given by \(f(x)=5+\sin 2x\).
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Detailed Solution

Step 1: Understand the range of the sine function

The sine function, \(\sin(u)\), always has a range between -1 and 1, inclusive. That is, \(-1 \leq \sin(u) \leq 1\) for any real number \(u\).

Step 2: Apply the range to the given function

We have \(f(x) = 5 + \sin 2x\). Since \(-1 \leq \sin 2x \leq 1\), we can add 5 to all parts of the inequality: $$5 - 1 \leq 5 + \sin 2x \leq 5 + 1$$ $$4 \leq f(x) \leq 6$$

Step 3: Determine the maximum and minimum values

From the inequality \(4 \leq f(x) \leq 6\), we can see that the minimum value of \(f(x)\) is 4 and the maximum value is 6.

Final Answer: Maximum value = 6, Minimum value = 4

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Pedagogical Audit
Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand the range of the sine function and how it affects the given function to find the maximum and minimum values.
Knowledge Dimension: CONCEPTUAL
Justification: The question requires understanding the concept of the range of trigonometric functions and how transformations affect the range.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of trigonometric functions and their properties, specifically the range of the sine function.

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