Class CBSE Class 12 Mathematics Applications of Derivatives Q #1699
COMPETENCY BASED
APPLY
1 Marks 2026 AISSCE(Board Exam) MCQ SINGLE
Absolute minimum value of $f(x)=(x-2)^{2}+5$ in the interval [-3, 2] is :
(A) -3
(B) 2
(C) 5
(D) 30
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Correct Answer: C

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Detailed Solution

Step 1: Identify the function and interval

The given function is $f(x) = (x-2)^2 + 5$ defined on the closed interval $[-3, 2]$.

Step 2: Find the critical points

To find the critical points, we calculate the derivative $f'(x)$ and set it to zero: $$f'(x) = 2(x-2) = 0$$ This gives $x = 2$. Since $x = 2$ is an endpoint of the interval, we evaluate the function at the critical point and the boundaries.

Step 3: Evaluate the function at boundaries

We evaluate $f(x)$ at $x = -3$ and $x = 2$: $$f(-3) = (-3-2)^2 + 5 = (-5)^2 + 5 = 25 + 5 = 30$$ $$f(2) = (2-2)^2 + 5 = 0 + 5 = 5$$

Step 4: Determine the absolute minimum

Comparing the values $f(-3) = 30$ and $f(2) = 5$, the absolute minimum value is $5$.

Final Answer: 5

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to apply the Extreme Value Theorem to find the absolute extrema of a function on a closed interval.
Knowledge Dimension: PROCEDURAL
Justification: The student must follow a specific sequence of steps: differentiating, identifying critical points, and evaluating endpoints.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It tests the student's ability to handle optimization problems, which is a core component of the Application of Derivatives unit.
||KEY:C

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