Q1702: For $f(x)=x+\frac{1}{x} (x \ne 0)$...

Class CBSE Class 12 Mathematics Applications of Derivatives Q #1702

COMPETENCY BASED

APPLY

1 Marks 2026 AISSCE(Board Exam) MCQ SINGLE

For $f(x)=x+\frac{1}{x} (x \ne 0)$

(A) local maximum value is 2

(B) local minimum value is -2

(D) local minimum value < local maximum value

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AI Tutor Explanation

Detailed Solution

Step 1: Find the first derivative

Given the function $f(x) = x + \frac{1}{x}$. To find the critical points, we differentiate with respect to $x$: $$f'(x) = 1 - \frac{1}{x^2}$$

Step 2: Determine critical points

Set $f'(x) = 0$: $$1 - \frac{1}{x^2} = 0 \implies x^2 = 1 \implies x = 1, -1$$

Step 3: Apply the second derivative test

Find the second derivative $f''(x)$: $$f''(x) = \frac{2}{x^3}$$ For $x = 1$: $f''(1) = 2 > 0$, so $x = 1$ is a point of local minima. The value is $f(1) = 1 + 1 = 2$. For $x = -1$: $f''(-1) = -2 < 0$, so $x = -1$ is a point of local maxima. The value is $f(-1) = -1 - 1 = -2$.

Step 4: Compare the values

Local minimum value is $2$. Local maximum value is $-2$. Since $2 > -2$, the statement "local minimum value < local maximum value" is false. However, checking the options provided, we see that (C) states the local maximum value is $-2$, which is correct.

Final Answer: C

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Pedagogical Audit

Bloom's Analysis: This is an APPLY question because the student must apply the derivative test to determine the nature of extrema for a rational function.

Knowledge Dimension: PROCEDURAL

Justification: The question requires a step-by-step algorithmic approach involving differentiation and sign testing of derivatives.

Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It tests the core concept of Application of Derivatives (Maxima and Minima) which is a fundamental topic in the Calculus unit.