Q1270: If M and m denote the local maximum and local minimum values...

Class CBSE Class 12 Mathematics Applications of Derivatives Q #1270

KNOWLEDGE BASED

REMEMBER

2 Marks 2024 AISSCE(Board Exam) VSA

If M and m denote the local maximum and local minimum values of the function $f(x)=x+\frac{1}{x}(x\ne0)$ respectively, find the value of $(M-m)$

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Detailed Solution

Step 1: Find the first derivative of f(x)

Given the function $f(x) = x + \frac{1}{x}$, we need to find its first derivative, $f'(x)$. $$f'(x) = \frac{d}{dx}(x + \frac{1}{x}) = \frac{d}{dx}(x + x^{-1}) = 1 - x^{-2} = 1 - \frac{1}{x^2}$$

Step 2: Find the critical points

To find the critical points, we set $f'(x) = 0$ and solve for $x$. $$1 - \frac{1}{x^2} = 0$$ $$\frac{1}{x^2} = 1$$ $$x^2 = 1$$ $$x = \pm 1$$ So, the critical points are $x = 1$ and $x = -1$.

Step 3: Find the second derivative of f(x)

Now, we find the second derivative, $f''(x)$, to determine the nature of the critical points. $$f''(x) = \frac{d}{dx}(1 - \frac{1}{x^2}) = \frac{d}{dx}(1 - x^{-2}) = 0 - (-2)x^{-3} = \frac{2}{x^3}$$

Step 4: Determine local maxima and minima

We evaluate $f''(x)$ at the critical points: For $x = 1$, $f''(1) = \frac{2}{1^3} = 2 > 0$. Since $f''(1) > 0$, $x = 1$ is a local minimum. For $x = -1$, $f''(-1) = \frac{2}{(-1)^3} = -2 < 0$. Since $f''(-1) < 0$, $x = -1$ is a local maximum.

Step 5: Calculate the local maximum and minimum values

Now, we find the local maximum and minimum values by plugging the critical points into the original function: Local maximum value, $M = f(-1) = -1 + \frac{1}{-1} = -1 - 1 = -2$ Local minimum value, $m = f(1) = 1 + \frac{1}{1} = 1 + 1 = 2$

Step 6: Calculate M - m

Finally, we find the value of $M - m$: $M - m = -2 - 2 = -4$

Final Answer: -4

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Pedagogical Audit

Bloom's Analysis: This is an REMEMBER question because it requires recalling the concepts of local maxima and minima, finding derivatives, and applying the second derivative test.

Knowledge Dimension: CONCEPTUAL

Justification: The question requires understanding the concepts of derivatives, critical points, and the second derivative test to determine local maxima and minima.

Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. It directly tests the student's understanding of the application of derivatives in finding local maxima and minima, a core concept in the syllabus.