Q1261: It is given that function $f(x)=x^{4}-62x^{2}+ax+9$ attains ...

Class CBSE Class 12 Mathematics Applications of Derivatives Q #1261

COMPETENCY BASED

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5 Marks 2024 AISSCE(Board Exam) LA

It is given that function $f(x)=x^{4}-62x^{2}+ax+9$ attains local maximum value at $x=1$ Find the value of 'a', hence obtain all other points where the given function f(x) attains local maximum or local minimum values.

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Detailed Solution

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

Given $f(x) = x^4 - 62x^2 + ax + 9$. We need to find the first derivative $f'(x)$. $$f'(x) = 4x^3 - 124x + a$$

Step 2: Use the local maximum condition

Since $f(x)$ attains a local maximum at $x=1$, we have $f'(1) = 0$. $$f'(1) = 4(1)^3 - 124(1) + a = 0$$ $$4 - 124 + a = 0$$ $$a = 120$$

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

Now we have $f(x) = x^4 - 62x^2 + 120x + 9$. We need to find the second derivative $f''(x)$. First, we have $f'(x) = 4x^3 - 124x + 120$. $$f''(x) = 12x^2 - 124$$

Step 4: Check the second derivative at x=1

To confirm that $x=1$ is a local maximum, we check the sign of $f''(1)$. $$f''(1) = 12(1)^2 - 124 = 12 - 124 = -112$$ Since $f''(1) < 0$, $x=1$ is indeed a local maximum.

Step 5: Find other critical points

To find other local maximum or minimum points, we need to solve $f'(x) = 0$. $$4x^3 - 124x + 120 = 0$$ $$x^3 - 31x + 30 = 0$$ Since we know $x=1$ is a root, we can divide the polynomial by $(x-1)$. $$(x-1)(x^2 + x - 30) = 0$$ $$(x-1)(x+6)(x-5) = 0$$ So, the critical points are $x = 1, x = -6, x = 5$.

Step 6: Analyze the second derivative at other critical points

We already know $f''(1) = -112 < 0$, so $x=1$ is a local maximum. Now we check $f''(-6)$ and $f''(5)$. $$f''(-6) = 12(-6)^2 - 124 = 12(36) - 124 = 432 - 124 = 308 > 0$$ So, $x=-6$ is a local minimum. $$f''(5) = 12(5)^2 - 124 = 12(25) - 124 = 300 - 124 = 176 > 0$$ So, $x=5$ is a local minimum.

Step 7: Summarize the results

The value of $a$ is 120. The function has a local maximum at $x=1$. The function has local minima at $x=-6$ and $x=5$.

Final Answer: a = 120, Local maximum at x=1, Local minima at x=-6 and x=5

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Pedagogical Audit

Bloom's Analysis: This is an REMEMBER question because the student needs to recall the concepts of derivatives and their application in finding local maxima and minima. The question directly provides the function and the point where local maxima occurs, requiring the student to apply the condition f'(x) = 0 and f''(x) < 0.

Knowledge Dimension: PROCEDURAL

Justification: The question requires a series of steps like finding the first and second derivatives, setting the first derivative to zero, and analyzing the sign of the second derivative to determine local maxima and minima. These are all procedural steps.

Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. The question assesses the student's ability to apply the concepts of differentiation to solve a problem involving local maxima and minima, which is a key competency in calculus.