Let A = [aij] be a 3 × 3 matrix, where aij = { 1, if i = j; -x, if |i - j| = 1; 2x + 1, otherwise }.

Q: Why did we only consider local extrema? A cubic function goes to infinity!

Great observation! In the context of JEE problems involving polynomials like this, 'maximum and minimum values' refers to the local extrema. Since the domain is $\mathbb{R}$, the global range is $(-\infty, \infty)$, so the question implicitly asks for the local peak and valley.

Q: Is there a property of determinants I could have used to avoid the long expansion?

Yes! You could try row or column operations to create zeros. For example, adding rows or columns might simplify the matrix, but for a $3 \times 3$, direct expansion is often safer and faster if you are careful with signs.

Q: I got a different derivative. Where did I go wrong?

Check your expansion of the determinant again. The term $(2x+1)(x^2 - (2x+1))$ is the most common place for a sign error. Ensure you distribute the negative signs correctly.

Question

Let $A = [a_{ij}]$ be a $3 	imes 3$ matrix, where $a_{ij} = \begin{cases} 1, & 	ext{if } i = j \ -x, & 	ext{if } |i - j| = 1 \ 2x + 1, & 	ext{otherwise} \end{cases}$. Let a function $f : \mathbb{R} 	o \mathbb{R}$ be defined as $f(x) = \det(A)$. Then the sum of maximum and minimum values of $f$ on $\mathbb{R}$ is equal to:

Options:

(A) $-\frac{20}{27}$

(B) $\frac{88}{27}$

(C) $\frac{20}{27}$

(D) $-\frac{88}{27}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a blank $3 	imes 3$ grid. The problem provides a set of rules, a piecewise definition that acts as our architectural blueprint. We are not just filling in numbers; we are building a mathematical object.

The diagonal elements, where $i = j$, are all $1$. The elements adjacent to this diagonal, where the absolute difference $|i - j| = 1$, are all $-x$. Finally, the corners, where the conditions do not apply, are $2x + 1$.

When you lay this out, you get the following matrix:

$$A = \begin{bmatrix} 1 & -x & 2x+1 \ -x & 1 & -x \ 2x+1 & -x & 1 \end{bmatrix}$$

This matrix is symmetric, elegant, and ready for us to unlock its secrets.

### The Heartbeat: Expanding the Determinant

Now, we need to find the function $f(x) = \det(A)$. This is the heartbeat of the problem. We expand along the first row, being systematic to avoid errors.

We have:

$$f(x) = 1(1 - x^2) - (-x)(-x - (-x)(2x+1)) + (2x+1)(x^2 - (2x+1))$$

This looks like a mess, but watch how it cleans up. The first part is $(1 - x^2)$. The second part involves the product of $-x$ and $(-x + 2x^2 + x)$, which simplifies significantly.

The third part, $(2x+1)(x^2 - 2x - 1)$, is where you must be vigilant with your signs. After grouping all the like terms, we arrive at a clean, powerful cubic polynomial:

$$f(x) = 4x^3 - 4x^2 - 4x$$

This is the function whose peaks and valleys we must now conquer.

### The Search for Peaks: Calculus to the Rescue

We have our function, but we need its extrema. This is where calculus becomes our compass. We need to find where the slope of the tangent is zero.

We differentiate $f(x)$ with respect to $x$:

$$f'(x) = \frac{d}{dx}(4x^3 - 4x^2 - 4x) = 12x^2 - 8x - 4$$

Setting $f'(x) = 0$ gives us the critical points. We factor out the $4$:

$$4(3x^2 - 2x - 1) = 0$$

This factors into $(3x + 1)(x - 1) = 0$. Our critical points are $x = 1$ and $x = -\frac{1}{3}$. These are the points where our function turns around.

### The Final Calculation: Bringing it Home

Now, we evaluate the function at these critical points to find the local minimum and maximum. For $x = 1$, we find:

$$f(1) = 4(1)^3 - 4(1)^2 - 4(1) = -4$$

This is our local minimum. For $x = -\frac{1}{3}$, we calculate:

$$f\left(-\frac{1}{3}ight) = 4\left(-\frac{1}{27}ight) - 4\left(\frac{1}{9}ight) - 4\left(-\frac{1}{3}ight) = -\frac{4}{27} - \frac{12}{27} + \frac{36}{27} = \frac{20}{27}$$

This is our local maximum. The question asks for the sum of these values:

$$	ext{Sum} = \frac{20}{27} + (-4) = \frac{20 - 108}{27} = -\frac{88}{27}$$

And there it is. We have navigated the matrix, conquered the algebra, and used calculus to find the final answer of **$-\frac{88}{27}$**.

Select Answer:

Visualized Solution

Understanding Matrix $A$

Constructing the Diagonal

Constructing Adjacent Elements

Constructing Corner Elements

Expanding the Determinant

Simplifying $f (x)$

Finding the Derivative $f^{'} (x)$

Solving for Critical Points

Evaluating Local Minimum

Evaluating Local Maximum

Calculating the Sum

Conclusion

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Heartbeat

The Search for Peaks

The Final Calculation

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