Suppose f(x) is a function satisfying the following conditions: (a) f(0) = 2, f(1) = 1, (b) f has a

Q: Wait, why can't I just expand the determinant directly? It's only a 3x3 matrix.

You absolutely *could*, but you would be walking into a trap. Expanding it directly leads to a massive algebraic expression where terms like $2axb$ and $2ax$ get tangled. In competitive exams, the 'brute force' method is often the path to a calculation error. Always pause for 10 seconds to look for row operations that create zeros—it's the elegant, 'JEE-proof' way to solve it.

Q: Why does the minimum at $x = 5/2$ mean $f'(5/2) = 0$?

This is a fundamental concept from the Application of Derivatives. A local minimum (or maximum) occurs where the slope of the tangent line is horizontal. Since the derivative $f'(x)$ represents the slope of the tangent, setting $f'(5/2) = 0$ is the mathematical way of saying 'the curve has flattened out at this point'.

Q: How do we know for sure that $f(x)$ is a quadratic function?

Great question! We deduced it by integrating the derivative. Since our simplified derivative $f'(x) = 2ax + b$ is a linear function (a polynomial of degree 1), integrating it with respect to $x$ naturally yields a polynomial of degree 2, which is a quadratic. The calculus tells us the shape of the function!

Question

Suppose $f(x)$ is a function satisfying the following conditions: (a) $f(0) = 2, f(1) = 1$, (b) $f$ has a minimum value at $x = 5/2$, and (c) for all $x$, $f'(x) = \begin{vmatrix} 2ax & 2ax-1 & 2ax+b+1 \ b & b+1 & -1 \ 2(ax+b) & 2ax+2b+1 & 2ax+b \end{vmatrix}$ where $a, b$ are some constants. Determine the constants $a, b$ and the function $f(x)$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The Art of Mathematical Elegance: Unmasking the Determinant. Welcome, future engineer. Today, we are going to tackle a problem that, at first glance, looks like a chaotic mess of variables and matrices.

You see a $3 	imes 3$ determinant, and your instinct might be to panic or start a long, tedious expansion. But in the world of JEE Advanced, the most intimidating problems often hide the most elegant solutions. Let us walk through this together.

### Phase 1: The Matrix Surgery

We are given the derivative $f'(x)$ defined by the determinant:

$$f'(x) = \begin{vmatrix} 2ax & 2ax-1 & 2ax+b+1 \ b & b+1 & -1 \ 2(ax+b) & 2ax+2b+1 & 2ax+b \end{vmatrix}$$

If you try to expand this directly, you will find yourself drowning in a sea of $a$, $b$, and $x$ terms. Instead, let us perform some 'matrix surgery.'

We look at the third row, $R_3$. Notice how it relates to the first two rows. If we take $R_1$ and add twice $R_2$ to it, we get $2ax + 2b$. This is exactly the first element of $R_3$!

By applying the row operation $R_3 	o R_3 - (R_1 + 2R_2)$, we perform a clean sweep. The first element becomes $2(ax+b) - (2ax + 2b) = 0$. The second element becomes $(2ax+2b+1) - (2ax-1 + 2b+2) = 0$. And the third element simplifies to $1$.

Suddenly, our terrifying determinant has a row with two zeros and a one. This is the moment where the problem shifts from 'impossible' to 'trivial.'

### Phase 2: The Reveal

Now that we have a row of zeros, expanding along the third row is a breeze. We only need to evaluate the minor for the element $1$. The determinant collapses into:

$$f'(x) = 1 \cdot [2ax(b+1) - b(2ax-1)]$$

Expanding this, we see the magic happen: $2axb + 2ax - 2axb + b$. The $2axb$ terms cancel out perfectly, leaving us with the beautifully simple linear expression:

$$f'(x) = 2ax + b$$

This is the heart of the problem. We have stripped away the complexity to find the linear slope of our function.

### Phase 3: The Calculus Bridge

Now that we have $f'(x) = 2ax + b$, we need to find the original function $f(x)$. We know that the derivative is the rate of change, so to go back to the function, we must integrate:

$$f(x) = \int (2ax + b) dx = ax^2 + bx + c$$

Here, $c$ is our constant of integration. We have successfully identified that our function is a parabola. Now, we just need to solve for the constants $a$, $b$, and $c$.

### Phase 4: The Detective Work

We have three clues, and we need to use them like a detective. First, $f(0) = 2$. Substituting $x=0$ into $f(x) = ax^2 + bx + c$, we get $c = 2$.

Second, $f(1) = 1$. Substituting $x=1$ and $c=2$, we get $a + b + 2 = 1$, or $a + b = -1$.

Third, the minimum value at $x = 5/2$. This tells us that the slope at this point is zero, so $f'(5/2) = 0$. Substituting $x = 5/2$ into $f'(x) = 2ax + b$, we get $2a(5/2) + b = 0$, which simplifies to $5a + b = 0$.

Now we have a system of two linear equations: $a + b = -1$ and $5a + b = 0$. Subtracting the first from the second, we get $4a = 1$, so $a = 1/4$. Plugging this back into $a + b = -1$, we find $b = -5/4$.

### Conclusion: The Final Function

We have done it. We have navigated the determinant, integrated the derivative, and solved for the constants. Our final function is:

$$f(x) = \frac{1}{4}x^2 - \frac{5}{4}x + 2$$

Take a moment to appreciate this. We started with a complex matrix and ended with a clean, elegant parabola. This is the beauty of physics and mathematics—no matter how scary the problem looks, there is always a path to simplicity if you look for the patterns.

Visualized Solution

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

Simplifying the Determinant

The Simplified Determinant

Expanding the Determinant

Integrating to Find $f (x)$

Applying the First Condition

Applying the Second Condition

Using the Minimum Condition

Solving for $a$ and $b$

Finalizing the Function

The Sigma Insight: Maxima and Minima

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

Similar Questions

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Comprehension Passage

Which of the following is true for $0 < x < 1$ ?

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Comprehension Passage

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Which of the following is true?

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$f (x)$ is cubic polynomial with $f (2) = 18$ and $f (1) = - 1$ . Also $f (x)$ has local maxima at $x = - 1$ and $f^{'} (x)$ has local minima at $x = 0$ , then

The minimum value of the twice differentiable function $f (x) = \int_{0}^{x} e^{x - t} f^{'} (t) d t - (x^{2} - x + 1) e^{x}, x \in R$ , is :

Let $f : R \to R$ be defined by $f (x) = {k - 2 x, 2 x + 3, if x \leq - 1 if x > - 1$ . If $f$ has a local minimum at $x = - 1$ , then a possible value of $k$ is

Visualized Solution

Analyzing the Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Simplifying the Determinant

The Simplified Determinant

Expanding the Determinant

Integrating to Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Applying the First Condition

Applying the Second Condition

Using the Minimum Condition

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Finalizing the Function

The Sigma Insight: Maxima and Minima

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

Similar Questions

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

Integrating to Find $f (x)$

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Let $f (x) = {- x^{3} + \frac{b ^{3} - b ^{2} + b - 1}{b ^{2} + 3 b + 2}, 2 x - 3, 0 \leq x < 1 1 \leq x \leq 3$ . Find all possible real values of $b$ such that $f (x)$ has the smallest value at $x = 1$ .

Which of the following is true for $0 < x < 1$ ?

If the function $e^{- x} f (x)$ assumes its minimum in the interval $[0, 1]$ at $x = 1/4$ , which of the following is true?

Let $g (x) = \int_{0}^{e^{x}} \frac{f ^{'} ( t )}{1 + t ^{2}} d t$ . Which of the following is true?

$f (x)$ is cubic polynomial with $f (2) = 18$ and $f (1) = - 1$ . Also $f (x)$ has local maxima at $x = - 1$ and $f^{'} (x)$ has local minima at $x = 0$ , then

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