Let f(x) be a polynomial of degree 5 such that x = ±1 are its critical points. If lim x→0 (2 + f(x)/

Q: Why can't $f(x)$ have terms like $x^2$ or $x$?

Great question! If $f(x)$ contained $x^2$, then $\frac{f(x)}{x^3}$ would contain a term like $\frac{x^2}{x^3} = \frac{1}{x}$. As $x \to 0$, $\frac{1}{x}$ approaches infinity. Since our limit is a finite value (4), all terms with degree less than 3 must have coefficients of zero.

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

Check the powers of $x$ in your final polynomial $f(x) = -\frac{6}{5}x^5 + 2x^3$. Both 5 and 3 are odd integers. Mathematically, if $f(-x) = -f(x)$, the function is odd. Since every term is an odd power, the entire function inherits this symmetry.

Q: Is there a faster way to check the maxima and minima without the first derivative test?

You could use the second derivative test, but be careful! At $x = 1$ and $x = -1$, the second derivative might be zero (inflection point check). The first derivative test is the most robust method here because it clearly shows the sign change of the slope around the critical points.

Question

Let $f(x)$ be a polynomial of degree 5 such that $x = \pm1$ are its critical points. If $\lim_{x 	o 0} (2 + \frac{f(x)}{x^3}) = 4$, then which one of the following is not true?

Options:

(A) $f(x)$ has minima at $x = 1$ & maxima at $x = -1$

(B) $f(1) - 4f(-1) = 4$

(C) $f(x)$ is maxima at $x = 1$ and minima at $x = -1$

(D) $f(x)$ is odd

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a mathematical puzzle, a polynomial of degree 5, shrouded in mystery. We know it exists, but we don't know its coefficients.

Our only clues are a limit and two critical points. This is the essence of JEE Advanced problems—they don't give you the function; they give you the DNA of the function, and it is your job to reconstruct it.

### Decoding the Limit

We start with the condition:

$$\lim_{x 	o 0} \left( 2 + \frac{f(x)}{x^3} ight) = 4$$

If we subtract 2 from both sides, we get:

$$\lim_{x 	o 0} \frac{f(x)}{x^3} = 2$$

Think about what this implies. If $f(x)$ had a constant term, an $x$ term, or an $x^2$ term, the division by $x^3$ would leave us with terms like $\frac{1}{x^3}$, $\frac{1}{x^2}$, or $\frac{1}{x}$.

As $x$ approaches zero, these terms explode to infinity. But our limit is a finite 2!

This forces the coefficients of $x^0, x^1,$ and $x^2$ to be zero. Thus, our polynomial must take the form:

$$f(x) = ax^5 + bx^4 + cx^3$$

When we divide this by $x^3$, we get $ax^2 + bx + c$. As $x 	o 0$, this simplifies to just $c$. Therefore, $c = 2$. We have successfully stripped away the mystery of the lower-order terms!

### The Critical Point Hunt

Now, we turn to the critical points at $x = 1$ and $x = -1$. In the language of calculus, a critical point is where the slope of the tangent is zero.

This means $f'(1) = 0$ and $f'(-1) = 0$. Let's differentiate our polynomial:

$$f'(x) = 5ax^4 + 4bx^3 + 6x^2$$

Now, we plug in our critical points. For $x = 1$, we get:

$$5a + 4b + 6 = 0$$

For $x = -1$, we get:

$$5a - 4b + 6 = 0$$

This is a beautiful system of linear equations. By adding them, we eliminate $b$ and find $10a + 12 = 0$, which gives $a = -\frac{6}{5}$.

By subtracting them, we eliminate $a$ and find $8b = 0$, which gives $b = 0$. Our polynomial is fully revealed:

$$f(x) = -\frac{6}{5}x^5 + 2x^3$$

### The Symmetry and the Verdict

Look at the powers of $x$ in our final function. They are 5 and 3—both odd. This confirms that $f(x)$ is an odd function, meaning $f(-x) = -f(x)$.

Now, let's check the nature of the critical points using the first derivative test. We have:

$$f'(x) = -6x^4 + 6x^2 = 6x^2(1 - x^2)$$

The sign of $f'(x)$ depends on $(1 - x^2)$. For $x < -1$, the derivative is negative (decreasing). For $-1 < x < 1$, it is positive (increasing). For $x > 1$, it is negative (decreasing).

This tells us that $x = -1$ is a local minimum and $x = 1$ is a local maximum. You have successfully navigated the logic, solved the system, and unmasked the polynomial. **Well done!**

Animated Solution for Mathematics - Differentiation: Let $f (x)$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4$ , then which one of the following is not true?

Select Answer:

Visualized Solution

The Limit Condition

Form of the Polynomial

Evaluating the Limit

Critical Points Condition

Differentiating $f (x)$

Setting up Equations

Solving for $a$ and $b$

The Final Polynomial

First Derivative Test

Maxima and Minima

Checking Option 2

Conclusion

The Sigma Insight: Maxima and Minima

Analyzing the Setup

Decoding the Limit

The Critical Point Hunt

The Symmetry and the Verdict

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