Let the function f(x) = 2x³ + (2p - 7)x² + 3(2p - 9)x - 6 have a maxima for some value of x

Q: Wait, why do we only check the product of the roots? Shouldn't we check the discriminant $D > 0$ as well?

That is a brilliant observation! In many problems, you would indeed need $D > 0$ to ensure the roots are real. However, if the product of the roots $\frac{c}{a}$ is negative, it mathematically guarantees that the roots are real and distinct. Since $c/a < 0$, the discriminant $b^2 - 4ac$ will automatically be positive because $-4ac$ becomes a positive term. You've just discovered a shortcut!

Q: Why does the smaller root correspond to the maxima and the larger to the minima?

Think about the shape of a cubic function with a positive leading coefficient. It looks like an 'N' shape. It climbs, hits a peak (maxima), drops into a valley (minima), and climbs again. The peak must occur at the smaller x-value, and the valley must occur at the larger x-value. It is the geometric nature of the curve!

Q: Can I use the second derivative test here instead?

You absolutely could, but it would be a long, winding road! You would find the roots in terms of $p$, plug them into $f''(x)$, and check the signs. While valid, it is computationally heavy. In JEE, we look for the most elegant path—using Vieta's formulas is like taking the express train.

Q: What if the leading coefficient was negative? Would the answer change?

Yes, it would flip! If the leading coefficient were negative, the cubic would start from positive infinity and end at negative infinity. The 'valley' would come before the 'peak'. Always check the sign of the leading coefficient before you start—it dictates the entire 'story' of the graph.

Q: Is it possible for the roots to be zero?

The problem specifies the maxima is at $x 0$. This strictly excludes $x = 0$. If a root were zero, the product of the roots would be zero, not negative. Our inequality $c/a < 0$ naturally handles this by excluding zero.

Question

Let the function $f(x) = 2x^3 + (2p - 7)x^2 + 3(2p - 9)x - 6$ have a maxima for some value of $x < 0$ and a minima for some value of $x > 0$. Then, the set of all values of $p$ is

Options:

(A) $(\frac{9}{2}, \infty)$

(B) $(0, \frac{9}{2})$

(C) $(-\infty, \frac{9}{2})$

(D) $(-\frac{9}{2}, \frac{9}{2})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine you are standing before a vast, rolling landscape. You are holding a mathematical function, $f(x) = 2x^3 + (2p - 7)x^2 + 3(2p - 9)x - 6$, and your task is to shape this landscape. You want to create a mountain peak (a local maximum) on the left side of the $y$-axis and a deep valley (a local minimum) on the right side. This geometric requirement is the heart of the problem. ### The Derivative: Our Compass To find where the landscape peaks and valleys, we must look at the rate of change. We calculate the derivative: $$f'(x) = 6x^2 + 2(2p - 7)x + 3(2p - 9)$$ This quadratic expression is our compass. The points where $f'(x) = 0$ are the critical points—the exact moments where the landscape stops climbing and starts falling, or vice versa. Because the leading coefficient of our cubic is positive ($2 > 0$), we know the general 'N' shape of the curve. It rises, falls, and rises again. For the peak to be at $x < 0$ and the valley at $x > 0$, the smaller root of our quadratic derivative must be negative, and the larger root must be positive. ### The Elegance of Vieta's Formulas Now, we could solve for the roots using the quadratic formula, but that would lead us into a thicket of square roots and messy algebra. Instead, let us use the elegance of Vieta's formulas. If we have a quadratic equation $ax^2 + bx + c = 0$, the product of its roots is given by $\frac{c}{a}$. If one root is negative and the other is positive, their product must be negative. This is the 'Aha!' moment. We don't need to know the exact values of the roots; we only need to know their signs. By setting the product of the roots to be less than zero, we effectively trap the parameter $p$ in a cage of our own making. ### Solving the Inequality We identify our coefficients from $f'(x) = 6x^2 + 2(2p - 7)x + 3(2p - 9)$: $$a = 6$$ $$c = 3(2p - 9)$$ The condition for roots of opposite signs is: $$\frac{c}{a} < 0 \implies \frac{3(2p - 9)}{6} < 0$$ Simplifying this, we get: $$\frac{1}{2}(2p - 9) < 0$$ Since $\frac{1}{2}$ is just a positive scaling factor, it does not change the direction of our inequality. We multiply by $2$ and find: $$2p - 9 < 0 \implies 2p < 9 \implies p < \frac{9}{2}$$ ### The Final Victory We have arrived at our destination: $p < \frac{9}{2}$. In interval notation, this is $(-\infty, \frac{9}{2})$. By understanding the geometric requirement of the cubic function and leveraging the power of Vieta's formulas, we bypassed the tedious calculations and went straight to the core of the problem. Remember, in JEE Advanced, the most powerful tool is your ability to see the underlying structure of the math. You have successfully constrained the landscape, ensuring the peak and valley fall exactly where you intended. The final answer is **$p < \frac{9}{2}$**.

Animated Solution for Mathematics - Differentiation: Let the function $f (x) = 2 x^{3} + (2 p - 7) x^{2} + 3 (2 p - 9) x - 6$ have a maxima for some value of $x < 0$ and a minima for some value of $x > 0$ . Then, the set of all values of $p$ is

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Visualized Solution

Analyze the Function $f (x)$

The Role of the Derivative $f^{'} (x)$

Differentiating $f (x)$

Calculating $f^{'} (x)$

Roots of $f^{'} (x) = 0$

Condition for Roots with Opposite Signs

Substituting Coefficients into Condition

Simplifying the Inequality

Solving for $p$

Final Range of $p$

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Derivative

The Elegance of Vieta's Formulas

Solving the Inequality

The Final Victory

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Visualized Solution

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

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

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

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

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