If the set of all values of a, for which the equation 5x³-15x-a=0 has three distinct real roots, is

Q: Wait, why do we reject the negative root here?

We actually don't! In step 3, we solve $x^2 - 1 = 0$, which gives $x = 1$ and $x = -1$. Both are valid critical points. We must evaluate the function at both to find the local maximum and local minimum. Forgetting the negative root would lead to an incomplete picture of the cubic curve's behavior.

Q: Why can't we just use the discriminant of a cubic equation?

While a cubic discriminant exists, it is often cumbersome and prone to calculation errors in a high-pressure exam setting. The graphical approach—analyzing the local extrema—is much more intuitive, robust, and aligns perfectly with how JEE expects you to visualize function behavior.

Q: Why is the interval open $(\alpha, \beta)$ and not closed?

Great observation! If $a$ were equal to the local maximum or minimum value, the line $y = a$ would be tangent to the curve at that point. A tangent line at an extremum results in a repeated root (a double root), which means we would have only two distinct roots instead of three. Thus, the interval must be strictly open.

Question

If the set of all values of a, for which the equation $5x^{3}-15x-a=0$ has three distinct real roots, is the interval $(\alpha,\beta)$, then $\beta-2\alpha$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a vast, undulating landscape defined by the function $f(x) = 5x^3 - 15x$. This is not just a collection of numbers; it is a dynamic entity that rises to a peak and falls into a valley.

Our mission is to find the values of $a$ such that the horizontal line $y = a$ intersects this landscape at exactly three distinct points. This represents a geometric dance between a cubic curve and a flat, unyielding line.

### The Microscope of Calculus

To understand where our curve turns, we employ the power of calculus. We calculate the derivative of the function:

$$f'(x) = \frac{d}{dx}(5x^3 - 15x) = 15x^2 - 15$$

This derivative represents the slope of the tangent at any point. When the slope is zero, the curve is momentarily flat, indicating our turning points.

Setting $f'(x) = 0$, we obtain:

$$15(x^2 - 1) = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$$

These are the $x$-coordinates of our local maximum and local minimum. The curve changes direction exactly at these two points.

### The Peaks and Valleys

Next, we determine the vertical height of these turning points. By substituting $x = -1$ into our original function, we find the local maximum:

$$f(-1) = 5(-1)^3 - 15(-1) = -5 + 15 = 10$$

Substituting $x = 1$ into the function, we find the local minimum:

$$f(1) = 5(1)^3 - 15(1) = 5 - 15 = -10$$

We now know the exact vertical bounds of our curve's turning points: $y = 10$ and $y = -10$.

### The Goldilocks Zone

For the line $y = a$ to intersect the curve at three distinct points, it must pass through the region strictly between the peak and the valley.

If the line is above $10$ or below $-10$, it intersects the curve only once. If it is exactly at $10$ or $-10$, it touches the curve at two points, creating a double root.

Therefore, the line must be trapped in the "Goldilocks zone" defined by the inequality:

$$-10 < a < 10$$

This gives us our interval $(\alpha, \beta)$, where $\alpha = -10$ and $\beta = 10$.

### Final Calculation

We have navigated the geometry and the calculus to identify our bounds. Now, we perform the final calculation for $\beta - 2\alpha$:

$$\beta - 2\alpha = 10 - 2(-10) = 10 + 20 = 30$$

The final result of this journey is **30**. Remember, in JEE Advanced, the math is the language, but the true skill lies in visualizing the story the equations tell.

Animated Solution for Mathematics - Differentiation: If the set of all values of a, for which the equation $5 x^{3} - 15 x - a = 0$ has three distinct real roots, is the interval $(α, β)$ , then $β - 2 α$ is equal to

Enter Numerical Value:

Visualized Solution

Analyzing the Equation $5 x^{3} - 15 x - a = 0$

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

Locating Critical Points

Solving for $x$

Evaluating Local Maximum

Evaluating Local Minimum

Visualizing the Intersections

Condition for Three Roots

Defining the Interval $(α, β)$

Final Calculation for $β - 2 α$

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Microscope of Calculus

The Peaks and Valleys

The Goldilocks Zone

Final Calculation

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Enter Numerical Value:

Visualized Solution

Analyzing the Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 5x3−15x−a=0

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

Locating Critical Points

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