Let α(a) and β(a) be the roots of the equation (∛(1+a)-1)x² + (√(1+a)-1)x + (⁶√(1+a)-1) = 0 where a

Q: Wait, why can't I just plug in $a=0$ directly into the equation?

That is a great question! If you plug in $a=0$, you get $0 \cdot x^2 + 0 \cdot x + 0 = 0$, which simplifies to $0=0$. This is an indeterminate form that tells you nothing about the roots $x$. We need to find the limiting behavior of the equation as $a$ approaches zero, which requires us to look at the ratio of the coefficients to $a$.

Q: Why are we allowed to divide the entire equation by $a$?

We are looking for the limit as $a \to 0^+$. This means $a$ is getting arbitrarily close to zero, but it is never actually equal to zero. Since $a \neq 0$, dividing the entire equation by $a$ is a perfectly valid algebraic operation that preserves the roots of the equation.

Q: How do I know which standard limit to use?

Whenever you see expressions like $(1+a)^n - 1$ in a limit problem, your mind should immediately jump to the binomial expansion or the standard limit $\lim_{h \to 0} \frac{(1+h)^n - 1}{h} = n$. It is the most efficient way to handle these types of indeterminate forms without getting bogged down in complex algebra.

Question

Let $\alpha(a)$ and $\beta(a)$ be the roots of the equation $(\sqrt[3]{1+a}-1)x^2 + (\sqrt{1+a}-1)x + (\sqrt[6]{1+a}-1) = 0$ where $a > -1$. Then $\lim_{a 	o 0^+} \alpha(a)$ and $\lim_{a 	o 0^+} \beta(a)$ are

Options:

(A) $-5/2$ and 1

(B) $-1/2$ and $-1$

(C) $-7/2$ and 2

(D) $-9/2$ and 3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex-looking quadratic equation: 
$$(\sqrt[3]{1+a}-1)x^2 + (\sqrt{1+a}-1)x + (\sqrt[6]{1+a}-1) = 0$$

At first glance, it looks intimidating. You see cube roots, square roots, and sixth roots, all tangled up with a parameter $a$.

Your instinct might be to panic or to try and plug in $a=0$ immediately. But hold on! If you do that, you get $0=0$. The equation vanishes.

This is the first trap of JEE Advanced. It is not a dead end; it is an invitation to look deeper. We are not looking for the roots at $a=0$; we are looking for the limit of the roots as $a$ approaches zero. This is a subtle but crucial distinction.

### The Surgical Strike: Normalization

Since we are dealing with a limit as $a 	o 0^+$, we know that $a$ is not zero. This gives us the freedom to perform a surgical strike on the equation.

We can divide the entire expression by $a$. When we divide by $a$, the equation transforms into:

$$\frac{(1+a)^{\frac{1}{3}} - 1}{a}x^2 + \frac{(1+a)^{\frac{1}{2}} - 1}{a}x + \frac{(1+a)^{\frac{1}{6}} - 1}{a} = 0$$

Now, look at those coefficients. They are no longer just vanishing; they are begging to be evaluated using the standard limit:

$$\lim_{h 	o 0} \frac{(1+h)^n - 1}{h} = n$$

This is the moment where the complexity collapses into elegance.

### The Symphony of Limits

Let us evaluate each coefficient one by one. For the $x^2$ term, we have $n = \frac{1}{3}$, so the limit is $\frac{1}{3}$.

For the $x$ term, $n = \frac{1}{2}$, so the limit is $\frac{1}{2}$. For the constant term, $n = \frac{1}{6}$, so the limit is $\frac{1}{6}$.

Suddenly, the terrifying equation has become a simple, clean quadratic:

$$\frac{1}{3}x^2 + \frac{1}{2}x + \frac{1}{6} = 0$$

To make it even friendlier, we multiply the entire equation by $6$, giving us:

$$2x^2 + 3x + 1 = 0$$

### Final Calculation

This is the kind of equation you have solved a thousand times. We split the middle term: $2x^2 + 2x + x + 1 = 0$.

This factors beautifully into:

$$(2x+1)(x+1) = 0$$

The roots are **$x = -\frac{1}{2}$** and **$x = -1$**.

### The Takeaway

What have we learned? We learned that in JEE Advanced, complexity is often a mask.

When you see a parameter causing an equation to vanish, do not retreat. Instead, normalize the equation, find the rate of change, and watch as the chaos organizes itself into a simple, solvable form.

You have the tools; you just need the courage to use them. Keep practicing, keep questioning, and most importantly, keep falling in love with the process of discovery.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $α (a)$ and $β (a)$ be the roots of the equation $(3 1 + a - 1) x^{2} + (1 + a - 1) x + (6 1 + a - 1) = 0$ where $a > - 1$ . Then $lim_{a \to 0^{+}} α (a)$ and $lim_{a \to 0^{+}} β (a)$ are

Select Answer:

Visualized Solution

Analyze the Given Equation

Transforming the Equation

The Standard Limit Tool

Limit of $x^{2}$ Coefficient

Limit of $x$ Coefficient

Limit of Constant Term

The Limiting Equation

Simplify the Quadratic

Factorize the Equation

Final Roots

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Surgical Strike

The Symphony of Limits

Final Calculation

The Takeaway

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