For t -1, let αt and βt be the roots of the equation ((t+2)^(1/7) - 1)x² + ((t+2)^(1/6) - 1)x + ((

Q: Wait, why can't we just solve for the roots using the quadratic formula?

That is a classic trap! If you try to use the quadratic formula, you will end up with a massive, unmanageable expression involving square roots of functions of $t$. The goal of this problem is to test your ability to use the properties of roots (Vieta's formulas) to bypass the algebra. Always look for the shortcut!

Q: Why does $t \to -1^+$ imply $y \to 1^+$?

This is a simple coordinate shift. We defined $y = t+2$. If $t$ approaches $-1$ from the right (meaning $t$ is slightly larger than $-1$), then $t+2$ approaches $-1+2 = 1$ from the right. It is just a way to simplify the expression to a standard limit form.

Q: How do I know to divide by $y-1$ in the limit?

The expression $\frac{y^n - 1}{y^m - 1}$ is an indeterminate form ($0/0$) at $y=1$. Dividing both the numerator and the denominator by $y-1$ is the standard calculus technique to transform it into the fundamental limit $\lim_{y \to 1} \frac{y^n - 1}{y - 1} = n$. It is a powerful tool in your JEE arsenal!

Question

For $t > -1$, let $\alpha_t$ and $\beta_t$ be the roots of the equation $((t+2)^{\frac{1}{7}} - 1)x^2 + ((t+2)^{\frac{1}{6}} - 1)x + ((t+2)^{\frac{1}{21}} - 1) = 0$. If $\lim_{t 	o -1^+} \alpha_t = a$ and $\lim_{t 	o -1^+} \beta_t = b$, then $72(a+b)^2$ is equal to ________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before this intimidating quadratic equation:

$$((t+2)^{\frac{1}{7}} - 1)x^2 + ((t+2)^{\frac{1}{6}} - 1)x + ((t+2)^{\frac{1}{21}} - 1) = 0$$

At first glance, it looks like a nightmare. The coefficients are not just numbers; they are complex functions of $t$.

If you were to dive in and try to solve for $x$ using the quadratic formula, you would be walking straight into a trap. The algebra would explode, and you would be lost in a sea of fractional powers.

In JEE Advanced, when you see a problem that looks like it requires brute force, there is almost always an elegant, hidden path.

### The Elegant Shortcut: Vieta's Wisdom

The key to this problem lies in the relationship between the roots and the coefficients of a quadratic equation. For any quadratic $Ax^2 + Bx + C = 0$, we know that the sum of the roots is $\alpha + \beta = -\frac{B}{A}$.

We do not need to know what $\alpha_t$ and $\beta_t$ are individually; we only need their sum! By focusing on the sum, we bypass the need to solve for the roots entirely.

Our target is $a+b = \lim_{t 	o -1^+} (\alpha_t + \beta_t)$. Using Vieta's formulas, this becomes:

$$\lim_{t 	o -1^+} -\frac{(t+2)^{\frac{1}{6}} - 1}{(t+2)^{\frac{1}{7}} - 1}$$

### The Calculus Transformation

Now, the problem has shifted from algebra to calculus. We have a limit to evaluate. Let us simplify the expression by introducing a new variable.

Let $y = t+2$. As $t 	o -1^+$, $y$ approaches $1^+$. Our expression now becomes:

$$-\lim_{y 	o 1^+} \frac{y^{\frac{1}{6}} - 1}{y^{\frac{1}{7}} - 1}$$

This is a classic indeterminate form of the type $0/0$. To resolve this, we use the standard limit formula: $\lim_{x 	o 1} \frac{x^n - 1}{x - 1} = n$.

By dividing both the numerator and the denominator by $y-1$, we get:

$$-\frac{\lim_{y 	o 1^+} \frac{y^{\frac{1}{6}} - 1}{y - 1}}{\lim_{y 	o 1^+} \frac{y^{\frac{1}{7}} - 1}{y - 1}}$$

Applying the formula, the numerator becomes $\frac{1}{6}$ and the denominator becomes $\frac{1}{7}$. Thus, $a+b = -\frac{1/6}{1/7} = -\frac{7}{6}$.

### The Final Victory

We have arrived at the value of $a+b$. The question asks for $72(a+b)^2$.

Substituting our value, we get:

$$72 	imes \left(-\frac{7}{6}ight)^2 = 72 	imes \frac{49}{36}$$

Notice how the numbers align perfectly? $72$ divided by $36$ is $2$.

So, we are left with $2 	imes 49 = 98$. The **final answer is 98**.

Enter Numerical Value:

Visualized Solution

Analyze the Quadratic Equation

Relation Between Roots and Coefficients

Applying Sum of Roots Formula

Defining the Limit $a + b$

Variable Substitution for Simplification

Applying the Standard Limit Formula

Evaluating the Limit

Final Calculation Setup

Final Computation

Conclusion and Key Takeaway

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

Analyzing the Setup

The Elegant Shortcut

The Calculus Transformation

The Final Victory

Similar Questions

Let $a > 0$ be a root of the equation $2 x^{2} + x - 2 = 0$ . If $lim_{x \to \frac{1}{a}} \frac{16 ( 1 - c o s ( 2 + x - 2 x ^{2} ))}{( 1 - a x ) ^{2}} = α + β 17$ , where $α, β \in Z$ , then $α + β$ is equal to _______

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

If $α > β > 0$ are the roots of the equation $a x^{2} + b x + 1 = 0$ , and $lim_{x \to \frac{1}{α}} (\frac{1 - c o s ( x ^{2} + b x + a )}{2 ( 1 - α x ) ^{2}})^{\frac{1}{2}} = \frac{1}{k} (\frac{1}{β} - \frac{1}{α})$ , then $k$ is equal to

If $α, β$ are the distinct roots of $x^{2} + b x + c = 0$ , then $lim_{x \to β} \frac{e ^{2 (x^{2} + b x + c)} - 1 - 2 ( x ^{2} + b x + c )}{( x - β ) ^{2}}$ is equal to:

If $α = lim_{x \to 0^{+}} (\frac{e ^{t a n x} - e ^{x}}{t a n x - x})$ and $β = lim_{x \to 0} (1 + sin x)^{\frac{1}{2} c o t x}$ are the roots of the quadratic equation $a x^{2} + b x - e = 0$ , then $12 lo g_{e} (a + b)$ is equal to__________

Let $α$ and $β$ be the distinct roots of $a x^{2} + b x + c = 0$ , then $lim_{x \to α} \frac{1 - c o s ( a x ^{2} + b x + c )}{( x - α ) ^{2}}$ is equal to

If $lim_{x \to 0} \frac{e ^{(a - 1) x} + 2 c o s b x + ( c - 2 ) e ^{- x}}{x c o s x - l o g _{e} ( 1 + x )} = 2$ , then $a^{2} + b^{2} + c^{2}$ is equal to :

$lim_{x \to 0} \frac{a x ^{2} e ^{x} - b l o g _{e} ( 1 + x ) + c x e ^{- x}}{x ^{2} s i n x} = 1$ then $16 (a^{2} + b^{2} + c^{2})$ is equal to _______.

If $lim_{x \to 0} \frac{e ^{a x} - c o s ( b x ) - \frac{c x e ^{- c x}}{2}}{1 - c o s ( 2 x )} = 17$ , then $5 a^{2} + b^{2}$ is equal to

If $a$ is the positive root of the equation, $p (x) = x^{2} - x - 2 = 0$ , then $lim_{x \to a^{+}} \frac{1 - c o s ( p ( x ))}{x + a - 4}$ is equal to :