Let a 0 be a root of the equation 2x² + x - 2 = 0. If lim(x→1/a) 16(1 - cos(2+x-2x²))/(1 - ax)² =

Q: Wait, why do we discard the negative root of the quadratic equation $2x^2 + x - 2 = 0$?

That is a fantastic question! The problem statement explicitly gives us the condition $a > 0$. When we solve the quadratic equation using the formula $x = \frac{-1 \pm \sqrt{17}}{4}$, we get two roots: one positive and one negative. Since the problem defines $a$ as a root that is strictly positive, we must reject the negative root to satisfy the given constraint. Always keep an eye on these domain restrictions; they are often the first place where students lose marks!

Q: Can I use Taylor series expansion instead of L'Hopital's Rule for the inner limit?

Absolutely! In fact, using the Taylor series expansion for $\cos \theta$ is a very elegant and powerful approach. Since $\cos \theta \approx 1 - \frac{\theta^2}{2}$ for small $\theta$, you can substitute this directly into the numerator. It often leads to the same result faster and with less risk of differentiation errors. Both methods are perfectly valid in JEE Advanced, so choose the one you are most comfortable with.

Q: Why does the limit of the inner part squared work? Can we just move the limit inside the square?

Yes, we can! Because the function $f(x) = x^2$ is a continuous function, the limit of the square of a function is equal to the square of the limit of that function, provided the limit exists. Mathematically, $\lim_{x \to c} [f(x)]^2 = [\lim_{x \to c} f(x)]^2$. This property is a lifesaver in complex limit problems because it allows us to break a massive, intimidating expression into smaller, manageable pieces.

Question

Let $a > 0$ be a root of the equation $2x^2 + x - 2 = 0$. If $\lim_{x 	o \frac{1}{a}} \frac{16(1 - \cos(2+x-2x^2))}{(1 - ax)^2} = \alpha + \beta\sqrt{17}$, where $\alpha, \beta \in \mathbb{Z}$, then $\alpha + \beta$ is equal to _______

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of radicals and trigonometric functions.

As we peel back the layers, you will see that it is actually a beautifully choreographed dance of limits and algebra. Let us begin by finding our anchor: the root $a$.

We are given the quadratic equation $2x^2 + x - 2 = 0$. Using the quadratic formula, we find the roots to be:

$$x = \frac{-1 \pm \sqrt{17}}{4}$$

The problem demands $a > 0$, so we discard the negative root and embrace:

$$a = \frac{\sqrt{17} - 1}{4}$$

This value is our compass; keep it close.

### The Indeterminate Trap

Now, look at our limit:

$$L = \lim_{x 	o \frac{1}{a}} \frac{16(1 - \cos(2+x-2x^2))}{(1 - ax)^2}$$

If you try to plug in $x = \frac{1}{a}$ immediately, you will see the denominator vanish into $0$. Inside the cosine, we have $2 + \frac{1}{a} - 2(\frac{1}{a})^2$.

If we find a common denominator, this becomes $\frac{2a^2 + a - 2}{a^2}$. Since $a$ is a root of $2x^2 + x - 2 = 0$, the numerator $2a^2 + a - 2$ is exactly zero.

Thus, we have a $\frac{0}{0}$ indeterminate form. This is not a wall; it is a doorway.

### The Magic of the Standard Limit

We invoke the most powerful identity in our limit toolkit:

$$\lim_{	heta 	o 0} \frac{1 - \cos 	heta}{	heta^2} = \frac{1}{2}$$

Let $	heta = 2 + x - 2x^2$. As $x 	o \frac{1}{a}$, $	heta 	o 0$. We multiply and divide our expression by $	heta^2$, effectively creating the standard limit form.

The expression becomes:

$$16 \cdot \frac{1}{2} \cdot \lim_{x 	o \frac{1}{a}} \left( \frac{2 + x - 2x^2}{1 - ax} ight)^2$$

We have successfully reduced the problem to evaluating the limit of the inner fraction, which we will call $M$.

### Taming the Inner Limit with L'Hopital

Now we face:

$$M = \lim_{x 	o \frac{1}{a}} \frac{2 + x - 2x^2}{1 - ax}$$

It is still $\frac{0}{0}$, so we call upon L'Hopital's Rule. We differentiate the numerator to get $1 - 4x$ and the denominator to get $-a$.

Substituting $x = \frac{1}{a}$, we get:

$$M = \frac{1 - 4/a}{-a} = \frac{4 - a}{a^2}$$

We have simplified the entire limit $L$ to:

$$L = 8 \cdot M^2 = 8 \cdot \left( \frac{4 - a}{a^2} ight)^2 = \frac{8(4 - a)^2}{a^4}$$

### The Final Algebraic Ascent

Now, the heavy lifting begins. We substitute $a = \frac{\sqrt{17} - 1}{4}$ into our expression. Calculating $(4 - a)^2$ and $a^4$ requires patience and precision.

After careful expansion and simplification, we find:

$$(4 - a)^2 = \frac{153 - 17\sqrt{17}}{8}$$

$$a^4 = \frac{49 - 9\sqrt{17}}{32}$$

When we plug these back into our expression for $L$, the terms begin to cancel in a way that feels almost magical. We are left with:

$$L = \frac{32(153 - 17\sqrt{17})}{49 - 9\sqrt{17}}$$

By factoring out $17$ and rationalizing the denominator, the expression collapses into $153 + 17\sqrt{17}$.

Comparing this to $\alpha + \beta\sqrt{17}$, we find $\alpha = 153$ and $\beta = 17$. The sum $\alpha + \beta = 170$.

We have reached the summit. The complexity has vanished, leaving behind only the elegant truth of the answer: **170**.

Enter Numerical Value:

Visualized Solution

Find the Positive Root $a$

Analyze the Limit Form

Apply Standard Limit Identity

Evaluate the Inner Limit

Simplify the Inner Limit

Calculate the Numerator Term $(4 - a)^{2}$

Calculate the Denominator Term $a^{4}$

Substitute into the Limit Expression

Rationalize the Denominator

Final Simplification and Comparison

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

Analyzing the Setup

The Indeterminate Trap

The Magic of the Standard Limit

Taming the Inner Limit with L'Hopital

The Final Algebraic Ascent

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