Let x=2 be a root of the equation x² + px + q = 0 and f(x) = { (1 - cos(x² - 4px + q² + 8q + 16))/(x

Q: Wait, why can't we just say the limit is 1/2 and the answer is [1/2] = 0?

That is a very common trap! The property $\lim_{x \to a} [f(x)] = [\lim_{x \to a} f(x)]$ is only true if the limit of $f(x)$ is not an integer. However, even if the limit is not an integer, you must be absolutely certain about the *direction* of approach. If $f(x)$ approaches $1/2$ from the left (i.e., $f(x) 1/2$), the GIF would still be $0$ in this specific case, but if the limit were $1$, the direction would change everything. Always check the sign of the next term in the Taylor expansion!

Q: Is there a faster way to solve this than Taylor series?

You could use L'Hopital's Rule, but you would have to apply it twice. The first application would give you a limit of $0/0$, and the second would give you the result. However, Taylor series is generally safer and more elegant for these types of problems because it explicitly shows you the 'correction term' (the $-t^2/24$ part), which tells you exactly how the function behaves near the limit point.

Q: Why did we substitute $t = (x - 2p)^2$?

We substituted $t$ to transform a complex expression into a standard limit form. In JEE Advanced, whenever you see $1 - \cos(\text{something})$, your brain should immediately think of the standard limit $\lim_{u \to 0} \frac{1 - \cos u}{u^2} = \frac{1}{2}$. By setting $t = (x - 2p)^2$, we mapped the complex argument directly to $u$, making the calculus trivial.

Question

Let $x=2$ be a root of the equation $x^2 + px + q = 0$ and $f(x) = \begin{cases} \frac{1 - \cos(x^2 - 4px + q^2 + 8q + 16)}{(x - 2p)^4}, & x 
eq 2p \ 0, & x = 2p \end{cases}$. Then $\lim_{x 	o 2p} [f(x)]$ where $[.]$ denotes greatest integer function, is

Options:

(A) 2

(B) 1

(C) 0

(D) -1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We begin with the quadratic equation $x^2 + px + q = 0$. We are given that $x=2$ is a root, which implies it must satisfy the equation:

$$2^2 + p(2) + q = 0$$

Simplifying this, we obtain $4 + 2p + q = 0$. From this, we extract our golden ticket:

$$q = -(2p + 4)$$

Keep this relation locked in your mind; we will need it to simplify the complex expression later.

### The Beast Within the Cosine

Now, consider the function $f(x)$ where the argument of the cosine function is $x^2 - 4px + q^2 + 8q + 16$. At first glance, this appears to be a chaotic collection of variables. However, notice that the tail end, $q^2 + 8q + 16$, is a perfect square:

$$q^2 + 8q + 16 = (q + 4)^2$$

Recall our golden ticket: $q = -2p - 4$, which implies $q + 4 = -2p$. Substituting this into our perfect square, we get:

$$(q + 4)^2 = (-2p)^2 = 4p^2$$

The argument of the cosine function now becomes $x^2 - 4px + 4p^2$. This is another perfect square, allowing the entire expression to collapse into:

$$(x - 2p)^2$$

### The Limit and the Taylor Insight

Our function is now simplified to:

$$f(x) = \frac{1 - \cos((x - 2p)^2)}{(x - 2p)^4}$$

To make this manageable, let us define a new variable $t = (x - 2p)^2$. As $x$ approaches $2p$, $t$ approaches $0$ from the positive side. Our limit expression becomes:

$$\lim_{t 	o 0^+} \frac{1 - \cos t}{t^2}$$

We know the standard limit $\lim_{t 	o 0} \frac{1 - \cos t}{t^2} = \frac{1}{2}$. However, the question asks for the limit of the Greatest Integer Function $[f(x)]$. We must determine if $f(t)$ approaches $1/2$ from above or below.

Using the Taylor series expansion $\cos t \approx 1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \dots$, we substitute this into our function:

$$f(t) \approx \frac{1 - (1 - \frac{t^2}{2} + \frac{t^4}{24})}{t^2} = \frac{1}{2} - \frac{t^2}{24}$$

Because $t^2$ is always positive, we are subtracting a tiny positive value from $1/2$. This confirms that $f(t)$ is strictly less than $1/2$.

### The Final Stroke

We have established that as $t$ approaches $0$, $f(t)$ approaches $1/2$ from below. This implies that in the immediate neighborhood of the limit, the values of $f(t)$ satisfy the inequality:

$$0 < f(t) < \frac{1}{2}$$

When we apply the Greatest Integer Function to any value in this interval, the result is always $0$. The graph of this function shows a curve creeping up toward the line $y = 1/2$ but never reaching it.

Therefore, the final answer is **0**.

Select Answer:

Visualized Solution

The Root Condition

Finding the Relation

Analyzing the Cosine Argument

Perfect Square Identification

Simplifying the Argument

Rewriting the Limit

Variable Substitution

Evaluating the Standard Limit

The JEE Trap: Greatest Integer Function

Taylor Series Expansion

Analyzing the Approach Direction

Visualizing the Approach

Applying the GIF

Final Conclusion

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

Analyzing the Setup

The Beast Within the Cosine

The Limit and the Taylor Insight

The Final Stroke

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