If a is the positive root of the equation, p(x) = x² -x - 2 = 0, then lim x→a⁺ (√(1-cos(p(x))))/(x+a

Q: Why do we need the absolute value sign when taking the square root? Can't we just write $\sin(\frac{p(x)}{2})$?

That is a dangerous assumption! Remember that $\sqrt{u^2}$ is defined as $|u|$, not $u$. If $u$ were negative, $\sqrt{u^2}$ would be $-u$. In JEE, precision is everything. We must check the sign of $p(x)$ as $x \to 2^+$ to see if the modulus opens as positive or negative.

Q: How do we know $p(x) > 0$ as $x \to 2^+$?

Great question. We factored $p(x)$ into $(x-2)(x+1)$. As $x$ approaches $2$ from the right, $x$ is slightly larger than $2$. Thus, $(x-2)$ is a tiny positive number, and $(x+1)$ is approximately $3$. The product of two positive numbers is positive. Therefore, $p(x) > 0$.

Q: Could we have used L'Hopital's rule instead of the standard limit?

Yes, you could, but it would be a nightmare. Differentiating the square root of a cosine function involves the chain rule, which leads to a very messy expression. The standard limit approach is elegant, faster, and much less prone to calculation errors. Always look for the algebraic path first!

Question

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

Options:

(A) $\frac{1}{\sqrt{2}}$

(B) $\frac{1}{2}$

(C) $\frac{3}{\sqrt{2}}$

(D) $\frac{3}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Quadratic Foundation

Welcome, future engineer! Today, we are going to dissect a limit problem that might look intimidating at first glance, but it is actually a beautiful exercise in precision.

We start with the quadratic equation $p(x) = x^2 - x - 2 = 0$. Our first mission is to find the positive root $a$.

By splitting the middle term, we factor this into $(x - 2)(x + 1) = 0$. The roots are $x = 2$ and $x = -1$. Since the problem specifies $a$ is the positive root, we have $a = 2$. This is our anchor point.

### The Trigonometric Transformation

Now, let us look at the limit:
$$\lim_{x 	o 2^+} \frac{\sqrt{1 - \cos(p(x))}}{x - 2}$$

The term $1 - \cos(p(x))$ is a classic signal in calculus. It screams for the half-angle identity: $1 - \cos	heta = 2\sin^2(\frac{	heta}{2})$.

Applying this, our expression transforms into:
$$\frac{\sqrt{2\sin^2(\frac{p(x)}{2})}}{x - 2}$$

### The Modulus Trap

Here is the crucial moment: $\sqrt{2\sin^2(\frac{p(x)}{2})}$ is not simply $\sqrt{2}\sin(\frac{p(x)}{2})$. It is $\sqrt{2}|\sin(\frac{p(x)}{2})|$.

Why? Because $\sqrt{u^2} = |u|$. We must determine the sign of $\sin(\frac{p(x)}{2})$ as $x 	o 2^+$.

Since $x > 2$, $p(x) = (x-2)(x+1)$ is positive. Thus, $\frac{p(x)}{2}$ is a small positive angle, and the sine of a small positive angle is positive. The modulus bars drop away, leaving us with $\sqrt{2}\sin(\frac{p(x)}{2})$.

### The Standard Limit

Finally, we use the standard limit $\lim_{	heta 	o 0} \frac{\sin	heta}{	heta} = 1$. We multiply and divide by $\frac{p(x)}{2}$ to get:

$$\sqrt{2} \cdot \lim_{x 	o 2^+} \left( \frac{\sin(\frac{p(x)}{2})}{\frac{p(x)}{2}} ight) \cdot \frac{p(x)}{2(x - 2)}$$

The first part goes to $1$. The second part becomes:
$$\frac{(x-2)(x+1)}{2(x-2)} = \frac{x+1}{2}$$

Substituting $x=2$, we get $\frac{3}{2}$. Multiplying by our $\sqrt{2}$ from earlier, we arrive at the final result:

**$$\frac{3\sqrt{2}}{2}$$**

You have successfully navigated the trap and solved the problem with mathematical elegance!

Animated Solution for Mathematics - Limits, Continuity and Differentiability: 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 :

Select Answer:

Visualized Solution

Analyze the Quadratic $p (x)$

Factorize $p (x)$

Find the Roots & Identify $a$

Substitute $a$ in the Limit

Apply Trigonometric Identity

Simplify the Square Root

Analyze the Sign of $p (x)$

Remove Absolute Value

Prepare for Standard Limit

Evaluate the Limit Components

Final Calculation

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

Analyzing the Quadratic Foundation

The Trigonometric Transformation

The Modulus Trap

The Standard Limit

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Visualized Solution

Analyze the Quadratic .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p(x)

Factorize .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p(x)

Find the Roots & Identify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

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