If the function
f(x) = {
(2/x){sin(k_1+1)x + sin(k_2-1)x}, x

Q: Wait, why do we equate both limits to 4? Shouldn't it just be one of them?

Great question! Continuity is a strict condition. For a function to be continuous at a point $x=c$, the limit from the left, the limit from the right, and the value of the function at that point must all be identical. If any one of these differs, the graph would have a 'jump' or a 'hole', breaking the continuity. Thus, $LHL = RHL = f(0)$ is non-negotiable.

Q: I got stuck on the log limit. Why did we factor out a 2 inside the log?

The standard limit $\lim_{x \to 0} \frac{\log(1+u)}{u} = 1$ requires the argument of the log to be in the form $(1+u)$. Our expression was $\log(2+k_1x)$. By factoring out the 2, we get $\log(2(1 + \frac{k_1}{2}x))$, which splits into $\log(2) + \log(1 + \frac{k_1}{2}x)$. This allows us to isolate the part that depends on $x$ and apply the limit property cleanly.

Q: Can I use L'Hôpital's Rule instead of standard limits?

Absolutely! L'Hôpital's Rule is a powerful tool for $0/0$ indeterminate forms. Since both limits result in $0/0$ at $x=0$, you can differentiate the numerator and denominator with respect to $x$. However, be careful with the chain rule when differentiating the log and sine terms. Standard limits are often faster, but L'Hôpital's is a perfectly valid path to the same result.

Q: Why did the $\log(2)$ term disappear in the RHL calculation?

When we expanded $\log(2+k_1x) - \log(2+k_2x)$, it became $(\log 2 + \log(1+\frac{k_1}{2}x)) - (\log 2 + \log(1+\frac{k_2}{2}x))$. The $\log 2$ terms cancel each other out perfectly! This is a beautiful piece of algebraic symmetry that simplifies the problem significantly.

Q: Is there any scenario where $k_1$ or $k_2$ could be complex numbers?

In the context of JEE Advanced, unless specified otherwise, we assume $k_1$ and $k_2$ are real constants. The continuity condition leads to a system of linear equations with real coefficients, which naturally yields real values for our variables. Stick to the real number domain, and you will be on the right track!

Question

If the function
$f(x)=\begin{cases}\frac{2}{x}\{\sin(k_{1}+1)x+\sin(k_{2}-1)x\},&x<0\ 4,&x=0\ \frac{2}{x}\log_{e}(\frac{2+k_{1}x}{2+k_{2}x}),&x>0\end{cases}$
is continuous at $x=0$, then $k_{1}^{2}+k_{2}^{2}$ is equal to

Options:

(A) 20

(B) 5

(C) 8

(D) 10

Sigma Percentile · Accepted Answer

### The Architecture of Continuity: Bridging the Gap

Imagine you are an architect tasked with building a bridge across a chasm at $x=0$. The bridge must be perfectly smooth; there can be no sudden jumps or disjointed sections. In the language of calculus, this is the essence of continuity.

We are given a function $f(x)$ defined in three different ways depending on whether we are to the left of the chasm, exactly at the chasm, or to the right of it. Our goal is to find the parameters $k_1$ and $k_2$ that ensure this bridge is perfectly connected.

### Phase 1: The LHL - Taming the Trigonometric Beast

To the left of the chasm ($x < 0$), our function is defined as $f(x) = \frac{2}{x} \{\sin(k_1+1)x + \sin(k_2-1)x\}$. As we approach $x=0$ from the left, we are looking for the limit:

$$LHL = \lim_{x 	o 0^-} \frac{2}{x} \{\sin(k_1+1)x + \sin(k_2-1)x\}$$

We can distribute the $\frac{2}{x}$ term to each sine function. Recall the fundamental limit $\lim_{x 	o 0} \frac{\sin(ax)}{x} = a$.

By applying this to each term, the $x$ in the denominator effectively 'cancels' the $x$ inside the sine functions, leaving us with the coefficients. Thus, the limit simplifies to:

$$2[(k_1+1) + (k_2-1)] = 2(k_1 + k_2)$$

Since the function must be continuous at $x=0$ and $f(0)=4$, we set $2(k_1 + k_2) = 4$. This gives us our first elegant equation:

$$k_1 + k_2 = 2$$

### Phase 2: The RHL - Unlocking the Logarithmic Secret

Now, let's look to the right ($x > 0$). Here, the function takes the form $f(x) = \frac{2}{x} \log_e(\frac{2+k_1x}{2+k_2x})$. Again, we approach $x=0$.

Using the property of logarithms, $\log(\frac{a}{b}) = \log(a) - \log(b)$, we split this into two parts:

$$RHL = \lim_{x 	o 0^+} \frac{2}{x} [\log_e(2+k_1x) - \log_e(2+k_2x)]$$

To use the standard limit $\lim_{u 	o 0} \frac{\log(1+u)}{u} = 1$, we need the argument of the log to be $1+u$. By factoring out a $2$ from the arguments, we get $\log(2(1 + \frac{k_1}{2}x)) = \log 2 + \log(1 + \frac{k_1}{2}x)$.

When we subtract the two logarithmic expressions, the $\log 2$ terms vanish into thin air! We are left with:

$$RHL = 2 \left[ \lim_{x 	o 0^+} \frac{\log_e(1+\frac{k_1}{2}x)}{x} - \lim_{x 	o 0^+} \frac{\log_e(1+\frac{k_2}{2}x)}{x} ight]$$

Applying the limit property, this simplifies to $2[\frac{k_1}{2} - \frac{k_2}{2}]$, which is simply $k_1 - k_2$. Setting this equal to $4$, we obtain our second equation:

$$k_1 - k_2 = 4$$

### Phase 3: The Final Synthesis

We now have a simple system of linear equations:

1. $k_1 + k_2 = 2$
2. $k_1 - k_2 = 4$

Adding these two equations, we get $2k_1 = 6$, which means $k_1 = 3$. Substituting this back into the first equation, $3 + k_2 = 2$, we find $k_2 = -1$.

The problem asks for the value of $k_1^2 + k_2^2$. Substituting our values, we get:

$$(3)^2 + (-1)^2 = 9 + 1 = 10$$

**The final result is 10.**

Select Answer:

Visualized Solution

Understanding Continuity at $x = 0$

Setting up the Left Hand Limit (LHL)

Applying Trigonometric Limit Property

Simplifying the LHL Equation

Setting up the Right Hand Limit (RHL)

Manipulating the Logarithmic Terms

Applying Logarithmic Limit Property

Simplifying the RHL Equation

Solving the System of Equations

Final Calculation

The Sigma Insight: Continuity at a Point and in an Interval

The Architecture of Continuity

Phase 1

Phase 2

Phase 3

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