For some a, b, let
f(x) = | [a+(sin x)/x, 1, b], [a, 1+(sin x)/x, b], [a, 1, b+(sin x)/x] |
x≠0, lim

Q: Wait, why can't I just expand the determinant first and then take the limit? Isn't that safer?

You absolutely can, but it is algebraically exhausting! Expanding a $3 \times 3$ determinant with variable expressions will lead to a massive, messy polynomial. By simplifying the matrix *before* expanding, you reduce the complexity significantly. It is the 'pro' way to handle these problems—work smarter, not harder!

Q: Why are we allowed to use row operations here? Does it change the limit?

Great question! Row operations like $R_i \to R_i - R_j$ are fundamental tools in linear algebra. They do not change the value of the determinant itself. Since the limit is just a number, and the determinant is a function of $x$, simplifying the determinant expression for any $x \neq 0$ is perfectly valid. The limit of the simplified expression will be the same as the limit of the original.

Q: How do I know which row operations to pick? I always get stuck choosing.

The goal of row operations is to create zeros. Look at the matrix: we have $1$s and $b$s in the columns. By subtracting rows, we can eliminate these terms. Always look for patterns—if you see identical terms in a column, subtracting those rows is almost always the right path to create the zeros you need for an easy expansion.

Question

For some a, b, let
$f(x)=|\begin{matrix}a+\frac{sin~x}{x}&1&b\ a&1+\frac{sin~x}{x}&b\ a&1&b+\frac{sin~x}{x}\end{matrix}|$
$x
e0$, $lim_{xightarrow0}f(x)=\lambda+\mu a+
u b$. Then
$(\lambda+\mu+
u)^{2}$ is equal to:

Options:

(A) 16

(B) 25

(C) 9

(D) 36

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex, intimidating $3 	imes 3$ determinant. It looks like a fortress of variables and trigonometric functions, designed to block your path. But in the world of JEE Advanced, every fortress has a secret door.

Our function is defined as:

$$f(x) = \begin{vmatrix} a+\frac{\sin x}{x} & 1 & b \ a & 1+\frac{\sin x}{x} & b \ a & 1 & b+\frac{\sin x}{x} \end{vmatrix}$$

At first glance, the term $\frac{\sin x}{x}$ might seem like a nuisance. However, we are looking for the limit as $x 	o 0$. This is the moment where the chaos settles.

We know the fundamental standard limit:

$$\lim_{x 	o 0} \frac{\sin x}{x} = 1$$

This is our magic key. By substituting this value, the entire matrix transforms into a simple, constant-filled structure. Let's call this simplified determinant $L$:

$$L = \begin{vmatrix} a+1 & 1 & b \ a & 2 & b \ a & 1 & b+1 \end{vmatrix}$$

### The Elegance of Simplification

Now, we could expand this determinant directly, but that would be like trying to cut down a tree with a butter knife. Instead, let's use the power of row operations. Our goal is to create zeros, which make the expansion process trivial.

First, let's apply $R_1 	o R_1 - R_2$. The first row becomes $(a+1)-a = 1$, $1-2 = -1$, and $b-b = 0$. Our matrix now looks like this:

$$L = \begin{vmatrix} 1 & -1 & 0 \ a & 2 & b \ a & 1 & b+1 \end{vmatrix}$$

See how much cleaner that is? But we can do better. Let's apply $R_2 	o R_2 - R_3$. The second row becomes $a-a = 0$, $2-1 = 1$, and $b-(b+1) = -1$.

Now we have:

$$L = \begin{vmatrix} 1 & -1 & 0 \ 0 & 1 & -1 \ a & 1 & b+1 \end{vmatrix}$$

### The Final Expansion

Now, the expansion is a breeze. Expanding along the first row ($R_1$), we get:

$$L = 1 \cdot \begin{vmatrix} 1 & -1 \ 1 & b+1 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 0 & -1 \ a & b+1 \end{vmatrix} + 0$$

Let's calculate those minors carefully. The first minor is $1(b+1) - (-1)(1) = b+2$. The second minor is $0(b+1) - (-1)(a) = a$.

Adding them together, we get:

$$L = (b+2) + a = a + b + 2$$

### The Comparison

The problem tells us that $\lim_{x 	o 0} f(x) = \lambda + \mu a + 
u b$. We have found that our limit is $2 + 1 \cdot a + 1 \cdot b$. By comparing the coefficients, we find:

$$\lambda = 2, \mu = 1, 
u = 1$$

Finally, we need to calculate $(\lambda + \mu + 
u)^2$. Substituting our values, we get:

$$(2 + 1 + 1)^2 = 4^2 = \mathbf{16}$$

And there it is! The fortress has fallen. By using the standard limit and the elegance of row operations, we turned a terrifying problem into a simple, satisfying victory.

Animated Solution for Mathematics - Matrices and Determinants: For some a, b, let $f (x) = ∣ a + \frac{s in x}{x} a a 1 1 + \frac{s in x}{x} 1 b b b + \frac{s in x}{x} ∣$ $x \neq = 0$ , $l i m_{x \to 0} f (x) = λ + μ a + ν b$ . Then $(λ + μ + ν)^{2}$ is equal to:

Select Answer:

Visualized Solution

The Function $f (x)$

Standard Limit $lim_{x \to 0} \frac{s i n x}{x}$

Substituting the Limit Value

Row Operation $R_{1} \to R_{1} - R_{2}$

Row Operation $R_{2} \to R_{2} - R_{3}$

Expanding along $R_{1}$

Simplifying the Expression

Comparing with $λ + μ a + ν b$

Finding $λ, μ, ν$

Final Numerical Value

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Elegance of Simplification

The Final Expansion

The Comparison

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Visualized Solution

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

Standard Limit .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→0​xsinx​

Substituting the Limit Value

Row Operation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​→R1​−R2​

Row Operation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​→R2​−R3​

Expanding along .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​

Simplifying the Expression

Comparing with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ+μa+νb

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ,μ,ν