If f(x) = det[cos x, x, 1; 2sin x, x², 2x; tan x, x, 1], then lim[x→0] f'(x)/x
(1) exists and is equ

Q: Why can't I just expand the determinant directly using the standard formula?

You absolutely could, but it would be a nightmare of algebraic terms! In JEE Advanced, time is your most precious resource. By performing row operations first, you create zeros, which makes the expansion trivial. Always look for ways to simplify before you calculate.

Q: Is it necessary to use L'Hopital's rule for this limit?

Not at all! While L'Hopital's rule is a powerful tool, it often leads to messy derivatives of derivatives. By simplifying the function $f(x)$ first, we turned a complex limit into a simple algebraic evaluation. Always simplify the function expression before jumping to calculus tools.

Q: Why did we choose $R_1 \to R_1 - R_3$ specifically?

Look at the elements: $R_1$ has $x$ and $1$ in the second and third columns, and $R_3$ also has $x$ and $1$ in those same columns. Subtracting them creates two zeros in the first row. Any row operation that creates zeros is a 'good' operation. You could have also done $R_3 \to R_3 - R_1$, which would just change the sign of the determinant.

Question

If $f(x) = \begin{vmatrix} \cos x & x & 1 \ 2 \sin x & x^2 & 2x \ 	an x & x & 1 \end{vmatrix}$, then $\lim_{x 	o 0} \frac{f'(x)}{x}$

Options:

(A) exists and is equal to 0

(B) exists and is equal to -2

(C) exists and is equal to 2

(D) does not exist

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given the function:

$$f(x) = \begin{vmatrix} \cos x & x & 1 \ 2 \sin x & x^2 & 2x \ 	an x & x & 1 \end{vmatrix}$$

We are tasked with finding the limit:

$$\lim_{x 	o 0} \frac{f'(x)}{x}$$

Many students see a $3 	imes 3$ determinant and immediately start expanding along the first row. Do not be that student. The secret to JEE Advanced is not brute force; it is observation.

### The Power of Observation

Look at row 1 ($R_1$) and row 3 ($R_3$). They are almost identical in the second and third columns, as the second element in both is $x$ and the third element is $1$.

Whenever you see identical or proportional elements in a determinant, you should immediately think of row operations to simplify it. We will apply the operation $R_1 	o R_1 - R_3$.

This creates zeros in the first row, making expansion much easier. After the subtraction, our determinant becomes:

$$f(x) = \begin{vmatrix} \cos x - 	an x & 0 & 0 \ 2 \sin x & x^2 & 2x \ 	an x & x & 1 \end{vmatrix}$$

### The Expansion

Now, we expand the determinant along the first row. Because of the two zeros, we only need to multiply the first element, $(\cos x - 	an x)$, by its corresponding $2 	imes 2$ minor.

The minor is formed by hiding the first row and the first column:

$$	ext{Minor} = \begin{vmatrix} x^2 & 2x \ x & 1 \end{vmatrix}$$

Calculating the value of this $2 	imes 2$ minor via cross-multiplication:

$$x^2(1) - (2x)(x) = x^2 - 2x^2 = -x^2$$

Multiplying this with our first term, we get $f(x) = -x^2(\cos x - 	an x)$. To simplify, we absorb the negative sign:

$$f(x) = x^2(	an x - \cos x)$$

### The Derivative and the Limit

Now that we have a clean expression for $f(x)$, we find its derivative $f'(x)$ using the product rule:

$$f'(x) = 2x(	an x - \cos x) + x^2(\sec^2 x + \sin x)$$

We need to evaluate the limit of $\frac{f'(x)}{x}$. Dividing our derivative expression by $x$:

$$\frac{f'(x)}{x} = \frac{2x(	an x - \cos x) + x^2(\sec^2 x + \sin x)}{x}$$

Since $x 	o 0$ (but $x 
eq 0$), we cancel $x$ from the numerator and denominator:

$$\frac{f'(x)}{x} = 2(	an x - \cos x) + x(\sec^2 x + \sin x)$$

### Final Calculation

Finally, we apply the limit as $x 	o 0$:

$$\lim_{x 	o 0} [2(	an x - \cos x) + x(\sec^2 x + \sin x)]$$

Substituting $x = 0$, where $	an 0 = 0$ and $\cos 0 = 1$:

$$2(0 - 1) + 0 \cdot (\sec^2 0 + \sin 0) = 2(-1) + 0$$

The limit exists and is exactly equal to **$-2$**.

Animated Solution for Mathematics - Matrices and Determinants: If $f (x) = cos x 2 sin x tan x x x^{2} x 1 2 x 1$ , then $lim_{x \to 0} \frac{f ^{'} ( x )}{x}$

Select Answer:

Visualized Solution

Defining the Function $f (x)$

Analyzing the Determinant Structure

Applying Row Operation $R_{1} \to R_{1} - R_{3}$

The Simplified Determinant

Expanding Along the First Row

Simplifying the Function $f (x)$

Finding the Derivative $f^{'} (x)$

Setting up the Limit $\frac{f ^{'} ( x )}{x}$

Simplifying the Expression

Evaluating the Limit as $x \to 0$

Final Calculation and Result

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Power of Observation

The Expansion

The Derivative and the Limit

Final Calculation

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=​cosx2sinxtanx​xx2x​12x1​​, then limx→0​xf′(x)​

Select Answer:

Visualized Solution

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

Analyzing the Determinant Structure

Applying Row Operation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​→R1​−R3​

The Simplified Determinant

Expanding Along the First Row

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

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

Setting up the Limit .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xf′(x)​