Determinants JEE 2024: Row Operations and Chain Rule

What does the determinants JEE 2024 question ask you to calculate?

The determinants JEE 2024 question below has numerical answer 0. The decisive step is the chain rule: expanding the determinant alone does not give the requested derivative.

f (x) = | \begin{matrix} 2 \cos^{4} x & 2 \sin^{4} x & 3 + \sin^{2} (2 x) \\ 3 + 2 \cos^{4} x & 2 \sin^{4} x & \sin^{2} x \\ 2 \cos^{4} x & 3 + 2 \sin^{4} x & \sin^{2} (2 x) \end{matrix} | .

Calculate one-fifth of its derivative at zero:

\frac{1}{5} f^{'} (0) .

This is a numerical-answer question, with no supplied answer options. The question bank labels it Hard, with a target time of 300 seconds, or 5 minutes.

Here, 2024 is the question-bank year label. Its date metadata conflicts with that label, so no exact paper date or shift is asserted.

Which row operations should you use before expanding?

Subtract the first row from each lower row. This creates two zeros, leaving simpler minors than a direct expansion of the original trigonometric entries.

Adding a multiple of another row leaves a determinant unchanged. Apply:

R_{2} \leftarrow R_{2} - R_{1}, R_{3} \leftarrow R_{3} - R_{1} .

The transformed determinant is:

f (x) = | \begin{matrix} 2 \cos^{4} x & 2 \sin^{4} x & 3 + \sin^{2} (2 x) \\ 3 & 0 & \sin^{2} x - 3 - \sin^{2} (2 x) \\ 0 & 3 & - 3 \end{matrix} | .

Name the longer entry before expanding. This keeps the signs visible without repeatedly copying the same expression:

D = \sin^{2} x - 3 - \sin^{2} (2 x) .

Expand along the first row, keeping the positive, negative, positive cofactor signs:

f (x) = & 2 \cos^{4} x [0 (- 3) - 3 D] & - 2 \sin^{4} x [3 (- 3) - 0 D] & + [3 + \sin^{2} (2 x)] [3 \cdot 3 - 0 \cdot 0] .

The middle contribution is positive because its minor and cofactor sign are both negative. The last minor equals nine, giving:

f (x) = - 6 \cos^{4} x D + 18 \sin^{4} x + 27 + 9 \sin^{2} (2 x) .

Use these row operations before differentiating. They remove repeated entries while preserving the determinant’s value.

How do you turn the trigonometric determinant into a polynomial?

Use sine squared as the new variable. Every trigonometric term then becomes a polynomial, so you can check the expansion coefficient by coefficient.

The substitutions are:

t & = \sin^{2} x, \cos^{4} x & = (1 - t)^{2}, \sin^{4} x & = t^{2}, \sin^{2} (2 x) & = 4 t (1 - t), D & = t - 3 - 4 t (1 - t) = 4 t^{2} - 3 t - 3 .

Insert them into the simplified determinant. Keep the final sine-double-angle contribution separate until coefficient collection:

f (x) = & - 6 (1 - 2 t + t^{2}) (4 t^{2} - 3 t - 3) & + 18 t^{2} + 27 + 36 t - 36 t^{2} .

Multiply the second bracket by each term of the first bracket in turn:

& (1 - 2 t + t^{2}) (4 t^{2} - 3 t - 3) & = (4 t^{2} - 3 t - 3) & + (- 8 t^{3} + 6 t^{2} + 6 t) & + (4 t^{4} - 3 t^{3} - 3 t^{2}) & = 4 t^{4} - 11 t^{3} + 7 t^{2} + 3 t - 3 .

Now distribute negative six and collect every contribution. The quadratic coefficient receives three separate terms:

f (x) = & - 24 t^{4} + 66 t^{3} & + (- 42 + 18 - 36) t^{2} & + (- 18 + 36) t + (18 + 27) .

This gives the official polynomial:

f (x) = - 24 t^{4} + 66 t^{3} - 60 t^{2} + 18 t + 45 .

Audit the constant using the simplified determinant and the polynomial, respectively:

f (0) & = - 6 (1) (- 3) + 0 + 27 + 0 = 45, f (0) & = 45 .

Matching constants is a check, not proof of every coefficient. Errors in higher-power terms are invisible when the substituted variable is zero.

How does the chain rule give the answer at zero?

The derivative is zero because the polynomial derivative must be multiplied by the derivative of sine squared. That multiplier vanishes at zero, even though the polynomial derivative itself does not.

Define the composition explicitly:

P (t) = - 24 t^{4} + 66 t^{3} - 60 t^{2} + 18 t + 45, f (x) = P (t (x)) .

Differentiate each polynomial term, then differentiate the inner function:

P^{'} (t) = - 96 t^{3} + 198 t^{2} - 120 t + 18 .

\frac{d t}{d x} = 2 \sin x \cos x = \sin (2 x) .

Therefore:

f^{'} (x) = [- 96 t^{3} + 198 t^{2} - 120 t + 18] \sin (2 x) .

At the required input:

x = 0, t = 0, \sin (2 x) = 0,

$f^{'} (0) = 18 \cdot 0 = 0 .$ Final numerical answer:

\frac{f^{'} (0)}{5} = 0 .

Keep these three quantities separate:

Determinant value: $f (0) = 45 .$
Polynomial derivative: $P^{'} (0) = 18 .$
Derivative with respect to the original variable: $f^{'} (0) = 0 .$

Only the last supplies the derivative requested in the question.

Why is differentiating only the polynomial incorrect?

The polynomial derivative is taken with respect to the substituted variable, not the original variable. Changing how the function is written does not change which derivative the question requests.

Differentiating the polynomial, substituting zero and dividing by five gives this incorrect calculation:

P^{'} (0) = 18 ⟹ incorrect required value = \frac{18}{5} .

The missing distinction is:

t = \sin^{2} x, \frac{d f}{d x} = \frac{d P}{d t} \frac{d t}{d x} .

Restore the omitted multiplier:

\frac{d t}{d x} = \sin (2 x), {\frac{d f}{d x} |}_{x = 0} = 18 \sin 0 = 0 .

The fraction above is an illustrative incorrect result, not an actual answer option. This numerical question has no option list.

Before substituting zero, write the variable of differentiation explicitly. Check that the chain-rule multiplier is present before dividing by five.

Which two related determinant questions should you try next?

Try one question that isolates row operations, followed by one that separates a determinant’s value from its derivative. Both are original related practice from Matrices and Determinants, not verified past-paper questions; attempt each before reading its worked answer.

Question 1: Simplify this determinant.

Δ = | \begin{matrix} a & b & c \\ a + 3 & b & d \\ a & b + 3 & e \end{matrix} | .

Subtract the first row from both lower rows. This isolates the main problem’s row-operation structure without trigonometric algebra:

R_{2} \leftarrow R_{2} - R_{1}, R_{3} \leftarrow R_{3} - R_{1},

Δ = | \begin{matrix} a & b & c \\ 3 & 0 & d - c \\ 0 & 3 & e - c \end{matrix} | .

Expanding along the first row gives:

Δ & = - 3 a (d - c) - 3 b (e - c) + 9 c & = 3 a (c - d) + 3 b (c - e) + 9 c .

Question 2: Find the value and derivative at zero.

g (x) = | \begin{matrix} 2 & 0 & 3 \\ 5 & 0 & \sin^{2} x \\ 2 & 3 & 0 \end{matrix} |; find g (0) and g^{'} (0) .

Again, subtract the first row from each lower row. Expand the resulting determinant before differentiating:

g (x) = | \begin{matrix} 2 & 0 & 3 \\ 3 & 0 & \sin^{2} x - 3 \\ 0 & 3 & - 3 \end{matrix} | .

g (x) = - 6 [\sin^{2} x - 3] + 27 = 45 - 6 \sin^{2} x .

$g^{'} (x) = - 6 \sin (2 x) .$ Hence:

g (0) = 45, g^{'} (0) = 0 .

This checks whether you distinguish the determinant’s value from its rate of change. On your next attempt, keep the two answers on separate lines and write the chain-rule multiplier before substituting zero.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Keep going with The p-Block Elements JEE 2017: Finding Y and Z.

Frequently asked questions

What is the answer to the determinants JEE 2024 question?

For the trigonometric determinant in this article, the requested value f'(0)/5 is 0. Writing f(x) as a polynomial in t = sin²x gives f'(0) = P'(0) × sin(0) = 18 × 0.

Which row operations simplify this trigonometric determinant?

Apply R₂ → R₂ − R₁ and R₃ → R₃ − R₁. These operations preserve the determinant and create two zeros, making expansion along the first row simpler.

Why use t = sin²x in the determinant?

This substitution turns every trigonometric entry into a polynomial: cos⁴x = (1 − t)², sin⁴x = t² and sin²(2x) = 4t(1 − t). After expanding, the determinant becomes P(t) = −24t⁴ + 66t³ − 60t² + 18t + 45.

Why is 18/5 the wrong answer for f'(0)/5?

The value 18 is P'(0), the derivative with respect to t, not x. Since t = sin²x, the chain rule requires multiplication by dt/dx = sin(2x), which is zero at x = 0. Therefore f'(0)/5 = 0, not 18/5.