JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Derivative of a Trigonometric Determinant

If f(x)=|[2 cos ^4 x 2 sin ^4 x 3+sin ^2 2 x; 3+2 cos ^4 x 2 sin ^4 x sin ^2 2 x; 2 cos ^4 x 3+2 sin ^4 x sin ^2 2 x]| then 1/5 f'(0) is equal to

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A0B1C2D6

Step-by-step solutionView

Correct answer

Simplifying the determinant by adding all columns yields a constant function $f (x) = 45$ , so its derivative is identically zero and $\frac{1}{5} f^{'} (0) = 0$ .

Option analysis

Why each option works or fails

A · 0

This is the correct value. Evaluating the determinant shows $f (x)$ is a constant independent of $x$ , hence $f^{'} (x) = 0$ for all $x$ , giving $\frac{1}{5} f^{'} (0) = 0$ .

B · 1

Confusing the value of $f^{'} (0)$ with an erroneous linear coefficient or improperly differentiating $\sin^{2} (2 x)$ without completing the determinant reduction. Fully simplify the determinant using column operations before differentiating, or notice all row sums are identical.

C · 2

Applying the derivative directly to individual elements at $x = 0$ and incorrectly combining terms without accounting for row cancellations. Use column operations $C_{1} \to C_{1} + C_{2} + C_{3}$ first to reveal that the sum of each row is a constant.

D · 6

Evaluating $\frac{1}{5} f (0)$ instead of $\frac{1}{5} f^{'} (0)$ , since $f (0) = 45$ leading to $\frac{1}{5} \times 45 = 9$ , or making an arithmetic slip with the factor of 5. Notice the question asks for the derivative $f^{'} (0)$ , not the function value $f (0)$ .

Reviewed route

Solution

StepWorking

01Given

$f (x) = ∣ \begin{array}{ccc} 2 \cos^{4} x & 2 \sin^{4} x & 3 + \sin^{2} 2 x \\ 3 + 2 \cos^{4} x & 2 \sin^{4} x & \sin^{2} 2 x \\ 2 \cos^{4} x & 3 + 2 \sin^{4} x & \sin^{2} 2 x \end{array} ∣$

02Goal

Find the value of $\frac{1}{5} f^{'} (0)$ .

03Approach

Apply elementary row operations $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ to create zeros and constant entries, simplify $f (x)$ , and then compute $f^{'} (x)$ .

04Execute

Applying $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ gives: $f (x) = ∣ \begin{array}{ccc} 2 \cos^{4} x & 2 \sin^{4} x & 3 + \sin^{2} 2 x \\ 3 & 0 & - 3 \\ 0 & 3 & - 3 \end{array} ∣$ Expanding along the rows or columns: $f (x) = 2 \cos^{4} x (0 - (- 9)) - 2 \sin^{4} x (- 9 - 0) + (3 + \sin^{2} 2 x) (9 - 0)$ $f (x) = 18 \cos^{4} x + 18 \sin^{4} x + 9 (3 + \sin^{2} 2 x)$ Recall that $\cos^{4} x + \sin^{4} x = 1 - 2 \sin^{2} x \cos^{2} x = 1 - \frac{1}{2} \sin^{2} 2 x$ . Therefore: $f (x) = 18 (1 - \frac{1}{2} \sin^{2} 2 x) + 27 + 9 \sin^{2} 2 x = 18 - 9 \sin^{2} 2 x + 27 + 9 \sin^{2} 2 x = 45$ Since $f (x) = 45$ is a constant function for all $x$ , its derivative is $f^{'} (x) = 0$ .

✓Verify

Since $f^{'} (x) = 0$ identically, $\frac{1}{5} f^{'} (0) = \frac{1}{5} (0) = 0$ , which is completely independent of the value of $x$ .

Hints that build this answer step by step

What is the most efficient first step to simplify $f (x)$ ?
Apply the column operation $C_{1} \to C_{1} + C_{2} + C_{3}$
What is the simplified value of $2 \cos^{4} x + 2 \sin^{4} x + \sin^{2} 2 x + 3$ ?
$5$
After factoring out $5$ from $C_{1}$ and applying $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ , what is the resulting function $f (x)$ ?
$f (x) = 45$
Given that $f (x) = 45$ is a constant function, what is the value of $\frac{1}{5} f^{'} (0)$ ?
$0$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 8 September 2026

Quick checks

Students also ask

Could we differentiate the determinant column by column at

x = 0

instead?

Yes, differentiating directly using the determinant differentiation rule at $x = 0$ is also valid, but simplifying the matrix via row operations first is much faster because the determinant reduces to a constant immediately.

Answer

Simplifying the determinant by adding all columns yields a constant function $f (x) = 45$ , so its derivative is identically zero and $\frac{1}{5} f^{'} (0) = 0$ .

Why each option works or fails

A · correct: 0 - This is the correct value. Evaluating the determinant shows $f (x)$ is a constant independent of $x$ , hence $f^{'} (x) = 0$ for all $x$ , giving $\frac{1}{5} f^{'} (0) = 0$ .
B: 1 - Confusing the value of $f^{'} (0)$ with an erroneous linear coefficient or improperly differentiating $\sin^{2} (2 x)$ without completing the determinant reduction. Fully simplify the determinant using column operations before differentiating, or notice all row sums are identical.
C: 2 - Applying the derivative directly to individual elements at $x = 0$ and incorrectly combining terms without accounting for row cancellations. Use column operations $C_{1} \to C_{1} + C_{2} + C_{3}$ first to reveal that the sum of each row is a constant.
D: 6 - Evaluating $\frac{1}{5} f (0)$ instead of $\frac{1}{5} f^{'} (0)$ , since $f (0) = 45$ leading to $\frac{1}{5} \times 45 = 9$ , or making an arithmetic slip with the factor of 5. Notice the question asks for the derivative $f^{'} (0)$ , not the function value $f (0)$ .

Step-by-step solution

Given: $f (x) = ∣ \begin{array}{ccc} 2 \cos^{4} x & 2 \sin^{4} x & 3 + \sin^{2} 2 x \\ 3 + 2 \cos^{4} x & 2 \sin^{4} x & \sin^{2} 2 x \\ 2 \cos^{4} x & 3 + 2 \sin^{4} x & \sin^{2} 2 x \end{array} ∣$
Goal: Find the value of $\frac{1}{5} f^{'} (0)$ .
Approach: Apply elementary row operations $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ to create zeros and constant entries, simplify $f (x)$ , and then compute $f^{'} (x)$ .
Execute: Applying $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ gives: $f (x) = ∣ \begin{array}{ccc} 2 \cos^{4} x & 2 \sin^{4} x & 3 + \sin^{2} 2 x \\ 3 & 0 & - 3 \\ 0 & 3 & - 3 \end{array} ∣$ Expanding along the rows or columns: $f (x) = 2 \cos^{4} x (0 - (- 9)) - 2 \sin^{4} x (- 9 - 0) + (3 + \sin^{2} 2 x) (9 - 0)$ $f (x) = 18 \cos^{4} x + 18 \sin^{4} x + 9 (3 + \sin^{2} 2 x)$ Recall that $\cos^{4} x + \sin^{4} x = 1 - 2 \sin^{2} x \cos^{2} x = 1 - \frac{1}{2} \sin^{2} 2 x$ . Therefore: $f (x) = 18 (1 - \frac{1}{2} \sin^{2} 2 x) + 27 + 9 \sin^{2} 2 x = 18 - 9 \sin^{2} 2 x + 27 + 9 \sin^{2} 2 x = 45$ Since $f (x) = 45$ is a constant function for all $x$ , its derivative is $f^{'} (x) = 0$ .
Verify: Since $f^{'} (x) = 0$ identically, $\frac{1}{5} f^{'} (0) = \frac{1}{5} (0) = 0$ , which is completely independent of the value of $x$ .

Derivative of a Trigonometric Determinant

What feels right?

Simplifying the determinant by adding all columns yields a constant function $f (x) = 45$ , so its derivative is identically zero and $\frac{1}{5} f^{'} (0) = 0$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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