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Matrices and Determinants: Mathematics | JEE Main

f(x)=\left|\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}\right|

then

\frac{1}{5} f^{\prime}(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

None. 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 $\sin^2(2x)$ 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

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

02goal

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

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: $f(x) = \left|\begin{array}{ccc}2 \cos ^{4} x & 2 \sin ^{4} x & 3+\sin ^{2} 2 x \\ 3 & 0 & -3 \\ 0 & 3 & -3\end{array}\right|$ Expanding along the rows or columns: $f(x) = 2\cos^4 x(0 - (-9)) - 2\sin^4 x(-9 - 0) + (3+\sin^2 2x)(9 - 0)$ $f(x) = 18\cos^4 x + 18\sin^4 x + 9(3 + \sin^2 2x)$ Recall that $\cos^4 x + \sin^4 x = 1 - 2\sin^2 x\cos^2 x = 1 - \frac{1}{2}\sin^2 2x$ . Therefore: $f(x) = 18\left(1 - \frac{1}{2}\sin^2 2x\right) + 27 + 9\sin^2 2x = 18 - 9\sin^2 2x + 27 + 9\sin^2 2x = 45$ 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 $2\cos^4 x + 2\sin^4 x + \sin^2 2x + 3$ ?
$5$
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
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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 - None. 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 $\sin^2(2x)$ 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: $f(x)=\left|\begin{array}{ccc}2 \cos ^{4} x & 2 \sin ^{4} x & 3+\sin ^{2} 2 x \\ 3+2 \cos ^{4} x & 2 \sin ^{4} x & \sin ^{2} 2 x \\ 2 \cos ^{4} x & 3+2 \sin ^{4} x & \sin ^{2} 2 x\end{array}\right|$
goal: Find the value of $\frac{1}{5} f^{\prime}(0)$ .
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: $f(x) = \left|\begin{array}{ccc}2 \cos ^{4} x & 2 \sin ^{4} x & 3+\sin ^{2} 2 x \\ 3 & 0 & -3 \\ 0 & 3 & -3\end{array}\right|$ Expanding along the rows or columns: $f(x) = 2\cos^4 x(0 - (-9)) - 2\sin^4 x(-9 - 0) + (3+\sin^2 2x)(9 - 0)$ $f(x) = 18\cos^4 x + 18\sin^4 x + 9(3 + \sin^2 2x)$ Recall that $\cos^4 x + \sin^4 x = 1 - 2\sin^2 x\cos^2 x = 1 - \frac{1}{2}\sin^2 2x$ . Therefore: $f(x) = 18\left(1 - \frac{1}{2}\sin^2 2x\right) + 27 + 9\sin^2 2x = 18 - 9\sin^2 2x + 27 + 9\sin^2 2x = 45$ 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 $x$ .

Matrices and Determinants: Mathematics | JEE Main

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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Matrices and Determinants: Mathematics | JEE Main

Simplifying the determinant by adding all columns yields a constant function f(x)=45f(x) = 45f(x)=45, so its derivative is identically zero and 15f′(0)=0\frac{1}{5}f'(0) = 051​f′(0)=0.

01given

02goal

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04execute

✓verify

Let…

Consider the system of linear equation…

Let S1\mathrm{S}_1S1​ and S2\mathrm{S}_2S2​ be respectively the sets of all…

Students also ask

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

Let $\mathrm{S}_1$ and $\mathrm{S}_2$ be respectively the sets of all…