Consider the system of linear equations in x, y, z: x + 2y + tz = 0, 6x + y + 5tz = 0, 3x + t^2y + f(t)z = 0, where f: R -> R is a differentiable function. If this system has infinitely many solutions for all t ∈ R, then f

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q4 – JEE-MAIN 2026

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Home Maths MatricesQuestion #289

Consider the system of linear equations in

x, y, z

x + 2 y + t z = 0

6 x + y + 5 t z = 0

3 x + t^{2} y + f (t) z = 0

where

f : R \to R

is a differentiable function. If this system has infinitely many solutions for all

t \in R

, then

f

Clue 1 of 3: System of Linear Equations

For a homogeneous system of linear equations $A x = 0$ to have infinitely many solutions, the determinant of the coefficient matrix $A$ must be zero.

🧠 Full Solution Path

Step 1: Formulate the Coefficient Matrix and Determinant Condition✦ Active

The given system of linear equations is homogeneous. For it to have infinitely many solutions for all $t \in R$ , the determinant of its coefficient matrix must be zero. The coefficient matrix $A$ is:

A = (\begin{matrix} 1 & 2 & t \\ 6 & 1 & 5 t \\ 3 & t^{2} & f (t) \end{matrix})

The condition for infinitely many solutions is $det (A) = 0$ .

Step 2: Calculate the Determinant and Solve for

f (t)

○ Expand

Calculate the determinant of $A$ :

det (A) = 1 (1 \cdot f (t) - 5 t \cdot t^{2}) - 2 (6 \cdot f (t) - 5 t \cdot 3) + t (6 \cdot t^{2} - 1 \cdot 3)

det (A) = f (t) - 5 t^{3} - 12 f (t) + 30 t + 6 t^{3} - 3 t

det (A) = - 11 f (t) + t^{3} + 27 t

Setting $det (A) = 0$ for all $t \in R$ :

- 11 f (t) + t^{3} + 27 t = 0

11 f (t) = t^{3} + 27 t

f (t) = \frac{t^{3} + 27 t}{11}

💡 Teacher's Secret Hint

Ensure careful calculation of the determinant to avoid algebraic errors.

Step 3: Analyze the Function

f (t)

using its Derivative○ Expand

To determine the nature of $f (t)$ , we find its derivative $f^{'} (t)$ :

f^{'} (t) = \frac{d}{d t} (\frac{t^{3} + 27 t}{11}) = \frac{1}{11} (3 t^{2} + 27) = \frac{3}{11} (t^{2} + 9)

Since $t^{2} \geq 0$ for all $t \in R$ , it follows that $t^{2} + 9 \geq 9$ . Therefore, $f^{'} (t) = \frac{3}{11} (t^{2} + 9) > 0$ for all $t \in R$ . Since $f^{'} (t) > 0$ for all $t \in R$ , the function $f (t)$ is strictly increasing on $R$ . It also has no critical points as $f^{'} (t)$ is never zero. Thus, option 2 is correct.

💡 Teacher's Secret Hint

Recall that a function is strictly increasing if its derivative is always positive.

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Maths Question 4 – JEE-MAIN 2026

Clue 1 of 3: System of Linear Equations

🧠 Full Solution Path