JEE Main 2026 — Apr 5, Shift 1 — Mathematics solution

JEE Main2026Apr 5, Shift 1Mathematics

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 $f$

A
is a constant function
B
is strictly increasing on $R$
C
is strictly decreasing on $R$
D
has two critical points

Solution

For infinitely many solutions, the coefficient determinant must vanish for every $t$ :

$12 t 615 t 3 t^{2} f (t) = 0$ .

Expanding along the first row: $(f (t) - 5 t^{3}) - 2 (6 f (t) - 15 t) + t (6 t^{2} - 3) = 0$ , which simplifies to $- 11 f (t) + t^{3} + 12 t = 0$ , giving $f (t) = \frac{t ^{3} + 12 t}{11}$ .

Differentiating, $f^{'} (t) = \frac{3 t ^{2} + 12}{11} > 0$ for every $t \in R$ , so $f$ is strictly increasing on $R$ .

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