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

(B) is strictly increasing on R

(C) is strictly decreasing on R

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Matrices

Consider the system of linear equations in

x, y, z

x + 2 y + tz = 0

6 x + y + 5 tz = 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

Answer: (B)

Infinitely many solutions for every

t

force the coefficient determinant to vanish identically:

163 21 t^{2} t 5 t f (t) = - 11 f (t) + t^{3} + 27 t = 0

, so

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

. Differentiating,

f^{'} (t) = \frac{3 t ^{2} + 27}{11} > 0

for every real

t

. Therefore

f

is strictly increasing on

R

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Single correct option